381 free, high-quality past-paper-style questions on Waves for Cambridge IGCSE Physics, each with a full mark scheme. Covers: Sound and ultrasound; Light, refraction, lenses and dispersion; The electromagnetic spectrum; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,368 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IGPC-037-03-01Sound and ultrasoundPaper CEasy1 mark
An electric bell is ringing inside a sealed glass jar. A pump slowly removes all the air from the jar. The bell can still be seen vibrating. What happens to the sound heard outside the jar?
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Sound is a mechanical wave carried by vibrating particles. As the air is removed, fewer particles are left to pass the vibrations on, so the sound fades and then stops, even though the bell still vibrates. Light needs no medium, which is why the bell can still be seen.
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Answer A
Answer: A
Common error: Do not assume sound behaves like light. Light can cross a vacuum but sound cannot, because sound must be passed on by particles.
2IGPC-037-03-02Sound and ultrasoundPaper CEasy1 mark
A musician plays the same note on a trumpet twice. The second time it is louder, but the pitch is the same. How has the sound wave changed?
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Loudness depends on amplitude and pitch depends on frequency. The pitch is the same, so the frequency has not changed. The sound is louder, so the amplitude is larger. The speed of sound depends on the medium, not on loudness.
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Answer B
Answer: B
Exam tip: Learn the pairs: loudness goes with amplitude, and pitch goes with frequency.
3IGPC-037-03-03Sound and ultrasoundPaper CEasy1 mark
Which of these sound frequencies can be heard by a person with normal hearing?
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The human audible range is about 20 Hz to 20 000 Hz (20 kHz). 15 kHz = 15 000 Hz, which is inside this range. 5 Hz is too low, and 35 kHz and 2 MHz are ultrasound, so they are too high to hear.
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Answer B
Answer: B
Common error: Check the prefixes: 15 kHz is 15 000 Hz, not 15 Hz. Convert to hertz before comparing with the 20 Hz – 20 000 Hz range.
4IGPC-037-03-12Sound and ultrasoundPaper CEasy4 marks
Students measure the speed of sound on a playing field. One student fires a starting pistol. A second student stands 400 m away, starts a stopwatch on seeing the smoke from the pistol, and stops it on hearing the bang. The time recorded is 1.18 s.
Diagram not to scale
(a)
Explain why the second student sees the smoke before hearing the bang.
(1)
(b)
Calculate the speed of sound from these results.
(2)
(c)
Suggest one way the students could make their result more reliable.
(1)
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Part (a)
Light travels much faster than sound, so it arrives almost instantly .
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Part (b)
speed = distance ÷ time = 400 ÷ 1.18
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= 339 m/s
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Part (c)
Any one: repeat the measurement and calculate an average; use a larger distance so that reaction time is a smaller fraction of the time measured; use several timers and average their times .
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Common error: In this method the sound travels only one way, so do not halve or double the distance. Halving is needed only for echoes.
5IGPC-037-01-01Light, refraction, lenses and dispersionPaper CEasy1 mark
A ray of light strikes a flat plane mirror. The ray makes an angle of 35° with the surface of the mirror. What is the angle of reflection?
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Angles of incidence and reflection are measured from the normal, not from the mirror surface. The normal is at 90° to the mirror, so the angle of incidence is 90° − 35° = 55°. The angle of reflection equals the angle of incidence, so it is 55°. 35° uses the surface instead of the normal. 70° is the full angle between the incident and reflected rays. 110° is twice the angle of reflection.
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Answer B
Answer: B
Common error: Always measure the angles of incidence, reflection and refraction from the normal. If an angle is given to the surface, subtract it from 90° first.
6IGPC-037-01-02Light, refraction, lenses and dispersionPaper CEasy1 mark
A student stands 1.5 m in front of a vertical plane mirror and looks at her image. Which description of the image is correct?
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A plane mirror forms a virtual image. No light actually passes through the image, so it cannot be shown on a screen. The image is upright and the same size as the object. It is as far behind the mirror as the object is in front, so it is 1.5 m behind the mirror. 3.0 m is the distance from the student to her image, not from the mirror to the image. The image is not smaller, even though distant objects look smaller.
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Answer C
Answer: C
Exam tip: The image distance is measured from the mirror. The distance between the object and its image is double this.
7IGPC-037-01-03Light, refraction, lenses and dispersionPaper CEasy1 mark
A ray of light in air enters a glass block along the normal (angle of incidence 0°). What happens to the speed and the direction of the light as it enters the glass?
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Light always slows down when it passes from air into glass, whatever the angle of incidence. A ray that travels along the normal meets the boundary at 0°, so it does not change direction. The change in speed still happens. 'Bends towards the normal' would only be true for a ray at an angle to the normal. Light does not speed up when it enters a denser medium.
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Answer A
Answer: A
Common error: A ray along the normal is not deviated, but it still changes speed. Refraction is caused by the change in speed.
8IGPC-037-01-06Light, refraction, lenses and dispersionPaper CEasy1 mark
A narrow beam of white light passes through a triangular glass prism and a spectrum forms on a screen. Which colour is refracted (deviated) the most?
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Violet light is refracted the most, so it appears at the end of the spectrum furthest from the original direction of the beam. Red light is refracted the least. Choosing red usually comes from thinking that a longer wavelength means more bending; the opposite is true for glass.
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Answer D
Answer: D
Common error: Red is deviated least and violet most. The order of the spectrum is red, orange, yellow, green, blue, indigo, violet.
9IGPC-037-01-11Light, refraction, lenses and dispersionPaper CEasy5 marks
A student shines a ray of light from air into one face of a rectangular glass block. The angle of incidence is 40°.
Diagram not to scale
(a)
State what is meant by the normal at the point where the ray enters the block.
(1)
(b)
State how the direction of the ray changes as it enters the glass.
(1)
(c)
State what happens to the speed of the light as it enters the glass.
(1)
(d)
Light inside glass can undergo total internal reflection at a glass–air boundary. State two conditions needed for total internal reflection.
(2)
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Part (a)
A line drawn at 90° (perpendicular) to the surface of the block at the point where the ray strikes it
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Part (b)
It bends towards the normal, so the angle of refraction is less than 40°
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Part (c)
The speed decreases
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Part (d)
The light must be travelling from the glass (denser medium) towards the air (less dense medium)
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The angle of incidence must be greater than the critical angle
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Exam tip: Both conditions are needed for total internal reflection. Saying only 'a large angle' does not earn the mark; compare the angle of incidence with the critical angle.
10IGPC-037-02-01The electromagnetic spectrumPaper CEasy1 mark
Which list shows four regions of the electromagnetic spectrum in order of increasing wavelength?
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Gamma rays have the shortest wavelength. Wavelength then increases through X-rays, ultraviolet and visible light. The radio-to-visible list is in order of decreasing wavelength, so it runs the wrong way. The third list puts X-rays before gamma rays. The fourth list puts visible light before ultraviolet, but ultraviolet has the shorter wavelength.
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Answer A
Answer: A
Exam tip: Learn one order (radio, microwave, infrared, visible, ultraviolet, X-ray, gamma). Wavelength decreases and frequency increases along it. Read the question to see which direction it asks for.
11IGPC-037-02-02The electromagnetic spectrumPaper CEasy1 mark
Which quantity is the same for all electromagnetic waves travelling in a vacuum?
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All electromagnetic waves travel at the same speed in a vacuum. Frequency and wavelength are what distinguish one region from another, so they differ across the spectrum. Only the high-frequency regions (some ultraviolet, X-rays and gamma rays) are ionising.
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Answer B
Answer: B
Common error: High-frequency waves such as gamma rays do not travel faster than radio waves. They carry more energy, but their speed in a vacuum is identical.
12IGPC-037-02-03The electromagnetic spectrumPaper CEasy1 mark
Sealed packs of surgical instruments are sterilised without opening them. Which type of electromagnetic radiation is used?
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Gamma rays are highly penetrating, so they pass through the sealed packaging. They kill bacteria on the instruments inside. Infrared and microwaves only heat materials and would not reliably sterilise sealed equipment. Radio waves have far too little energy to kill bacteria.
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Answer C
Answer: C
Common error: Do not assume sterilising means heating. Gamma sterilisation works by killing microorganisms with ionising radiation, not by making the instruments hot.
13IGPC-037-02-04The electromagnetic spectrumPaper CEasy1 mark
A firefighter uses a thermal imaging camera to find a person in a smoke-filled room. Which radiation does the camera detect?
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Warm objects such as the human body emit infrared radiation. A thermal imaging camera detects this infrared, even through smoke and in darkness. The body does not emit significant ultraviolet, microwaves or X-rays.
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Answer B
Answer: B
Exam tip: Thermal imaging, remote controls, electric grills and some optical fibres all use infrared.
14IGPC-037-00-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Water waves on a lake have a frequency of 2.5 Hz and a wavelength of 0.40 m. What is the speed of the waves?
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Wave speed = frequency × wavelength, so v = 2.5 × 0.40 = 1.0 m/s. The value 0.16 m/s comes from λ ÷ f. The value 6.25 m/s comes from f ÷ λ. The value 2.9 m/s comes from adding the two numbers.
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Answer B
Answer: B
Exam tip: Write v = fλ before substituting. Check that m/s comes out as (1/s) × m.
15IGPC-037-00-02Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Which of these is a longitudinal wave?
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In a sound wave the air particles vibrate parallel to the direction the wave travels. This makes sound longitudinal. Light and radio waves are electromagnetic waves, so they are transverse. Water ripples are also treated as transverse, because the surface moves up and down at right angles to the direction of travel.
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Answer C
Answer: C
Common error: Do not confuse the direction of vibration with the direction the energy travels. For a longitudinal wave the two directions are parallel.
16IGPC-037-00-05Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
A student stands 1.5 m in front of a vertical plane mirror. What is the distance between the student and the student's image?
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The image in a plane mirror is as far behind the mirror as the object is in front of it, so the image is 1.5 m behind the mirror. The distance from the student to the image is 1.5 + 1.5 = 3.0 m. The answer 1.5 m is only the distance from the mirror to the image.
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Answer C
Answer: C
Common error: Read whether the question asks for the distance to the mirror or the distance to the image.
17IGPC-037-00-11Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
A buoy floats in the sea near a harbour. Sea waves pass the buoy and make it rise and fall.
(a)
State what is meant by the wavelength of a wave.
(1)
(b)
The buoy rises and falls 12 times in 60 s. Calculate the frequency of the waves.
(1)
(c)
The wavelength of the waves is 15 m. Calculate the speed of the waves.
(2)
(d)
The buoy moves up and down but does not travel towards the shore. State what the waves transfer towards the shore.
(1)
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Part (a)
The distance between a point on one wave and the same point on the next wave, e.g. crest to crest.
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Part (b)
f = 12 ÷ 60 = 0.20 Hz
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Part (c)
v = fλ = 0.20 × 15
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v = 3.0 m/s
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Part (d)
Energy, without transferring the water itself.
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Exam tip: Frequency is the number of complete waves per second. Divide the number of oscillations by the time taken.
18IGPC-037-02-11The electromagnetic spectrumPaper CEasy5 marks
A hospital uses several types of electromagnetic radiation.
(a)
State the type of electromagnetic radiation used to produce images of broken bones.
(1)
(b)
State the type of electromagnetic radiation used to sterilise sealed packs of medical equipment.
(1)
(c)
The radiographer who takes the images stands behind a screen during each exposure. Explain why.
(2)
(d)
State how the speed of X-rays in a vacuum compares with the speed of visible light in a vacuum.
(1)
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Part (a)
X-rays
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Part (b)
gamma rays / gamma radiation / γ
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Part (c)
X-rays are ionising / can damage cells / cause mutations or cancer ; the screen (e.g. containing lead) absorbs X-rays, so the radiographer's exposure (dose) is reduced
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Part (d)
the speeds are the same (accept both 3.0 × 108 m/s)
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Exam tip: When explaining a safety precaution, state both the hazard and how the precaution reduces it. Each idea is worth a separate mark.
19IGPC-037-02-12The electromagnetic spectrumPaper CEasy4 marks
A kitchen contains a microwave oven and a television with a remote control.
(a)
State the type of electromagnetic radiation used by the television remote control.
(1)
(b)
State the main hazard of microwaves to the human body.
(1)
(c)
The microwave oven uses waves of frequency 2.45 × 109 Hz. The speed of the waves is 3.0 × 108 m/s. Calculate the wavelength of the microwaves.
(2)
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Part (a)
infrared
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Part (b)
internal heating of body cells / tissue (accept damage to or burning of internal tissue)
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Part (c)
λ = v / f = (3.0 × 108) / (2.45 × 109) ; λ = 0.12 m (accept 0.122 m)
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Common error: Rearrange v = fλ correctly. The wavelength is v divided by f, not v multiplied by f.
20IGPC-037-00-13Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
A dentist uses a small plane mirror to look at the back of a patient's teeth.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
A ray of light strikes the mirror with an angle of incidence of 25°. State the angle between the incident ray and the reflected ray.
(1)
(c)
Describe two properties of the image formed by a plane mirror.
(2)
(d)
A person stands 2.4 m in front of a large plane mirror and then walks 0.8 m towards it. State the new distance between the person and their image.
(1)
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Part (a)
The angle of incidence equals the angle of reflection, both measured from the normal.
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Part (b)
50°
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Part (c)
Any two from: virtual (cannot be formed on a screen); same size as the object; upright; laterally inverted; same distance behind the mirror as the object is in front.
—
First correct property
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Second correct property
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Part (d)
(2.4 − 0.8) × 2 = 3.2 m
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Exam tip: The distance from the person to their image is twice the distance from the person to the mirror.
21IGPC-037-01-15Light, refraction, lenses and dispersionPaper CEasy5 marks
A narrow beam of white light from a ray box passes through a triangular glass prism. A band of colours appears on a white screen beyond the prism.
(a)
State the name of the process that produces the band of colours.
(1)
(b)
State which colour is deviated least and which colour is deviated most.
(2)
(c)
Explain why white light is split into colours by the prism.
(2)
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Part (a)
Dispersion
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Part (b)
Least: red
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Most: violet
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Part (c)
White light is a mixture of different colours (of different frequencies / wavelengths)
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Each colour travels at a different speed in glass (accept: glass has a different refractive index for each colour), so each is refracted through a different angle and the colours separate
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Common error: The prism does not add colour to the light. The colours are already present in white light; the prism only separates them because each colour travels at a different speed in glass.
22IGPC-037-03-11Sound and ultrasoundPaper CEasy5 marks
A guitarist plucks a string on an acoustic guitar. A listener a few metres away hears the note.
(a)
Describe how the plucked string produces a sound that reaches the listener.
(2)
(b)
The guitarist plucks the same string harder. State how the sound heard changes, and state the change in the sound wave that causes this.
(2)
(c)
The note has a frequency of 196 Hz. State whether a person with normal hearing can hear this note. Give a reason.
(1)
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Part (a)
The string vibrates (accept: the string makes the guitar body or soundboard vibrate) . The vibration makes the air particles vibrate, and this passes through the air to the listener's ear as a (longitudinal) sound wave. Accept: compressions and rarefactions travel through the air to the ear .
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Part (b)
The sound is louder . The amplitude of the wave is larger .
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Part (c)
Yes, because 196 Hz is within the human audible range of about 20 Hz to 20 000 Hz .
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Exam tip: When describing how sound is produced, always start with something vibrating. Then say how the vibration is passed through the medium, for example as compressions and rarefactions in the air.
23IGPC3-3-03c-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
A buoy rises and falls 15 times in 30 s as ocean waves pass it. The distance between neighbouring crests is 6.0 m. What is the speed of the waves?
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Frequency = 15 / 30 = 0.50 Hz, so v = fλ = 0.50 × 6.0 = 3.0 m/s (A). B multiplies the wavelength by the period (2.0 s) instead of the frequency. C multiplies the wavelength by the number of oscillations (15 × 6.0) and ignores the time. D divides the time by the number of oscillations and the wavelength (30 / (15 × 6.0)).
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Answer A
Answer: A
Common error: Frequency is oscillations per second, so divide the count by the time before using v = fλ.
24IGPC3-3-03c-02Light, refraction, lenses and dispersionPaper CEasy1 mark
Light travelling inside a glass prism meets the glass–air surface. Which pair of conditions is needed for total internal reflection?
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Total internal reflection needs light going from a more optically dense medium (glass) towards a less dense one (air) with an angle of incidence above the critical angle (C). A: below the critical angle the light is refracted out. B: light entering glass from air bends towards the normal and cannot be totally internally reflected. D: a ray along the normal passes straight out with little reflection.
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Answer C
Answer: C
Exam tip: Check both conditions: which way the light is going and how the angle compares with the critical angle.
25IGPC3-3-03c-03The electromagnetic spectrumPaper CEasy1 mark
Which list shows electromagnetic waves in order of increasing frequency?
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Frequency rises from radio, through microwaves, infrared, visible, ultraviolet and X-rays to gamma rays, so B is correct. A lists the order of decreasing frequency. C puts visible light below infrared. D puts microwaves above infrared.
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Answer B
Answer: B
Exam tip: Learn the spectrum in one direction (radio to gamma) and remember that frequency rises while wavelength falls.
26IGPC3-3-03c-08Sound and ultrasoundPaper CEasy1 mark
A guitar string is plucked gently. The string is then shortened by pressing it against the neck and plucked harder. How do the amplitude and frequency of the new note compare with the first note?
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Plucking harder gives a larger amplitude (louder), and a shorter string vibrates faster, giving a higher frequency (higher pitch), so B is correct. A reverses the effect of string length. C assumes loudness changes nothing about pitch but ignores the shorter string. D assumes a shorter string is quieter.
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Answer B
Answer: B
Exam tip: Amplitude is linked to loudness and frequency to pitch; treat the two separately.
27IGPC3-3-03c-09Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
Waves in a harbour move at a steady speed of 2.4 m/s. The distance between one crest and the next crest is 3.0 m.
Diagram not to scale
(a)
State what is meant by the wavelength of a wave.
(1)
(b)
Calculate the frequency of the waves.
(2)
(c)
Calculate the time taken for 12 complete waves to pass a floating marker.
(2)
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Part (a)
the distance between two successive identical points on a wave, for example crest to crest
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Part (b)
f = v / λ = 2.4 / 3.0
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f = 0.80 Hz
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Part (c)
time for one wave = 1 / 0.80 = 1.25 s, ECF from (b)
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time for 12 waves = 12 × 1.25 = 15 s
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Exam tip: Write the equation, then the substitution, then the answer with its unit so that method marks are secure.
28IGPC3-3-03c-12The electromagnetic spectrumPaper CEasy5 marks
A ship's radio uses electromagnetic waves of frequency 150 MHz.
(a)
Name the region of the electromagnetic spectrum that lies between visible light and microwaves.
(1)
(b)
State one use of X-rays in hospitals.
(1)
(c)
State the speed of electromagnetic waves in a vacuum.
(1)
(d)
Calculate the wavelength of the radio waves. 1 MHz = 106 Hz.
(2)
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Part (a)
infrared
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Part (b)
to produce images of bones or other dense structures
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Part (c)
3.0 × 108 m/s
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Part (d)
λ = v / f = 3.0 × 108 / (150 × 106)
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λ = 2.0 m
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Common error: Convert megahertz to hertz before substituting into v = fλ.
29IGPC3-3-05r9-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Divers are surveying a flooded archaeological site. Water ripples on the surface above it have a frequency of 2.5 Hz and a distance of 0.80 m between neighbouring crests. What is the speed of the ripples?
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Correct: C. Speed = frequency × wavelength = 2.5 × 0.80 = 2.0 m/s. A divides wavelength by frequency (0.80 ÷ 2.5). B divides frequency by wavelength (2.5 ÷ 0.80). D adds the two quantities.
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Answer C
Answer: C
Common error: Check the units: a speed in m/s comes from multiplying hertz (per second) by metres, never from adding or dividing them the wrong way round.
30IGPC3-3-05r9-07Sound and ultrasoundPaper CEasy1 mark
A technician uses an ultrasonic leak detector working at 40 kHz on a data-centre cooling pipe, but cannot hear it directly. What is the reason?
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Correct: B. Humans hear roughly 20 Hz to 20 kHz, so 40 kHz is ultrasound. A is wrong because a high frequency means a short wavelength. C is wrong because the speed of sound in air does not depend on whether we can hear it. D is wrong because ultrasound can have a large amplitude; it is the frequency that makes it inaudible.
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Answer B
Answer: B
Exam tip: Remember the hearing range 20 Hz to 20 kHz; anything above is ultrasound and anything below is infrasound.
31IGPC3-3-05r9-08The electromagnetic spectrumPaper CEasy1 mark
A cyclist's satellite navigation unit receives microwave signals from a satellite. Compared with visible light, how do microwaves behave in a vacuum?
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Correct: A. All electromagnetic waves travel at 3.0 × 108 m/s in a vacuum; microwaves lie below visible light in frequency, so their wavelength is longer. B is wrong because speed does not differ in a vacuum. C is wrong because microwaves have a lower frequency than visible light. D is wrong because electromagnetic waves need no medium.
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Answer A
Answer: A
Common error: Different parts of the electromagnetic spectrum differ in frequency and wavelength but not in speed in a vacuum.
32IGPC3-3-05r9-09Sound and ultrasoundPaper CEasy5 marks
An archaeologist stands in a large underground chamber and claps her hands once. Take the speed of sound in air as 340 m/s.
Diagram not to scale
(a)
State what is meant by an echo.
(1)
(b)
She claps more loudly. State which property of the sound wave increases.
(1)
(c)
She hears the echo 0.50 s after the clap. Calculate the distance from her to the wall that reflects the sound.
(3)
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Part (a)
sound (wave) reflected from a surface / wall
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Part (b)
amplitude
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Part (c)
distance travelled by sound = 340 × 0.50 = 170 m
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halved because sound travels to the wall and back
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distance = 85 m
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Common error: In echo questions the sound travels to the wall and back, so the distance to the wall is half the total distance travelled.
33IGPC3-3-05r9-11The electromagnetic spectrumPaper CEasy5 marks
A data centre sends signals along an optical fibre using infrared radiation of wavelength 1500 nm (1 nm = 1 × 10−9 m). The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
State one property that all electromagnetic waves share.
(1)
(b)
Calculate the frequency of the infrared radiation.
(3)
(c)
Radio waves are used for short wireless links in the same building. State whether radio waves have a longer or shorter wavelength than infrared.
(1)
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Part (a)
transverse waves / travel at the same speed in a vacuum / transfer energy without transferring matter / can travel through a vacuum
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Part (b)
frequency = speed ÷ wavelength
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wavelength converted to 1.5 × 10−6 m
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frequency = 2.0 × 1014 Hz
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Part (c)
longer
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Exam tip: Convert nanometres to metres before using v = fλ; the powers of ten are where most slips happen.
34IGPC3-3-06r12-01The electromagnetic spectrumPaper CEasy1 mark
A gamer's wireless controller sends signals using infrared radiation. Which type of electromagnetic wave has a longer wavelength than infrared?
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In order of increasing wavelength the sequence is X-rays, ultraviolet, visible light, infrared, microwaves, radio waves. Only microwaves are longer than infrared: D is correct. A, B and C all have shorter wavelengths.
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Answer D
Answer: D
Common error: Wavelength increases as frequency decreases, so the order by wavelength is the reverse of the order by frequency.
35IGPC3-3-06r12-02Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Waves on a reservoir have frequency 0.50 Hz and wavelength 3.0 m. What is the speed of the waves?
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v = fλ = 0.50 × 3.0 = 1.5 m/s. 0.17 m/s comes from f ÷ λ. 3.5 m/s adds the two quantities. 6.0 m/s comes from λ ÷ f.
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Answer B
Answer: B
Exam tip: Check units: speed in m/s comes from multiplying Hz (1/s) by m.
36IGPC3-3-06r12-03Sound and ultrasoundPaper CEasy1 mark
A maintenance worker taps one end of a long steel water pipe. A colleague at the other end can hear the sound travelling through the steel wall, the water in the pipe and the air outside the pipe. Through which does the sound arrive first?
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Notes
Sound travels fastest in solids, where the particles are close together and strongly bonded, slower in liquids and slowest in gases: it arrives first through the steel, C. A is the slowest route. B is faster than air but slower than steel. D is wrong because the speed depends on the medium.
✓ 1
Answer C
Answer: C
Common error: Sound travels faster in denser, more rigid media, not slower.
37IGPC3-3-06r12-09The electromagnetic spectrumPaper CEasy5 marks
Ultraviolet lamps are used in a pharmacy to sterilise equipment.
(a)
State the speed of electromagnetic waves in a vacuum.
(1)
(b)
The ultraviolet radiation has a frequency of 1.2 × 1015 Hz. Calculate its wavelength.
(2)
(c)
State whether ultraviolet has a longer or shorter wavelength than visible light and give one hazard of exposure to it.
(2)
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Notes
Part (a)
3.0 × 108 m/s
✓ 1
Part (b)
λ = v / f
✓ 1
λ = 3.0 × 108 / 1.2 × 1015 = 2.5 × 10−7 m
✓ 1
Part (c)
Shorter
✓ 1
Any one from: skin burns, skin cancer, eye damage
✓ 1
Exam tip: Write the equation, substitute with standard form, then give the unit.
38IGPC3-3-06r12-10Light, refraction, lenses and dispersionPaper CEasy5 marks
A gamer uses a plane mirror to check a webcam position.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
A ray strikes the mirror with an angle of incidence of 35°. Calculate the angle between the incident ray and the reflected ray.
(2)
(c)
State two characteristics of the image formed by a plane mirror.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The angle of incidence equals the angle of reflection (both measured from the normal)
✓ 1
Part (b)
Angle of reflection = 35°
✓ 1
Angle between rays = 35° + 35° = 70°
✓ 1
Part (c)
Any one from: virtual, upright, same size as the object, laterally inverted
✓ 1
A second different characteristic, or: as far behind the mirror as the object is in front
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the mirror surface.
39IGPC3-3-06r14-01Sound and ultrasoundPaper CEasy1 mark
A train's horn sends a sound wave through the air to a person standing beside the track. Which statement about the sound wave is correct?
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Notes
A is correct: sound is a longitudinal wave, so particles oscillate parallel to the direction of travel, forming compressions and rarefactions. B describes a transverse wave. C is wrong because the wave transfers energy, not the particles themselves. D is wrong because sound needs a medium and cannot travel through a vacuum.
✓ 1
Answer A
Answer: A
Common error: Waves transfer energy without transferring matter; the particles only oscillate about fixed positions.
40IGPC3-3-06r14-03The electromagnetic spectrumPaper CEasy1 mark
An astronomer's telescope detects radio waves, infrared, ultraviolet and gamma rays from a distant nebula. Which row lists these in order of increasing frequency?
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Notes
In the electromagnetic spectrum, frequency increases from radio waves to infrared, visible, ultraviolet, X-rays and gamma rays, so A. B is the order of decreasing frequency. C and D swap neighbouring members of the spectrum.
✓ 1
Answer A
Answer: A
Exam tip: Learn the spectrum in order: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. Frequency rises along this list.
41IGPC3-3-06r14-04Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
An athlete in a gym stands 2.0 m in front of a large plane mirror. What is the distance between the athlete and their image?
Diagram not to scale
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Notes
The image is as far behind the mirror as the athlete is in front of it, so the athlete-to-image distance is 2.0 + 2.0 = 4.0 m (C). A halves the distance and B uses only the distance to the mirror. D doubles the correct total.
✓ 1
Answer C
Answer: C
Common error: The image distance is measured from the mirror, so the athlete-to-image separation is twice the athlete-to-mirror distance.
42IGPC3-3-06r14-09Light, refraction, lenses and dispersionPaper CEasy4 marks
A spectrometer attached to an astronomer's telescope uses a glass prism to split starlight into a spectrum.
(a)
State the name of this effect.
(1)
(b)
State which colour of visible light is refracted the most by the prism.
(1)
(c)
Explain why the different colours leave the prism in different directions.
(2)
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Notes
Part (a)
dispersion
✓ 1
Part (b)
violet
✓ 1
Part (c)
different colours (wavelengths) travel at different speeds in glass
✓ 1
so each colour is refracted through a different angle
✓ 1
Exam tip: Violet light has the shortest wavelength of visible light and is refracted most; red has the longest wavelength and is refracted least.
43IGPC3-3-06r14-16Light, refraction, lenses and dispersionPaper CEasy4 marks
An astronomer wants to estimate the focal length of a converging lens.
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
Describe how the astronomer can estimate the focal length using a distant light source and a screen.
(3)
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Notes
Part (a)
the point at which rays travelling parallel to the principal axis are brought together (converge) after passing through the lens
✓ 1
Part (b)
point the lens at a distant object or light source so the rays arriving are almost parallel
✓ 1
move a screen behind the lens until a sharp image forms
✓ 1
measure the distance from the lens to the screen, which is the focal length
✓ 1
Exam tip: Light from a very distant source is almost parallel, which is why the sharp image forms at the principal focus.
44IGPC3-3-06r19-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Surface waves in a marine aquarium have a frequency of 4.0 Hz and a wavelength of 0.15 m. What is the speed of the waves?
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Notes
Wave speed = frequency × wavelength = 4.0 × 0.15 = 0.60 m/s, so B is correct. A divides wavelength by frequency (0.15 ÷ 4.0). C adds the two quantities. D divides frequency by wavelength (4.0 ÷ 0.15).
✓ 1
Answer B
Answer: B
Common error: Mixing up which quantities to multiply and which to divide; check that the units of your answer, m/s, match a distance per time.
45IGPC3-3-06r19-02The electromagnetic spectrumPaper CEasy1 mark
A logistics depot uses radio tags to track pallets, a barcode scanner that emits red light, and an X-ray scanner to check sealed containers. Which statement about these three types of wave is correct?
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Notes
All electromagnetic waves travel at 3.0 × 108 m/s in a vacuum, so B is correct. A is wrong because the types differ in frequency and wavelength. C is wrong because X-rays have the highest frequency and radio waves the lowest. D is wrong because all electromagnetic waves are transverse.
✓ 1
Answer B
Answer: B
Exam tip: Remember the order of the spectrum: radio waves have the longest wavelength and lowest frequency, gamma rays the shortest wavelength and highest frequency.
46IGPC3-3-06r19-04Sound and ultrasoundPaper CEasy1 mark
A blue whale produces calls of frequency 15 Hz. Humans with normal hearing can detect sounds from 20 Hz to 20 kHz. Which of these frequencies is too low for a human to hear?
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Notes
Sounds below 20 Hz are infrasound and cannot be heard, so only 15 Hz (A) is outside the hearing range. 150 Hz, 1.5 kHz (1500 Hz) and 15 kHz (15 000 Hz) all lie between 20 Hz and 20 kHz.
✓ 1
Answer A
Answer: A
Common error: Forgetting that 1 kHz = 1000 Hz, so 15 kHz is 15 000 Hz and well inside the range of human hearing.
47IGPC3-3-06r19-10The electromagnetic spectrumPaper CEasy5 marks
A logistics company uses three kinds of electromagnetic radiation: radio waves of frequency 900 MHz for tracking tags, red light for barcode scanners and X-rays to inspect cargo containers.
(a)
State two properties that are common to all electromagnetic waves.
(2)
(b)
State one hazard of exposure to X-rays.
(1)
(c)
The speed of electromagnetic waves in air is 3.0 × 108 m/s. Calculate the wavelength of the 900 MHz radio waves. Give your answer to 2 significant figures.
(2)
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Notes
Part (a)
one valid property, e.g. they are transverse waves
✓ 1
a second, different valid property, e.g. they travel at the same speed in a vacuum (3.0 × 108 m/s), they transfer energy, they can travel through a vacuum
✓ 1
Part (b)
damage to cells / cancer / mutations / burns
✓ 1
Part (c)
900 MHz = 9.0 × 108 Hz and wavelength = speed ÷ frequency
✓ 1
0.33 m
✓ 1
Common error: Forgetting that 1 MHz = 1 000 000 Hz; convert to hertz before using v = fλ.
48IGPC3-3-08r23-01Sound and ultrasoundPaper CEasy1 mark
A retail store plays background music through a loudspeaker. Which statement about the sound waves from the loudspeaker is correct?
Show mark scheme
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Notes
Sound is a longitudinal wave, so the air particles vibrate back and forth along the direction of travel. Vibration at right angles describes a transverse wave. The particles only oscillate about fixed positions; the wave transfers energy, not the air itself. Sound needs a medium and cannot travel through a vacuum.
✓ 1
Answer A
Answer: A
Common error: A wave transfers energy without transferring matter; the particles only vibrate about their rest positions.
49IGPC3-3-08r23-02Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
A wave-energy device floats at sea. Water waves of frequency 0.25 Hz and wavelength 40 m pass it. What is the speed of these waves?
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Notes
Wave speed = frequency × wavelength = 0.25 × 40 = 10 m/s. The value 160 m/s comes from dividing wavelength by frequency. The value 0.0063 m/s comes from dividing frequency by wavelength. The value 40 m/s ignores the frequency.
✓ 1
Answer B
Answer: B
Exam tip: Write v = f × λ and check that the units are Hz, m and m/s before you substitute.
50IGPC3-3-08r23-08The electromagnetic spectrumPaper CEasy1 mark
A hospital sterilises surfaces using a lamp that emits radiation with a higher frequency than visible light and a lower frequency than X-rays. Which type of radiation is this?
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Notes
In the electromagnetic spectrum, the order of increasing frequency is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Ultraviolet lies between visible light and X-rays. The other three all have lower frequencies than visible light.
✓ 1
Answer D
Answer: D
Exam tip: Learn the order of the electromagnetic spectrum from radio waves to gamma rays and use frequency or wavelength consistently.
51IGPC3-3-08r23-09Wave properties, reflection, refraction and diffractionPaper CEasy6 marks
A wave-energy buoy bobs up and down as water waves pass it. 15 wave crests pass the buoy in 60 s. The distance between neighbouring crests is 24 m.
Diagram not to scale
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the speed of the waves.
(2)
(c)
State whether water waves are transverse or longitudinal and explain what this means.
(2)
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Notes
Part (a)
f = number of waves / time
✓ 1
= 15 / 60 = 0.25 Hz
✓ 1
Part (b)
v = f × λ
✓ 1
= 0.25 × 24 = 6.0 m/s
✓ 1
Part (c)
Transverse
✓ 1
The water vibrates at right angles to the direction in which the wave travels
✓ 1
Exam tip: The distance between neighbouring crests is the wavelength; do not confuse it with the amplitude.
52IGPC3-3-08r23-12Light, refraction, lenses and dispersionPaper CEasy5 marks
A shop mirror is used to reflect light from a window to a counter. A ray of light strikes the plane mirror at an angle of incidence of 35°.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
Calculate the angle between the incident ray and the reflected ray.
(2)
(c)
A ray of light enters a glass block from air at an angle and bends towards the normal. Explain why.
(2)
Show mark scheme
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Notes
Part (a)
The angle of incidence equals the angle of reflection
✓ 1
Part (b)
Angle of reflection = 35°
✓ 1
35° + 35° = 70°
✓ 1
Part (c)
The light travels more slowly in glass than in air
✓ 1
The change in speed makes the ray change direction at the boundary
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the mirror surface.
53IGPC3-3-06r26-01Sound and ultrasoundPaper CEasy1 mark
A volcano emits low-frequency sound waves of frequency 8.5 Hz. The speed of sound in the air is 340 m/s. What is the wavelength of these waves?
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Notes
Wavelength = speed ÷ frequency = 340 ÷ 8.5 = 40 m. Dividing frequency by speed gives 0.025 m, which inverts the equation. Multiplying speed by frequency gives about 2900 m, which uses the wrong operation. 4.0 m comes from a slip of a factor of ten in the division.
✓ 1
Answer C
Answer: C
Exam tip: Rearrange v = fλ before substituting, and check that your wavelength is a sensible size for the wave.
54IGPC3-3-06r26-02The electromagnetic spectrumPaper CEasy1 mark
In a theatre, a technician uses lamps that give out ultraviolet radiation, visible light, infrared radiation and microwaves. Which list shows these in order of increasing wavelength?
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Notes
In the electromagnetic spectrum, wavelength increases from ultraviolet to visible light to infrared to microwaves, so this order is correct. The reversed list is in order of decreasing wavelength, so it is wrong. The list starting with visible light places ultraviolet in the wrong position, since it has a shorter wavelength than visible light. The list starting with infrared places infrared before visible light and microwaves last, so it is not in order of wavelength.
✓ 1
Answer A
Answer: A
Common error: Do not confuse increasing frequency with increasing wavelength: they run in opposite directions across the spectrum.
55IGPC3-3-06r26-06Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Water waves of wavelength 12 m reach the entrance to a harbour. The entrance is a gap in the sea wall 10 m wide. Which change would make the diffraction at the gap smaller?
Diagram not to scale
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Notes
Increasing the frequency at constant wave speed decreases the wavelength, and the amount of diffraction is smaller when the wavelength is small compared with the gap. A narrower gap makes diffraction greater. Decreasing the frequency increases the wavelength and so increases diffraction. Moving the source further away does not change the wavelength or the gap, so it does not change the diffraction.
✓ 1
Answer D
Answer: D
Exam tip: Diffraction is greatest when the gap is about the same size as the wavelength or smaller.
56IGPC3-3-06r26-09Sound and ultrasoundPaper CEasy7 marks
A volcano erupts with a loud explosion. An observer is 6.8 km from the volcano. The speed of sound in the air is 340 m/s.
Diagram not to scale
(a)
Calculate the time between the observer seeing the eruption and hearing the explosion. Assume that the light arrives instantly.
(2)
(b)
Explain why the observer sees the eruption before hearing it.
(1)
(c)
The explosion also produces sound of frequency 12 Hz. State whether the observer can hear this sound, and give a reason.
(2)
(d)
Calculate the wavelength of the 12 Hz sound. Give your answer to 3 significant figures.
(2)
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Mark
Notes
Part (a)
distance = speed × time, so time = 6800 ÷ 340
✓ 1
20 s
✓ 1
Part (b)
light travels much faster than sound
✓ 1
Part (c)
no
✓ 1
12 Hz is below the lowest frequency a human can hear, about 20 Hz
✓ 1
Part (d)
λ = v ÷ f = 340 ÷ 12
✓ 1
28.3 m
✓ 1
Common error: Convert kilometres to metres before using the speed in m/s.
57IGPC3-3-06r26-10Light, refraction, lenses and dispersionPaper CEasy6 marks
A stage actor stands in front of a large plane mirror at the back of a theatre stage.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
The actor is 3.0 m from the mirror. Calculate the distance between the actor and his image.
(1)
(c)
A ray of light from a lamp hits the mirror with an angle of incidence of 35°. Calculate the angle between the incident ray and the reflected ray.
(2)
(d)
State two characteristics of the image formed by a plane mirror.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the angle of incidence equals the angle of reflection
✓ 1
Part (b)
6.0 m
✓ 1
Part (c)
angle of reflection = 35°, so the angle between the rays is 35° + 35°
✓ 1
70°
✓ 1
Part (d)
any valid characteristic, for example virtual
✓ 1
a second valid characteristic, for example upright / same size as the object / laterally inverted / image as far behind the mirror as the object is in front
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the surface of the mirror.
58IGPC3-3-07r29-01Sound and ultrasoundPaper CEasy1 mark
A mountaineer on a glacier shouts towards a rock face and hears the echo 2.4 s later. The speed of sound in the cold air is 340 m/s. How far away is the rock face?
Diagram not to scale
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Notes
The sound travels to the rock face and back, so the total distance is 340 × 2.4 = 816 m and the rock face is half of this, 408 m. 816 m forgets that the sound makes a round trip. 142 m comes from dividing speed by time instead of multiplying. 204 m halves the distance twice.
✓ 1
Answer B
Answer: B
Common error: An echo covers the distance to the reflector twice, so always halve speed × time.
59IGPC3-3-07r29-02The electromagnetic spectrumPaper CEasy1 mark
A telecoms company uses microwaves for a satellite link and infrared radiation in an optical fibre. Which statement about these two types of electromagnetic wave is correct?
Show mark scheme
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Mark
Notes
Microwaves lie below infrared in frequency, so they have a longer wavelength, and all electromagnetic waves travel at 3.0 × 108 m/s in a vacuum. The statements claiming higher frequency or shorter wavelength reverse the order of the spectrum, and the statement about faster travel wrongly gives different speeds to different electromagnetic waves.
✓ 1
Answer D
Answer: D
Exam tip: Learn the spectrum in order of wavelength, then remember that frequency goes the opposite way.
60IGPC3-3-07r29-06Sound and ultrasoundPaper CEasy1 mark
A mountain guide blows a whistle and then blows it again so that the note is louder but has the same pitch. Which row describes the change in the sound wave?
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Notes
Loudness depends on amplitude, so a louder note has a greater amplitude. Pitch depends on frequency, so the same pitch means the same frequency. The other rows change the frequency, which would change the pitch, or keep the amplitude the same, which would not make the note louder.
✓ 1
Answer B
Answer: B
Common error: Louder means larger amplitude; higher pitch means higher frequency. Do not mix them up.
61IGPC3-3-07r29-09Sound and ultrasoundPaper CEasy6 marks
A mountaineer at the foot of a cliff shouts and hears an echo 3.6 s later. The speed of sound in the air is 340 m/s.
Diagram not to scale
(a)
Explain why the mountaineer hears an echo.
(1)
(b)
Calculate the distance from the mountaineer to the cliff.
(3)
(c)
The shout has a frequency of 425 Hz. Calculate the wavelength of the sound.
(2)
Show mark scheme
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Mark
Notes
Part (a)
The sound waves are reflected by the cliff and return to the mountaineer
✓ 1
Part (b)
M1 total distance travelled = speed × time = 340 × 3.6
✓ 1
M1 distance to the cliff is half of the total distance
✓ 1
A1 612 m
✓ 1
Part (c)
M1 wavelength = speed ÷ frequency = 340 ÷ 425
✓ 1
A1 0.80 m
✓ 1
Exam tip: Write the equation v = fλ, substitute, then give the answer with its unit.
62IGPC3-3-07r29-12Light, refraction, lenses and dispersionPaper CEasy6 marks
A food scientist uses a thin converging lens of focal length 6.0 cm to inspect a grain of salt.
Diagram not to scale
(a)
State what is meant by the principal focus of a converging lens.
(2)
(b)
The grain of salt is placed 4.0 cm from the lens. State three features of the image seen.
(3)
(c)
State the condition on the distance between a grain and the lens for a real image to be formed.
(1)
Show mark scheme
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Notes
Part (a)
The point on the principal axis where rays parallel to the axis meet
✓ 1
after being refracted by the lens
✓ 1
Part (b)
Virtual
✓ 1
Upright
✓ 1
Magnified
✓ 1
Part (c)
The grain must be further from the lens than the focal length (more than 6.0 cm)
✓ 1
Exam tip: Compare the object distance with the focal length first; this decides whether the image is real or virtual.
63IGPC3-3-07r29-16Light, refraction, lenses and dispersionPaper CEasy6 marks
A narrow beam of white light passes through a glass prism and a spectrum is produced on a screen. Later, a mountaineer wears goggles that block ultraviolet radiation from the Sun at high altitude.
(a)
Name the effect that separates white light into different colours.
(1)
(b)
State which colour of visible light is refracted most by the prism.
(1)
(c)
Explain why the different colours are separated by the prism.
(2)
(d)
(i) State whether ultraviolet radiation has a longer or shorter wavelength than violet light. (ii) State one harmful effect of ultraviolet radiation on the human body.
(2)
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Notes
Part (a)
Dispersion
✓ 1
Part (b)
Violet
✓ 1
Part (c)
Different colours have different frequencies (or wavelengths)
✓ 1
so they are refracted by different amounts in the glass (they travel at different speeds)
✓ 1
Part (d)
Shorter
✓ 1
Skin damage, skin cancer or eye damage (any one)
✓ 1
Exam tip: The list of colours in order red to violet is also the order of increasing frequency and increasing amount of bending.
64IGPC3-3-10r32-01The electromagnetic spectrumPaper CEasy1 mark
Which list shows types of electromagnetic wave in order of increasing wavelength?
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Notes
Across the spectrum the wavelength increases from gamma rays and X-rays, through ultraviolet, visible and infrared, to microwaves and radio waves, so the key lists them correctly. The first wrong list is in order of increasing frequency (decreasing wavelength). The other two lists mix up the positions of the regions.
✓ 1
Answer C
Answer: C
Common error: Wavelength and frequency run in opposite directions across the spectrum; check which one the question asks you to order.
65IGPC3-3-10r32-02Sound and ultrasoundPaper CEasy1 mark
A guitarist plucks a string harder, without changing the length or tension of the string. How does the sound change?
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Notes
A harder pluck gives a larger amplitude, which makes the sound louder. The frequency (pitch) is fixed by the string's length and tension, so it is unchanged. Pitch rising confuses amplitude with frequency. The speed of sound in air depends on the air, not on how loud the sound is.
✓ 1
Answer A
Answer: A
Common error: Loudness depends on amplitude; pitch depends on frequency. Changing one does not change the other.
66IGPC3-3-10r32-09The electromagnetic spectrumPaper CEasy6 marks
Air traffic control radar emits pulses of microwaves of frequency 3.0 × 109 Hz. Microwaves travel through air at 3.0 × 108 m/s.
(a)
State the region of the electromagnetic spectrum that has longer wavelengths than microwaves.
(1)
(b)
Calculate the wavelength of the microwaves.
(2)
(c)
A pulse is reflected from an aircraft and the echo returns to the radar after 4.0 × 10−4 s. Calculate the distance from the radar to the aircraft.
(3)
Show mark scheme
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Notes
Part (a)
radio waves
✓ 1
Part (b)
M1 λ = v ÷ f = 3.0 × 108 ÷ 3.0 × 109
✓ 1
A1 0.10 m
✓ 1
Part (c)
M1 distance travelled by the pulse = speed × time = 3.0 × 108 × 4.0 × 10−4 = 1.2 × 105 m
✓ 1
M1 halve, as the pulse travels out and back
✓ 1
A1 6.0 × 104 m
✓ 1
Common error: All electromagnetic waves travel at the same speed in a vacuum; the echo distance is half of speed × time.
67IGPC3-3-10r32-10Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
Water waves travel along an irrigation channel. A student counts 15 waves passing a fixed point in 6.0 s. The distance between neighbouring crests is 0.80 m.
Diagram not to scale
(a)
State what is meant by the wavelength of a wave.
(1)
(b)
Calculate the frequency of the waves.
(2)
(c)
Calculate the speed of the waves.
(2)
Show mark scheme
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Notes
Part (a)
the distance between two adjacent crests (or the distance between the same point on two successive waves)
✓ 1
Part (b)
M1 f = number of waves ÷ time = 15 ÷ 6.0
✓ 1
A1 2.5 Hz
✓ 1
Part (c)
M1 v = fλ = 2.5 × 0.80
✓ 1
A1 2.0 m/s (ECF from (b))
✓ 1
Exam tip: Frequency is the number of waves per second; find it first and then use v = fλ.
68IGPC3-3-09r34-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
During an excavation, a student models ground vibrations by sending waves along a rope. The waves have frequency 4.0 Hz and wavelength 0.75 m. What is the speed of the waves?
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Notes
Wave speed = frequency × wavelength = 4.0 × 0.75 = 3.0 m/s. 0.19 m/s comes from dividing wavelength by frequency. 5.3 m/s comes from dividing frequency by wavelength. 4.8 m/s comes from adding the two quantities.
✓ 1
Answer B
Answer: B
Common error: Check the units of your answer: m/s only arises from Hz × m, never from adding or dividing the two quantities.
69IGPC3-3-09r34-02The electromagnetic spectrumPaper CEasy1 mark
Archaeologists examine a sealed jar using X-rays. Which statement about all electromagnetic waves travelling in a vacuum is correct?
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Notes
All electromagnetic waves travel at 3.0 × 108 m/s in a vacuum. Gamma rays are not faster than radio waves, since higher frequency does not mean higher speed. The waves differ in wavelength and frequency. They are transverse, not longitudinal.
✓ 1
Answer C
Answer: C
Exam tip: Frequency and wavelength change across the spectrum but the speed in a vacuum does not.
70IGPC3-3-09r34-08The electromagnetic spectrumPaper CEasy1 mark
A thermal imaging camera in a data centre finds hot server racks. Which type of electromagnetic radiation does the camera detect?
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Notes
Warm objects emit infrared radiation, which thermal cameras detect. Ultraviolet and X-rays are emitted mainly by much hotter or high-energy sources. Microwaves are used for communication and cooking.
✓ 1
Answer B
Answer: B
Exam tip: Link infrared with heat and thermal imaging.
71IGPC3-3-09r34-09Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
Divers at an underwater archaeological site watch water waves pass a marker buoy on the surface. The distance between neighbouring crests is 4.5 m.
Diagram not to scale
(a)
State what is meant by the wavelength of a wave.
(1)
(b)
The buoy completes 12 complete oscillations in 20 s. Calculate the frequency of the waves.
(2)
(c)
Calculate the speed of the waves. Use your answer to (b).
(2)
Show mark scheme
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Mark
Notes
Part (a)
The distance between two adjacent points in phase, for example crest to crest
✓ 1
Part (b)
M1 f = 12 ÷ 20
✓ 1
A1 0.60 Hz
✓ 1
Part (c)
M1 v = fλ = 0.60 × 4.5
✓ 1
A1 2.7 m/s, ECF from (b)
✓ 1
Exam tip: Write the equation, substitute with units, then state the answer.
72IGPC3-3-09r34-12The electromagnetic spectrumPaper CEasy4 marks
A technician in a data centre uses a range of electromagnetic waves. The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
Put these in order of wavelength, starting with the shortest: infrared, X-rays, microwaves, visible light.
(1)
(b)
State one use of microwaves.
(1)
(c)
A microwave signal has frequency 2.0 × 109 Hz. Calculate its wavelength.
(2)
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Notes
Part (a)
X-rays, visible light, infrared, microwaves (all four in the correct order)
✓ 1
Part (b)
Any one from: mobile phone communication; satellite communication; cooking
✓ 1
Part (c)
M1 λ = v ÷ f = 3.0 × 108 ÷ 2.0 × 109
✓ 1
A1 0.15 m
✓ 1
Common error: Higher frequency means shorter wavelength, so reverse the order if you start from frequency.
73IGPC3-3-10r34-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Engineers studying a bridge compare light waves and sound waves. Which row is correct?
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Notes
Light is an electromagnetic wave whose oscillations are at right angles to its direction of travel, so it is transverse. Sound is a series of compressions and rarefactions along the direction of travel, so it is longitudinal. The other rows swap or repeat the wave types.
✓ 1
Answer D
Answer: D
Exam tip: Link transverse to light and water ripples, and longitudinal to sound.
74IGPC3-3-10r34-02Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Water waves pass under a bridge. The vertical distance from the top of a crest to the bottom of the next trough is 1.2 m. What is the amplitude of the waves?
Diagram not to scale
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Notes
Amplitude is the maximum displacement from the undisturbed level, which is half the crest-to-trough distance: 1.2 ÷ 2 = 0.60 m. 1.2 m treats crest-to-trough as the amplitude. 2.4 m doubles it. 0.30 m halves it twice.
✓ 1
Answer A
Answer: A
Common error: Amplitude is measured from the middle level, not from trough to crest.
75IGPC3-3-10r34-06Sound and ultrasoundPaper CEasy1 mark
Which frequency of sound can be heard by a healthy young person?
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Notes
The audible range is about 20 Hz to 20 000 Hz, so 15 kHz (15 000 Hz) is within it. 15 Hz is below the range (infrasound). 25 kHz and 2.5 MHz are above it (ultrasound).
✓ 1
Answer D
Answer: D
Exam tip: Convert kHz and MHz to Hz before comparing with 20 Hz to 20 000 Hz.
76IGPC3-3-10r34-09Sound and ultrasoundPaper CEasy5 marks
A space probe sends a radio signal to Earth. The signal takes 1500 s to arrive. The speed of radio waves in space is 3.0 × 108 m/s.
(a)
State why radio waves, rather than sound waves, are used to communicate with the probe.
(1)
(b)
Calculate the distance from the probe to Earth.
(2)
(c)
The signal has frequency 6.0 × 109 Hz. Calculate its wavelength.
(2)
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Notes
Part (a)
Sound cannot travel through the vacuum of space, but radio waves can
✓ 1
Part (b)
M1 distance = speed × time = 3.0 × 108 × 1500
✓ 1
A1 4.5 × 1011 m
✓ 1
Part (c)
M1 λ = v ÷ f = 3.0 × 108 ÷ 6.0 × 109
✓ 1
A1 0.050 m
✓ 1
Exam tip: Write powers of ten carefully when using v = fλ.
77IGPC3-3-10r34-12Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
A cable on a bridge vibrates at right angles to its length, and a wave travels along it. The wave has period 0.40 s and wavelength 3.0 m.
(a)
State whether the wave is transverse or longitudinal.
(1)
(b)
Calculate the frequency of the wave.
(2)
(c)
Calculate the speed of the wave. Use your answer to (b).
(2)
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Notes
Part (a)
Transverse
✓ 1
Part (b)
M1 f = 1 ÷ T = 1 ÷ 0.40
✓ 1
A1 2.5 Hz
✓ 1
Part (c)
M1 v = fλ = 2.5 × 3.0
✓ 1
A1 7.5 m/s, ECF from (b)
✓ 1
Exam tip: Period and frequency are reciprocals: f = 1/T.
78IGPC3-3-09r36-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
A wave machine in a swimming pool makes waves of wavelength 0.50 m and frequency 1.6 Hz. What is the speed of the waves?
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Notes
Speed = frequency × wavelength = 1.6 × 0.50 = 0.80 m/s. 0.31 m/s comes from dividing wavelength by frequency, 3.2 m/s from dividing frequency by wavelength, and 2.1 m/s from adding the two quantities.
✓ 1
Answer B
Answer: B
Common error: Check that your answer has units of m/s: wavelength (m) multiplied by frequency (1/s) gives m/s, whereas dividing gives a different unit.
79IGPC3-3-09r36-02The electromagnetic spectrumPaper CEasy1 mark
Which list shows types of electromagnetic radiation in order of increasing wavelength, starting with the shortest?
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Notes
X-rays have a shorter wavelength than ultraviolet, which is shorter than infrared, which is shorter than microwaves, so this list rises in wavelength. The list starting with radio waves runs from the longest wavelength to the shortest. The list starting with gamma rays puts microwaves before ultraviolet, but microwaves have a much longer wavelength. The list starting with ultraviolet swaps ultraviolet and X-rays, but X-rays have the shorter wavelength.
✓ 1
Answer D
Answer: D
Exam tip: Learn the spectrum in one direction (radio to gamma) and then read the question carefully to see whether it asks for increasing or decreasing wavelength or frequency.
80IGPC3-3-09r36-09Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
A wave machine in a swimming pool produces water waves that travel along the pool.
Diagram not to scale
(a)
State what is meant by the wavelength of a wave.
(1)
(b)
A swimmer measures the distance from the first crest to the sixth crest in a line of waves as 2.5 m. Calculate the wavelength.
(2)
(c)
The wave machine produces 1.5 waves every second. Calculate the speed of the waves. Use your wavelength from (b).
(2)
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Notes
Part (a)
the distance between two adjacent crests (or between the same point on two successive waves)
✓ 1
Part (b)
there are 5 wavelengths between 6 crests
✓ 1
wavelength = 2.5 / 5 = 0.50 m
✓ 1
Part (c)
speed = frequency × wavelength
✓ 1
speed = 1.5 × 0.50 = 0.75 m/s (ECF from (b))
✓ 1
Common error: Six crests in a row enclose only five wavelengths, not six.
81IGPC3-3-09r36-10The electromagnetic spectrumPaper CEasy5 marks
A wind farm control room communicates with each turbine using radio waves.
(a)
State the speed of all electromagnetic waves in a vacuum.
(1)
(b)
The signal has a frequency of 150 MHz (1 MHz = 106 Hz). Calculate the wavelength of the signal.
(2)
(c)
Engineers check turbine blades for hot spots using a thermal imaging camera. State the type of electromagnetic radiation that the camera detects.
(1)
(d)
Technicians working outdoors on the turbines are exposed to ultraviolet radiation. State one harmful effect of ultraviolet radiation on the human body.
(1)
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Notes
Part (a)
3.0 × 108 m/s
✓ 1
Part (b)
wavelength = speed / frequency
✓ 1
wavelength = 3.0 × 108 / 1.5 × 108 = 2.0 m
✓ 1
Part (c)
infrared
✓ 1
Part (d)
skin cancer or sunburn or eye damage (any one)
✓ 1
Exam tip: Convert prefixes such as M (10⁶) to standard form before substituting into v = fλ.
82IGPC3-3-09r36-11Light, refraction, lenses and dispersionPaper CEasy5 marks
A library has a large plane mirror fixed to one wall.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
A ray of light strikes the mirror so that it makes an angle of 35° with the surface of the mirror. Calculate the angle of reflection.
(2)
(c)
State two characteristics of the image of a reader formed by a plane mirror.
(2)
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Notes
Part (a)
the angle of incidence equals the angle of reflection
✓ 1
Part (b)
angle of incidence = 90° − 35° = 55°
✓ 1
angle of reflection = 55°
✓ 1
Part (c)
any two from: virtual; upright; same size as the object; same distance behind the mirror as the object is in front; laterally inverted
✓ 2
Common error: Angles of incidence and reflection are measured between the ray and the normal, not between the ray and the mirror surface.
83IGPC3-3-09r37-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
A beekeeper taps the wall of a hive and the sound travels through the air to her ear. Which statement about the air particles as the sound wave passes is correct?
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Notes
Sound is a longitudinal wave, so the particles vibrate parallel to the direction of travel, forming compressions and rarefactions. Vibration at right angles describes a transverse wave such as light. The particles only oscillate about fixed positions and do not travel from hive to ear. Circular motion at right angles to the travel direction is not how sound moves the air.
✓ 1
Answer A
Answer: A
Common error: Do not say that sound carries air particles from the source to the listener; the wave transfers energy while the particles only oscillate.
84IGPC3-3-09r37-03The electromagnetic spectrumPaper CEasy1 mark
Four types of electromagnetic radiation are radio waves, microwaves, infrared and X-rays. Which list shows them in order of increasing wavelength?
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Notes
Along the spectrum from gamma rays to radio waves, wavelength increases while frequency decreases, so X-rays have the shortest wavelength and radio waves the longest, with infrared shorter than microwaves. The list starting with radio waves is in order of increasing frequency, not wavelength. Placing microwaves before infrared swaps two neighbours. The last list does not follow any order along the spectrum.
✓ 1
Answer B
Answer: B
Common error: Wavelength and frequency run in opposite directions along the electromagnetic spectrum, so check which one the question asks you to order.
85IGPC3-3-09r37-05Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Water waves of wavelength 0.40 m in a hospital hydrotherapy pool pass through a gap between two barriers. Which width of gap produces the most noticeable diffraction of the waves?
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Notes
Diffraction is most noticeable when the gap width is about the same as the wavelength, which is the case for a 0.40 m gap. The wider gaps of 1.6 m, 3.2 m and 6.4 m are 4, 8 and 16 times the wavelength, so most of the wave passes straight through with little spreading. The widest gaps show the least diffraction, not the most.
✓ 1
Answer C
Answer: C
Common error: A wider gap does not give more spreading; diffraction is greatest when the gap is comparable to the wavelength or narrower.
86IGPC3-3-09r37-09Sound and ultrasoundPaper CEasy6 marks
A beekeeper holds a microphone close to a hive and records a steady buzzing sound of frequency 200 Hz. The speed of sound in air is 340 m/s.
(a)
State what is meant by the frequency of a wave.
(1)
(b)
Calculate the wavelength of the sound waves in air.
(2)
(c)
A second hive produces a buzz of frequency 400 Hz. Calculate the wavelength of this sound in air and state how it compares with your answer to (b).
(3)
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Notes
Part (a)
the number of complete waves (oscillations) passing a point, or produced by the source, each second
✓ 1
Part (b)
wavelength = speed ÷ frequency = 340 ÷ 200
✓ 1
1.7 m
✓ 1
Part (c)
340 ÷ 400
✓ 1
0.85 m
✓ 1
the wavelength is half as long / shorter
✓ 1
Exam tip: Write the equation wavelength = speed ÷ frequency first, then substitute the numbers; this earns the method mark even if the arithmetic slips.
87IGPC3-3-09r37-10Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
A Formula 1 driver checks a flat rear-view mirror. Light from a following car reaches the mirror along a ray that makes an angle of 28° with the surface of the mirror.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
Calculate the angle of incidence of the ray.
(2)
(c)
Calculate the angle between the incident ray and the reflected ray.
(2)
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Notes
Part (a)
the angle of incidence equals the angle of reflection (both measured from the normal)
✓ 1
Part (b)
the normal is at 90° to the mirror surface, so angle from the normal = 90 − 28
✓ 1
62°
✓ 1
Part (c)
the angle of reflection is also 62°
✓ 1
angle between rays = 2 × 62 = 124° (ECF from (b))
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the surface of the mirror.
88IGPC3-3-08r40-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Ripples spread across a swimming pool. The source vibrates at 2.5 Hz and the distance between neighbouring crests is 0.40 m. What is the speed of the ripples?
Diagram not to scale
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Notes
Wave speed = frequency × wavelength = 2.5 × 0.40 = 1.0 m/s. 0.16 m/s comes from dividing wavelength by frequency. 2.9 m/s comes from adding the two quantities. 6.3 m/s comes from dividing frequency by wavelength.
✓ 1
Answer B
Answer: B
Common error: Check the units: frequency in Hz multiplied by wavelength in m gives m/s only when you multiply, never divide.
89IGPC3-3-08r40-02Sound and ultrasoundPaper CEasy1 mark
A librarian claps once in a large empty hall and hears an echo 0.40 s later. The speed of sound in air is 330 m/s. How far away is the wall that reflects the sound?
Diagram not to scale
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Notes
The sound travels to the wall and back, so the distance to the wall is 330 × 0.40 ÷ 2 = 66 m. 132 m is the full there-and-back distance, found by not halving. 33 m comes from halving twice. 825 comes from dividing speed by time, which does not give a distance.
✓ 1
Answer C
Answer: C
Exam tip: For echo questions the sound covers the distance twice, so halve speed × time.
90IGPC3-3-08r40-03The electromagnetic spectrumPaper CEasy1 mark
Radar at a wind farm uses microwaves. Which type of electromagnetic radiation has a higher frequency than microwaves but a lower frequency than visible light?
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Notes
In order of increasing frequency the spectrum runs radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Infrared lies between microwaves and visible light. Radio waves have a lower frequency than microwaves. Ultraviolet and X-rays have higher frequencies than visible light.
✓ 1
Answer A
Answer: A
Exam tip: Learn the electromagnetic spectrum as one ordered list, then read it forwards for increasing frequency and backwards for increasing wavelength.
91IGPC3-3-08r40-09The electromagnetic spectrumPaper CEasy6 marks
A wind farm uses radar with microwaves of frequency 9.4 GHz (9.4 × 109 Hz). The speed of electromagnetic waves in air may be taken as the same as in a vacuum.
(a)
State the speed of electromagnetic waves in a vacuum.
(1)
(b)
Calculate the wavelength of the microwaves. Give your answer to 2 significant figures.
(2)
(c)
State one danger of exposure of the human body to too much microwave radiation.
(1)
(d)
State two other properties that are common to all electromagnetic waves.
any two from: they are transverse waves; they transfer energy without transferring matter; they can travel through a vacuum; they can be reflected and refracted; they obey v = fλ
✓ 2
Exam tip: Write the equation in words or symbols first, then substitute: it earns the method mark even if the arithmetic slips.
92IGPC3-3-08r40-10Wave properties, reflection, refraction and diffractionPaper CEasy6 marks
A floating ball in a swimming pool bobs up and down 15 times in 10 s as ripples pass it. The distance between neighbouring crests of the ripples is 0.40 m. The ripples travel horizontally across the pool.
(a)
Explain why the ripples are transverse waves.
(2)
(b)
Calculate the frequency of the ripples.
(2)
(c)
Calculate the speed of the ripples.
(2)
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Notes
Part (a)
the ball (the water) vibrates vertically / up and down
✓ 1
this is perpendicular to the direction in which the wave travels
✓ 1
Part (b)
frequency = number of waves ÷ time = 15 ÷ 10
✓ 1
1.5 Hz
✓ 1
Part (c)
speed = frequency × wavelength = 1.5 × 0.40
✓ 1
0.60 m/s
✓ 1
Common error: A transverse wave vibrates at right angles to its direction of travel; do not say it just moves 'up and down' without saying what it is compared with.
93IGPC3-3-09r42-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
On a water-supply reservoir, a student counts 12 wave crests passing a fixed buoy in 4.0 s. The distance between neighbouring crests is 0.50 m. What is the speed of the waves?
Diagram not to scale
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Notes
Frequency = 12 ÷ 4.0 = 3.0 Hz, so speed = frequency × wavelength = 3.0 × 0.50 = 1.5 m/s. 6.0 m/s comes from multiplying the number of crests by the wavelength and ignoring the time. 0.17 m/s comes from dividing the wavelength by the frequency. 24 m/s comes from multiplying the crest count, the wavelength and the time together.
✓ 1
Answer B
Answer: B
Common error: Count waves per second to get the frequency before using speed = frequency × wavelength.
94IGPC3-3-09r42-02The electromagnetic spectrumPaper CEasy1 mark
Gamma rays are used to sterilise surgical instruments before they are packed. Which statement about gamma rays is correct?
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Mark
Notes
All electromagnetic waves travel at the same speed, 3.0 × 108 m/s, in a vacuum, so the first statement is correct. Gamma rays do not travel faster than visible light, because all electromagnetic waves share one speed in a vacuum. They have the shortest wavelengths in the spectrum, not longer than radio waves. Electromagnetic waves are transverse, not longitudinal.
✓ 1
Answer A
Answer: A
Exam tip: Remember that every electromagnetic wave has the same speed in a vacuum; frequency and wavelength are what differ.
95IGPC3-3-09r42-09Wave properties, reflection, refraction and diffractionPaper CEasy6 marks
Water waves of frequency 0.80 Hz and wavelength 2.5 m travel across a reservoir that supplies a town. A sensor sits 30 m from the shore.
(a)
Calculate the speed of the waves.
(2)
(b)
Calculate the time taken for a wave crest to travel 30 m to the shore. Use your answer to (a).
(2)
(c)
State whether water waves are transverse or longitudinal.
(1)
(d)
State what is transferred from the sensor to the shore by the waves.
(1)
Show mark scheme
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Notes
Part (a)
v = f × λ
✓ 1
= 0.80 × 2.5 = 2.0 m/s
✓ 1
Part (b)
t = distance ÷ speed
✓ 1
= 30 ÷ 2.0 = 15 s (ECF from (a))
✓ 1
Part (c)
transverse
✓ 1
Part (d)
energy
✓ 1
Exam tip: Write the equation, then substitute, then give the answer with its unit.
96IGPC3-3-09r42-10Wave properties, reflection, refraction and diffractionPaper CEasy6 marks
In an escape-room game, a ray from a laser pointer strikes a plane mirror. The angle between the ray and the surface of the mirror is 28°.
Diagram not to scale
(a)
Calculate the angle of incidence.
(1)
(b)
Calculate the angle between the incident ray and the reflected ray.
(1)
(c)
State two properties of the image formed by a plane mirror.
(2)
(d)
A player stands 1.8 m in front of the mirror. Calculate the distance between the player and the image.
(2)
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Mark
Notes
Part (a)
90° − 28° = 62°
✓ 1
Part (b)
2 × 62° = 124° (ECF from (a))
✓ 1
Part (c)
any two from: virtual / upright / same size as the object / same distance behind the mirror as the object is in front / laterally inverted
✓ 2
Part (d)
image is 1.8 m behind the mirror
✓ 1
distance = 1.8 + 1.8 = 3.6 m
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the mirror surface.
97IGPC3-3-09r42-16The electromagnetic spectrumPaper CEasy6 marks
A water-treatment works uses ultraviolet lamps to disinfect drinking water.
(a)
Microwaves, X-rays, infrared and radio waves are all electromagnetic waves. Put them in order of increasing frequency, starting with the lowest.
(2)
(b)
State why ultraviolet radiation is used to treat drinking water.
(1)
(c)
State one hazard to workers of excessive exposure to ultraviolet radiation.
(1)
(d)
Microwaves of frequency 2.45 GHz are used in the works to heat samples. Calculate their wavelength. The speed of electromagnetic waves in a vacuum is 3.00 × 108 m/s.
(2)
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Notes
Part (a)
radio waves lowest and X-rays highest
✓ 1
microwaves below infrared, giving the order radio waves, microwaves, infrared, X-rays
✓ 1
Part (b)
it kills bacteria / micro-organisms
✓ 1
Part (c)
skin cancer / skin burns / eye damage (any one)
✓ 1
Part (d)
λ = v ÷ f
✓ 1
= 3.00 × 108 ÷ 2.45 × 109 = 0.122 m
✓ 1
Exam tip: 1 GHz = 10⁹ Hz; convert prefixes before using the wave equation.
98IGPC3-3-08r43-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Waves in a swimming pool have a frequency of 2.0 Hz and a wavelength of 0.25 m. What is the speed of the waves?
Show mark scheme
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Notes
Wave speed = frequency × wavelength = 2.0 × 0.25 = 0.50 m/s. The value 8.0 m/s comes from dividing frequency by wavelength. The value 2.25 m/s comes from adding the two quantities. The value 0.125 m/s comes from dividing wavelength by frequency.
✓ 1
Answer C
Answer: C
Common error: Wave speed is frequency multiplied by wavelength; check the units give m/s.
99IGPC3-3-08r43-02The electromagnetic spectrumPaper CEasy1 mark
Four types of electromagnetic wave are listed. Which one has the shortest wavelength?
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Notes
In order of increasing wavelength the waves are ultraviolet, infrared, microwaves, radio waves, so ultraviolet is shortest. The other three all have longer wavelengths than ultraviolet, with radio waves the longest.
✓ 1
Answer A
Answer: A
Exam tip: Learn the spectrum in order from radio waves to gamma rays; wavelength decreases as frequency increases.
100IGPC3-3-08r43-08The electromagnetic spectrumPaper CEasy1 mark
A technician at a wind farm uses a thermal imaging camera to find an overheating gearbox. Which type of radiation does the camera detect?
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Notes
Hot objects emit infrared radiation, and thermal cameras detect it to show temperature differences. Ultraviolet, microwaves and X-rays are not the radiation emitted in useful amounts by an overheated machine part at these temperatures, and cameras do not use them for this purpose.
✓ 1
Answer D
Answer: D
Exam tip: Link thermal imaging, remote controls and heating to infrared radiation.
101IGPC3-3-08r43-09Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
A swimmer makes waves in a pool. The distance between neighbouring crests is 0.40 m. A float on the water is passed by 10 complete waves in 8.0 s.
Diagram not to scale
(a)
State what is meant by the wavelength of a wave.
(1)
(b)
Calculate the frequency of the waves.
(2)
(c)
Calculate the speed of the waves. Use the unrounded value of the frequency.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
distance between two adjacent crests (or the same point on two successive waves)
✓ 1
Part (b)
M1 frequency = number of waves ÷ time = 10 ÷ 8.0
✓ 1
A1 1.25 Hz
✓ 1
Part (c)
M1 speed = frequency × wavelength = 1.25 × 0.40
✓ 1
A1 0.50 m/s
✓ 1
Exam tip: Frequency is the number of complete waves per second; find it before using v = fλ.
102IGPC3-3-08r43-16Light, refraction, lenses and dispersionPaper CEasy5 marks
A mirror is fixed to the wall of a library corridor.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
A ray of light strikes the mirror and makes an angle of 35° with the surface of the mirror. Calculate the angle of reflection.
(2)
(c)
A student stands 1.5 m in front of the mirror. Calculate the distance between the student and the image they see.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
angle of incidence = angle of reflection
✓ 1
Part (b)
M1 angle of incidence = 90° − 35° (angles are measured from the normal)
✓ 1
A1 55°
✓ 1
Part (c)
M1 the image is 1.5 m behind the mirror, the same distance as the object is in front
✓ 1
A1 3.0 m
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the mirror surface.
103IGPC3-3-10r44-01The electromagnetic spectrumPaper CEasy1 mark
Archaeologists use ground-penetrating radar, which sends pulses of radio waves into the soil to find buried walls. Which statement about radio waves is correct?
Show mark scheme
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Mark
Notes
Radio waves are electromagnetic waves. All electromagnetic waves are transverse and need no medium, so they travel through a vacuum. The longitudinal options describe sound-like behaviour. The option saying they cannot travel through a vacuum confuses them with mechanical waves, which need a medium.
✓ 1
Answer C
Answer: C
Common error: Do not mix up the wave type with the need for a medium: sound is longitudinal and needs a medium, whereas electromagnetic waves are transverse and do not.
104IGPC3-3-10r44-02The electromagnetic spectrumPaper CEasy1 mark
A data centre uses Wi-Fi signals of frequency 2.4 GHz. These are electromagnetic waves travelling at 3.0 × 108 m/s in air. What is the wavelength of the signals?
Show mark scheme
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Mark
Notes
wavelength = speed ÷ frequency = 3.0 × 108 ÷ 2.4 × 109 = 0.125 m. 8.0 m comes from dividing frequency by speed. 0.0125 m is a power-of-ten error from entering 2.4 GHz incorrectly. 7.2 × 1017 m multiplies speed by frequency.
✓ 1
Answer A
Answer: A
Exam tip: Convert GHz to Hz (× 10⁹) before using v = fλ.
105IGPC3-3-10r44-08The electromagnetic spectrumPaper CEasy1 mark
A thermal-imaging camera in a data centre detects electromagnetic radiation to find overheating servers. Which list shows electromagnetic waves in order of increasing frequency?
Show mark scheme
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Mark
Notes
In order of increasing frequency the spectrum runs radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. The correct list is in that order. One distractor swaps infrared and microwaves. One places microwaves after ultraviolet. The last lists waves in order of decreasing frequency.
✓ 1
Answer B
Answer: B
Exam tip: Learn the spectrum in one direction, from radio waves to gamma rays, and note that frequency increases while wavelength decreases.
106IGPC3-3-10r44-09Wave properties, reflection, refraction and diffractionPaper CEasy5 marks
In a ripple tank, a vibrating bar makes straight water waves. A student counts 5 complete waves passing a fixed mark in 2.0 s. The wavelength is 0.040 m.
Diagram not to scale
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the speed of the waves.
(2)
(c)
A barrier with a gap about 0.04 m wide is placed in the tank. Describe what happens to the waves that pass through the gap.
(1)
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Part (a)
M1 f = number of waves ÷ time = 5 ÷ 2.0
✓ 1
A1 2.5 Hz
✓ 1
Part (b)
M1 v = fλ = 2.5 × 0.040 (ECF from (a))
✓ 1
A1 0.10 m/s
✓ 1
Part (c)
B1 the waves spread out (diffract) into the region behind the gap
✓ 1
Exam tip: Diffraction is greatest when the gap is about the same size as the wavelength.
107IGPC3-3-10r44-10The electromagnetic spectrumPaper CEasy5 marks
A data centre uses Wi-Fi for staff devices and thermal cameras to monitor equipment. Electromagnetic waves are used in both.
(a)
State the speed of electromagnetic waves in a vacuum.
(1)
(b)
The Wi-Fi uses microwaves of frequency 5.0 GHz. Calculate the wavelength. Use 3.0 × 108 m/s for the speed.
(2)
(c)
State the type of electromagnetic wave detected by a thermal camera.
(1)
(d)
Name the type of electromagnetic wave with the highest frequency.
(1)
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Notes
Part (a)
B1 3.0 × 108 m/s
✓ 1
Part (b)
M1 λ = v ÷ f = 3.0 × 108 ÷ 5.0 × 109
✓ 1
A1 0.060 m
✓ 1
Part (c)
B1 infrared
✓ 1
Part (d)
B1 gamma rays
✓ 1
Common error: Convert GHz to Hz (× 10⁹) before substituting into v = fλ.
108IGPC3-3-08r45-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Waves on the surface of a swimming pool have a wavelength of 0.40 m and a frequency of 1.5 Hz. What is the speed of the waves?
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Wave speed = frequency × wavelength = 1.5 × 0.40 = 0.60 m/s. Dividing wavelength by frequency gives 0.27, adding the two numbers gives 1.9, and dividing frequency by wavelength gives 3.8; none of these is the wave equation.
✓ 1
Answer B
Answer: B
Exam tip: Write the wave equation v = f × λ first, then substitute the numbers with their units.
109IGPC3-3-08r45-03The electromagnetic spectrumPaper CEasy1 mark
Which list shows electromagnetic waves in order of increasing wavelength?
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Wavelength increases from gamma rays through X-rays, ultraviolet, visible light, infrared and microwaves to radio waves, so gamma rays, ultraviolet, infrared, radio waves is correct. The first wrong list is in order of decreasing wavelength, and the other two place microwaves before infrared, but microwaves have the longer wavelength.
✓ 1
Answer A
Answer: A
Exam tip: Learn the spectrum in one order, such as gamma to radio, and note whether that order is of increasing wavelength or of increasing frequency.
110IGPC3-3-08r45-04Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
Light passes from air into the glass of a library display case. Which quantity is the same in the air and in the glass?
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The source sets the frequency, and it does not change at a boundary. The speed falls in glass, which makes the wavelength shorter. A ray bends towards the normal unless it arrives along the normal, so its direction does not stay the same for every angle of incidence.
✓ 1
Answer C
Answer: C
Exam tip: At a boundary, frequency stays fixed while speed and wavelength change together.
111IGPC3-3-08r45-07Sound and ultrasoundPaper CEasy1 mark
A pest-control device in a library store room emits sound of one frequency. Which frequency is ultrasound for a human listener?
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Ultrasound has a frequency above the upper limit of human hearing, about 20 kHz, so 45 kHz qualifies. 15 Hz is below the lower limit of about 20 Hz, 440 Hz is within the audible range, and 18 kHz is just inside the audible range.
✓ 1
Answer D
Answer: D
Exam tip: Humans hear roughly 20 Hz to 20 kHz; ultrasound is above this range.
112IGPC3-3-08r45-09Wave properties, reflection, refraction and diffractionPaper CEasy7 marks
Waves travel across the surface of a swimming pool.
Diagram not to scale
(a)
State what is meant by the wavelength of a wave.
(1)
(b)
18 wave crests pass a marker in 12 s. Calculate the frequency of the waves.
(2)
(c)
The distance between adjacent crests is 0.60 m. Calculate the speed of the waves.
(2)
(d)
The waves move into a shallower part of the pool, where they travel more slowly. The frequency stays the same. State what happens to the wavelength and give a reason.
(2)
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Part (a)
distance between two adjacent crests (or same point on two consecutive waves)
✓ 1
Part (b)
frequency = number of waves ÷ time, or 18 ÷ 12
✓ 1
1.5 Hz
✓ 1
Part (c)
v = f × λ
✓ 1
0.60 × 1.5 = 0.90 m/s (ECF from the frequency in the previous part)
✓ 1
Part (d)
wavelength decreases
✓ 1
v = f λ: f constant so a smaller v gives a smaller λ
✓ 1
Exam tip: Show the equation, the substitution and the unit in any calculation; each step can earn a mark.
113IGPC3-3-08r46-01Sound and ultrasoundPaper CEasy1 mark
A train driver sounds a horn that emits a note of frequency 440 Hz. The speed of sound in air is 330 m/s. What is the wavelength of the sound?
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Wavelength = speed ÷ frequency = 330 ÷ 440 = 0.75 m. Dividing frequency by speed gives 1.3, which inverts the equation. Multiplying speed by frequency gives 145 000, which uses the wrong operation. Halving the correct value gives 0.38 and comes from treating the distance between a crest and a trough as one wavelength.
✓ 1
Answer C
Answer: C
Common error: Rearrange v = fλ carefully; check that the answer has a sensible size for sound in air.
114IGPC3-3-08r46-02The electromagnetic spectrumPaper CEasy1 mark
A telescope detects radiation from a distant galaxy in several regions of the electromagnetic spectrum. Which list shows the radiation in order of increasing wavelength?
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Gamma rays have the shortest wavelength, then ultraviolet, then infrared, so this order is increasing wavelength. The list starting with radio waves ends at X-rays, so wavelength decreases along it. The infrared, visible, ultraviolet list is in order of decreasing wavelength. In the last list X-rays come after microwaves although X-rays have a much shorter wavelength.
✓ 1
Answer A
Answer: A
Exam tip: Radio waves have the longest wavelength and gamma rays the shortest; frequency changes in the opposite direction.
115IGPC3-3-08r46-03Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
An underwater camera in a swimming pool records a swimmer. Light travels from the air into the water. What happens to the frequency and the wavelength of the light as it enters the water?
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Notes
The frequency is set by the source and does not change at a boundary. Light slows down in water, and since v = fλ with f constant, the wavelength decreases. A decrease in frequency is wrong because frequency never changes at a boundary. An increase in wavelength would need the light to speed up. A change in frequency with constant wavelength would make the speed rise sharply, which is not what happens.
✓ 1
Answer B
Answer: B
Common error: Only the frequency stays the same when a wave changes medium; speed and wavelength both change.
116IGPC3-3-08r46-09Sound and ultrasoundPaper CEasy6 marks
A stadium announcer speaks into a microphone and the sound is played through loudspeakers. A spectator sits some distance from a loudspeaker. The speed of sound in air is 340 m/s.
(a)
State what is meant by the frequency of a sound wave.
(1)
(b)
A loudspeaker emits a note of frequency 500 Hz. The wavelength of the sound is 0.68 m. Calculate the speed of the sound wave.
(2)
(c)
The sound takes 0.30 s to reach the spectator. Calculate the distance from the loudspeaker to the spectator. Use 340 m/s.
(2)
(d)
State why the sound of the loudspeaker could not reach a spectator if the air were removed completely.
(1)
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Notes
Part (a)
the number of waves (vibrations) passing a point each second
✓ 1
Part (b)
speed = frequency × wavelength = 500 × 0.68
✓ 1
340 m/s
✓ 1
Part (c)
distance = speed × time = 340 × 0.30
✓ 1
102 m
✓ 1
Part (d)
sound needs a medium (particles) to travel through, and a vacuum has no particles
✓ 1
Exam tip: Write the equation you use and substitute the numbers; the method mark is awarded even if the arithmetic slips.
117IGPC3-3-08r46-10The electromagnetic spectrumPaper CEasy6 marks
A railway company uses an infrared camera to check for overheating wheel bearings. The train driver also communicates by radio waves of frequency 125 MHz (1.25 × 108 Hz). All electromagnetic waves travel at 3.0 × 108 m/s in a vacuum.
(a)
State why the infrared camera can detect an overheating bearing.
(1)
(b)
State one property that is the same for all electromagnetic waves.
(1)
(c)
Calculate the wavelength of the radio waves.
(2)
(d)
State whether infrared radiation has a higher or lower frequency than radio waves.
(1)
(e)
State one harmful effect of ultraviolet radiation on track workers who spend long periods outdoors.
(1)
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Part (a)
hot objects emit infrared radiation, and the hotter the bearing the more it emits
✓ 1
Part (b)
they travel at the same speed in a vacuum (3.0 × 108 m/s) or they are transverse waves
Common error: Remember to convert megahertz to hertz (× 10⁶) before using the wave equation.
118IGPC3-3-09r46-01Wave properties, reflection, refraction and diffractionPaper CEasy1 mark
A wave machine in a swimming pool makes 15 wave crests pass a floating ball in 10 s. The distance between neighbouring crests is 0.40 m. What is the speed of the waves?
Diagram not to scale
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Notes
Frequency = 15 ÷ 10 = 1.5 Hz and speed = frequency × wavelength = 1.5 × 0.40 = 0.60 m/s. 0.27 m/s divides the wavelength by the frequency. 3.8 m/s divides the frequency by the wavelength. 6.0 m/s multiplies the number of crests by the wavelength and ignores the time.
✓ 1
Answer C
Answer: C
Common error: Always convert a count of waves and a time into a frequency (waves per second) before using speed = frequency × wavelength.
119IGPC3-3-09r46-02The electromagnetic spectrumPaper CEasy1 mark
Microwaves and ultraviolet radiation both travel through a vacuum. Which statement compares them correctly?
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Microwaves lie at longer wavelengths and so lower frequencies than ultraviolet in the electromagnetic spectrum, and all electromagnetic waves travel at the same speed in a vacuum. The other options either reverse the frequency order or claim that speed depends on frequency in a vacuum.
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Answer A
Answer: A
Exam tip: All electromagnetic waves share one speed in a vacuum, so differences between them are differences in frequency and wavelength only.
120IGPC3-3-09r46-07The electromagnetic spectrumPaper CEasy1 mark
Which type of electromagnetic radiation is used in a television remote control?
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Notes
Remote controls send pulses of infrared radiation to a detector on the television. Ultraviolet is used for security marking and sterilising, microwaves for satellite links and cooking, and radio waves for broadcasting; none of these is used in a typical remote control.
✓ 1
Answer C
Answer: C
Exam tip: Learn one clear use for each region of the electromagnetic spectrum.
121IGPC3-3-09r46-09Wave properties, reflection, refraction and diffractionPaper CEasy6 marks
A wave machine in a swimming pool makes 24 waves in 20 s. The distance between neighbouring crests is 0.35 m. A small float on the surface bobs up and down as each wave passes.
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the speed of the waves.
(2)
(c)
State whether the wave is transverse or longitudinal and describe how the float moves relative to the direction of travel.
(2)
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Part (a)
M1 frequency = number of waves ÷ time
✓ 1
A1 24 ÷ 20 = 1.2 Hz
✓ 1
Part (b)
M1 speed = frequency × wavelength
✓ 1
A1 1.2 × 0.35 = 0.42 m/s (ECF from (a))
✓ 1
Part (c)
B1 transverse
✓ 1
B1 float moves up and down, at right angles to the direction the wave travels
✓ 1
Exam tip: Show the formula you are using before substituting; method marks are earned even if the arithmetic slips.
122IGPC3-3-09r46-11The electromagnetic spectrumPaper CEasy7 marks
Control signals are sent to the turbines of a wind farm using radio waves of frequency 150 MHz. The speed of electromagnetic waves in air is 3.0 × 108 m/s.
(a)
State two properties that are the same for all electromagnetic waves.
(2)
(b)
Radio waves, microwaves, infrared and X-rays are four regions of the electromagnetic spectrum. List them in order, starting with the longest wavelength.
(2)
(c)
Calculate the wavelength of the 150 MHz radio waves.
(3)
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Part (a)
B1 any one correct property, e.g. they are transverse waves
✓ 1
B1 any second correct property, e.g. they travel at the same speed in a vacuum / they transfer energy
✓ 1
Part (b)
B1 radio waves first (longest) and X-rays last (shortest)
✓ 1
B1 full order: radio waves, microwaves, infrared, X-rays
✓ 1
Part (c)
M1 wavelength = speed ÷ frequency
✓ 1
M1 converts 150 MHz to 1.5 × 108 Hz
✓ 1
A1 2.0 m
✓ 1
Common error: Mega means ×10⁶: write the frequency in hertz before substituting into v = fλ.
123IGPC3-3-09r48-01Sound and ultrasoundPaper CEasy1 mark
A humpback whale produces a sound of frequency 200 Hz. In sea water the wavelength of this sound is 7.5 m. What is the speed of sound in sea water?
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Notes
Speed = frequency × wavelength = 200 × 7.5 = 1500 m/s. 26.7 m/s comes from dividing frequency by wavelength. 0.0375 m/s comes from dividing wavelength by frequency. 207.5 m/s comes from adding the two quantities, which have different units and cannot be added.
✓ 1
Answer A
Answer: A
Common error: Wave speed is a product, v = f × λ. Check that your answer is sensible: sound in water travels much faster than in air.
124IGPC3-3-09r48-03The electromagnetic spectrumPaper CEasy1 mark
Microwaves and gamma rays both travel through a vacuum. Which quantity is the same for both?
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All electromagnetic waves travel at the same speed, 3.0 × 108 m/s, in a vacuum. Microwaves have a much longer wavelength and a much lower frequency than gamma rays. Gamma rays are strongly ionising but microwaves are not, so that is also different.
✓ 1
Answer B
Answer: B
Exam tip: When comparing regions of the electromagnetic spectrum, remember that speed in a vacuum is the one thing they share; frequency and wavelength change together.
125IGPC3-3-09r48-08Sound and ultrasoundPaper CEasy1 mark
A loudspeaker produces a note. The sound then becomes louder and its pitch becomes higher. What has happened to the amplitude and the frequency of the sound wave?
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Notes
Loudness depends on amplitude, so a louder sound has a larger amplitude. Pitch depends on frequency, so a higher pitch means a higher frequency. The other options each reverse at least one of these links.
✓ 1
Answer C
Answer: C
Common error: Do not mix up loudness and pitch: loudness is linked to amplitude, pitch to frequency.
126IGPC3-3-09r48-09Wave properties, reflection, refraction and diffractionPaper CEasy7 marks
A marine aquarium uses a wave machine to produce water waves in a tank. The machine makes 8 complete waves in 4.0 s. The distance between neighbouring crests is 0.30 m.
Diagram not to scale
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the speed of the waves.
(2)
(c)
State what is meant by the amplitude of a wave.
(1)
(d)
The frequency of the machine is doubled but the wave speed stays the same. Calculate the new wavelength.
(2)
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Part (a)
M1 frequency = number of waves ÷ time
✓ 1
A1 8 ÷ 4.0 = 2.0 Hz
✓ 1
Part (b)
M1 speed = frequency × wavelength
✓ 1
A1 2.0 × 0.30 = 0.60 m/s
✓ 1
Part (c)
maximum displacement of a point on the wave from its undisturbed (rest) position
✓ 1
Part (d)
M1 wavelength = speed ÷ frequency, with new frequency 4.0 Hz
✓ 1
A1 0.60 ÷ 4.0 = 0.15 m (accept ECF from (b))
✓ 1
Exam tip: If frequency doubles at constant speed, the wavelength halves. Use this to check your calculation.
127IGPC3-3-09r48-11The electromagnetic spectrumPaper CEasy7 marks
A logistics company tracks freight containers using radio waves and checks some of them using X-rays. In a vacuum, all electromagnetic waves travel at 3.0 × 108 m/s.
(a)
State two properties that are the same for all electromagnetic waves.
(2)
(b)
A tracking signal is a radio wave of frequency 6.0 × 108 Hz. Calculate its wavelength.
(2)
(c)
State one danger of exposure to X-rays and one way to reduce the risk.
(2)
(d)
Name the type of electromagnetic radiation that has a shorter wavelength than visible light but a longer wavelength than X-rays.
(1)
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Notes
Part (a)
one valid property, e.g. they are transverse / they transfer energy / they can travel through a vacuum
✓ 1
a second, different valid property, e.g. they travel at the same speed in a vacuum
✓ 1
Part (b)
M1 wavelength = speed ÷ frequency
✓ 1
A1 (3.0 × 108) ÷ (6.0 × 108) = 0.50 m
✓ 1
Part (c)
danger: damages or kills cells / causes cancer / mutation
✓ 1
precaution: shielding with lead / limit exposure time / keep distance
✓ 1
Part (d)
ultraviolet
✓ 1
Exam tip: When asked for a danger, name the harm to the body, such as cell damage, not just 'it is harmful'.
128IGPC-037-03-04Sound and ultrasoundPaper CMedium1 mark
A girl claps her hands once while standing in front of a tall cliff. She hears the echo 0.60 s later. The speed of sound in air is 330 m/s. How far is she from the cliff?
Diagram not to scale
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Notes
In 0.60 s the sound travels to the cliff and back: distance = 330 × 0.60 = 198 m. This is twice the distance to the cliff, so the distance is 198 ÷ 2 = 99 m. 198 m forgets to halve, 49.5 m halves twice, and 0.0018 m divides time by speed.
✓ 1
Answer C
Answer: C
Common error: An echo time covers the journey there and back. Halve the total distance to find how far away the reflecting surface is.
129IGPC-037-03-05Sound and ultrasoundPaper CMedium1 mark
A tuning fork produces a sound of frequency 440 Hz. The speed of sound in the air is 341 m/s. What is the wavelength of the sound?
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Notes
Using v = fλ, λ = v ÷ f = 341 ÷ 440 = 0.775 m. 1.29 m comes from f ÷ v, 150 000 m from multiplying v by f, and 0.00129 m is 1 ÷ 775, a wrong power of ten.
✓ 1
Answer B
Answer: B
Exam tip: Rearrange v = fλ before putting numbers in. A wavelength of under a metre is sensible for a musical note in air.
130IGPC-037-01-07Light, refraction, lenses and dispersionPaper CMedium1 mark
A thin converging lens has a focal length of 10 cm. A small object is placed 35 cm from the lens. Which row describes the image?
Diagram not to scale
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Notes
The object is further from the lens than twice the focal length (35 cm > 20 cm). A converging lens then forms a real, inverted image that is smaller than the object. The image lies between F and 2F on the other side of the lens. An enlarged real image needs the object between F and 2F. A virtual, upright, enlarged image needs the object closer than F. A converging lens never forms a virtual, inverted image.
✓ 1
Answer B
Answer: B
Exam tip: Compare the object distance with f and 2f first. Beyond 2f gives a diminished image, between f and 2f gives a magnified image, and inside f gives a virtual image.
131IGPC-037-01-08Light, refraction, lenses and dispersionPaper CMedium1 mark
A converging lens is used as a magnifying glass to read small print. Where must the print be placed, and what type of image is seen?
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A converging lens acts as a magnifying glass when the object is closer to the lens than the principal focus. The rays leaving the lens diverge, so the eye sees a virtual, upright, enlarged image on the same side as the object. Such an image cannot be real, because the rays never actually meet. An object beyond 2F gives a diminished real image. An object exactly at F gives parallel rays and no image at a finite distance.
✓ 1
Answer A
Answer: A
Exam tip: A magnifying glass always gives a virtual image. You can only see a virtual image by looking through the lens; it cannot be caught on a screen.
132IGPC-037-02-05The electromagnetic spectrumPaper CMedium1 mark
A person uses a sunbed for long periods. Which statement describes the main hazard of the ultraviolet radiation?
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Ultraviolet is absorbed by the surface layers of the body. It damages skin cells and eye cells, increasing the risk of skin cancer and eye conditions such as cataracts. Internal heating is the hazard of microwaves. Skin burns from heating are the hazard of infrared. Ultraviolet not penetrating deeply is exactly why the damage is concentrated at the surface, so it is not harmless.
✓ 1
Answer B
Answer: B
Exam tip: Match each hazard to its region: microwaves give internal heating, infrared gives skin burns, ultraviolet damages surface cells and eyes, and X-rays and gamma rays cause cell mutation.
133IGPC-037-00-04Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
Water waves travel from deep water into shallow water. Which row describes the changes to the waves?
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Waves slow down in shallower water. The frequency is fixed by the source, so it does not change. Since v = fλ and f is constant, the wavelength must decrease as the speed decreases.
✓ 1
Answer A
Answer: A
Exam tip: When a wave enters a new medium, its frequency never changes. Speed and wavelength change together.
134IGPC-037-00-09Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
A transverse wave travels along a rope. The horizontal distance between a crest and the next trough is 0.30 m. Four complete waves pass a point on the rope every second. What is the speed of the wave?
Diagram not to scale
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Notes
From a crest to the next trough is half a wavelength, so λ = 2 × 0.30 = 0.60 m. The frequency is 4.0 Hz. So v = fλ = 4.0 × 0.60 = 2.4 m/s. The value 1.2 m/s wrongly uses λ = 0.30 m.
✓ 1
Answer C
Answer: C
Common error: A crest-to-trough distance is half a wavelength. One full wavelength runs from crest to crest.
135IGPC-037-03-14Sound and ultrasoundPaper CMedium5 marks
A loudspeaker connected to a signal generator produces a sound of frequency 680 Hz. The speed of sound in the air is 340 m/s.
(a)
Calculate the wavelength of the sound.
(2)
(b)
The frequency is increased to 1360 Hz. (i) State how the pitch of the sound changes. (ii) Calculate the new wavelength.
(2)
(c)
The volume control is then turned down, without changing the frequency. State what happens to the amplitude of the sound wave.
(1)
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Part (a)
λ = v ÷ f = 340 ÷ 680
✓ 1
= 0.50 m
✓ 1
Part (b)
(i) The pitch becomes higher
✓ 1
(ii) λ = 340 ÷ 1360 = 0.25 m
✓ 1
Part (c)
The amplitude decreases
✓ 1
Common error: Turning the volume down does not change the frequency or the wave speed. Only the amplitude changes.
136IGPC-037-02-15The electromagnetic spectrumPaper CMedium7 marks
A student is given these seven regions of the electromagnetic spectrum: visible light, gamma rays, microwaves, X-rays, radio waves, ultraviolet, infrared
(a)
Write the seven regions in order of increasing frequency.
(2)
(b)
State which region has the longest wavelength.
(1)
(c)
State two properties shared by all electromagnetic waves.
(2)
(d)
A security guard can see an intruder on a thermal imaging camera in complete darkness, but not on an ordinary camera. Suggest why.
(2)
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Notes
Part (a)
radio waves, microwaves, infrared in the correct order at the start ; visible light, ultraviolet, X-rays, gamma rays in the correct order at the end
✓ 2
Part (b)
radio waves
✓ 1
Part (c)
first correct property from: they are transverse waves; they can travel through a vacuum; they all travel at the same speed (3.0×108 m/s) in a vacuum; they transfer energy ; a second, different correct property from the same list
✓ 2
Part (d)
the intruder's warm body emits infrared radiation even when there is no light ; the thermal camera detects infrared, but the ordinary camera needs visible light, which is absent
✓ 2
Exam tip: Frequency and wavelength run in opposite directions along the spectrum. Check which one the question asks for before writing the order.
137IGPC-037-00-12Wave properties, reflection, refraction and diffractionPaper CMedium6 marks
A student uses a ripple tank to study water waves. A vibrating bar produces straight wavefronts on the water surface.
Diagram not to scale
(a)
Water waves are transverse. Explain what is meant by a transverse wave.
(2)
(b)
The student counts ten consecutive crests. The distance from the first crest to the tenth crest is 27 cm. Calculate the wavelength of the waves.
(2)
(c)
The bar vibrates with a frequency of 8.0 Hz. Calculate the speed of the waves in m/s.
(2)
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Part (a)
The particles (or water surface) vibrate at right angles (perpendicular)
✓ 1
to the direction in which the wave travels or energy is transferred.
✓ 1
Part (b)
Ten consecutive crests span 9 wavelengths
✓ 1
λ = 27 ÷ 9 = 3.0 cm (0.030 m)
✓ 1
Part (c)
v = fλ = 8.0 × 0.030 (wavelength converted to metres)
✓ 1
v = 0.24 m/s
✓ 1
(Allow error carried forward from the wavelength calculated in the previous part.)
—
Common error: Dividing the distance by 10 instead of 9 gives the wrong wavelength. Count the gaps between crests, not the crests themselves. Convert cm to m before giving the speed in m/s.
138IGPC3-3-03c-04Sound and ultrasoundPaper CMedium1 mark
A research ship sends a pulse of ultrasound straight down. The echo from the seabed is detected 0.80 s later. The speed of sound in sea water is 1500 m/s. What is the depth of the water?
Diagram not to scale
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The pulse travels down and back, so the one-way distance is half of 1500 × 0.80 = 1200 m, which is 600 m (C). A forgets to halve the total distance. B divides speed by time. D divides time by speed.
✓ 1
Answer C
Answer: C
Common error: An echo covers the distance twice, so halve the total path to get the depth.
139IGPC3-3-03c-10Light, refraction, lenses and dispersionPaper CMedium5 marks
A person stands 2.5 m in front of a large plane mirror in a dance studio.
Diagram not to scale
(a)
State the distance between the person and their image.
(1)
(b)
State whether the image is real or virtual, and whether it is larger than, smaller than or the same size as the person.
(2)
(c)
A ray of light from a lamp strikes the mirror, making an angle of 35° with the mirror surface. Calculate the angle of reflection.
(2)
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Part (a)
5.0 m
✓ 1
Part (b)
virtual
✓ 1
same size
✓ 1
Part (c)
angle of incidence = 90° − 35° = 55°, measured from the normal
✓ 1
angle of reflection = 55°
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the mirror surface.
140IGPC3-3-03c-13Sound and ultrasoundPaper CMedium5 marks
An ultrasound scanner is used to examine tissue inside a patient. The probe emits a short pulse of ultrasound and detects the echo from a boundary inside the body.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
The pulse travels at 1500 m/s in the tissue and the echo returns 1.2 × 10−4 s after the pulse is sent. Calculate the depth of the boundary.
(2)
(c)
Explain why a gel is spread between the probe and the skin.
(2)
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Notes
Part (a)
sound with a frequency above 20 000 Hz, higher than humans can hear
✓ 1
Part (b)
distance travelled = 1500 × 1.2 × 10−4 = 0.18 m
✓ 1
depth = 0.18 / 2 = 0.090 m
✓ 1
Part (c)
air between the probe and the skin would reflect most of the ultrasound
✓ 1
the gel removes the air gap so more ultrasound enters the body
✓ 1
Exam tip: For echo questions, ask whether the time given is for the one-way or the round trip.
141IGPC3-3-05r9-02Light, refraction, lenses and dispersionPaper CMedium1 mark
Pulses of infrared radiation carry data along an optical fibre in a data centre. Which statement explains why the light stays inside the fibre, even when the cable is bent gently?
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Notes
Correct: D. Light in the core meets the boundary with the less dense cladding at an angle larger than the critical angle, so it is totally internally reflected each time. A is wrong because diffraction is spreading at gaps and edges and does not keep light in the glass. B is wrong because dispersion separates colours and does not confine light. C is wrong because absorption and re-emission would waste the signal and is not how fibres work.
✓ 1
Answer D
Answer: D
Exam tip: Total internal reflection needs two things: light travelling towards a less dense material, and an angle of incidence greater than the critical angle.
142IGPC3-3-05r9-03Sound and ultrasoundPaper CMedium1 mark
An archaeologist taps a steel rail, a water pipe full of water and a column of air, and compares how fast the sound travels in each. Which row lists the materials in order of sound speed, fastest first?
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Notes
Correct: B. Particles in a solid are closest together and pass vibrations on fastest; in gases they are furthest apart, so the speed is lowest. A reverses the order, wrongly treating sound like light moving faster through thin media. C wrongly puts water above steel. D wrongly puts air above water.
✓ 1
Answer B
Answer: B
Common error: Sound needs particles to pass on vibrations, so it is fastest in solids. Do not confuse this with light, which is fastest in a vacuum.
143IGPC3-3-05r9-04Light, refraction, lenses and dispersionPaper CMedium1 mark
An archaeologist uses a converging lens of focal length 10 cm as a magnifying glass. She holds an object 6 cm from the lens. Which row describes the image she sees?
Diagram not to scale
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Notes
Correct: D. An object closer to the lens than its principal focus gives a virtual, upright, magnified image seen on the same side as the object. A is the image for an object far beyond twice the focal length. B is the image for an object between one and two focal lengths. C is wrong because a lens used as a magnifier enlarges the object.
✓ 1
Answer D
Answer: D
Exam tip: For a converging lens, compare the object distance with the focal length first: inside the focal length means a virtual, magnified image.
144IGPC3-3-05r9-10Wave properties, reflection, refraction and diffractionPaper CMedium5 marks
A cyclist has a small plane mirror fixed to the handlebars. A car is 8.0 m from the mirror.
Diagram not to scale
(a)
State the distance between the mirror and the image of the car.
(1)
(b)
State two characteristics of the image formed by a plane mirror.
(2)
(c)
A ray from the car hits the mirror with an angle of incidence of 25°, measured from the normal. Calculate the angle between the incident ray and the reflected ray.
(2)
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Notes
Part (a)
8.0 m (behind the mirror)
✓ 1
Part (b)
virtual (not real)
✓ 1
same size as the object / laterally inverted / upright
✓ 1
Part (c)
angle of reflection equals angle of incidence, so also 25°
✓ 1
angle between rays = 25° + 25° = 50°
✓ 1
Common error: The angle between the incident and reflected rays is double the angle of incidence, not the angle of incidence itself.
145IGPC3-3-05r9-12Sound and ultrasoundPaper CMedium5 marks
A repair team uses ultrasound to check a cycle frame for cracks. A pulse is sent into the aluminium alloy, where its speed is 5000 m/s.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
Suggest why part of the pulse is reflected when it reaches a crack.
(1)
(c)
An echo from a crack returns to the surface 24 μs after the pulse is sent. Calculate the depth of the crack below the surface.
(3)
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Notes
Part (a)
sound with a frequency above 20 000 Hz (above the range of human hearing)
✓ 1
Part (b)
the crack contains air, so the pulse meets a boundary between two different materials
✓ 1
Part (c)
total distance = 5000 × 24 × 10−6 = 0.12 m
✓ 1
depth is half this distance
✓ 1
depth = 0.060 m
✓ 1
Common error: Convert microseconds to seconds (1 μs = 10⁻⁶ s) and remember the pulse crosses the depth twice.
146IGPC3-3-05r9-13Light, refraction, lenses and dispersionPaper CMedium6 marks
An archaeologist uses a converging lens of focal length 8.0 cm to examine a small carved symbol.
Diagram not to scale
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
The symbol is 5.0 cm from the lens. Describe the image that is formed.
(3)
(c)
The lens is moved so that it is 20 cm from the symbol. State whether the image is now real or virtual and whether it is upright or inverted.
(2)
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Notes
Part (a)
the point where rays parallel to the principal axis meet / converge after passing through the lens
✓ 1
Part (b)
virtual
✓ 1
upright
✓ 1
magnified (larger than the object)
✓ 1
Part (c)
real
✓ 1
inverted
✓ 1
Exam tip: When describing an image, give all three features: real or virtual, upright or inverted, and larger or smaller.
147IGPC3-3-06r12-04Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
In a games studio a performer stands 2.0 m in front of a plane mirror and then walks 0.50 m towards the mirror. What is the distance between the performer and their image now?
Diagram not to scale
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Notes
The performer is now 1.5 m from the mirror and the image is 1.5 m behind it, so the separation is 3.0 m. 1.5 m is only the distance to the mirror. 3.5 m adds the original distance to the new one. 4.0 m is the separation before the performer moved.
✓ 1
Answer B
Answer: B
Common error: The object-to-image distance is twice the object-to-mirror distance.
148IGPC3-3-06r12-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A projector at a gaming event uses a converging lens of focal length 10 cm. A small slide is placed 15 cm from the lens. Which description fits the image formed on the screen?
Diagram not to scale
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Notes
The slide is between one and two focal lengths from the lens, so a real, inverted, magnified image forms beyond twice the focal length: C. A describes a magnifying glass, where the object is inside the focal length. B applies to an object beyond twice the focal length. D cannot be produced by a converging lens.
✓ 1
Answer C
Answer: C
Exam tip: Compare the object distance with the focal length to decide the type of image.
149IGPC3-3-06r12-11Sound and ultrasoundPaper CMedium7 marks
An engineer checks the depth of a dry well, using sound. Take the speed of sound in air to be 330 m/s.
Diagram not to scale
(a)
State the range of frequencies that a human can hear.
(2)
(b)
A short sound pulse is made at the top of the well. Its echo from the bottom returns after 0.28 s. Calculate the depth of the well. Give your answer to 2 significant figures.
(3)
(c)
Explain why sound in air is described as a longitudinal wave.
(2)
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Notes
Part (a)
Lower limit about 20 Hz
✓ 1
Upper limit about 20 000 Hz
✓ 1
Part (b)
Distance travelled = 330 × 0.28 = 92.4 m
✓ 1
The sound travels down and back, so depth = 92.4 / 2
✓ 1
Depth = 46 m
✓ 1
Part (c)
The air particles vibrate parallel to the direction in which the wave travels
✓ 1
forming a series of compressions and rarefactions
✓ 1
Common error: The echo travels to the bottom and back, so halve the total distance.
150IGPC3-3-06r12-12Sound and ultrasoundPaper CMedium5 marks
An ultrasonic bath of water is used to clean glassware in a pharmaceutical laboratory. The ultrasound has a frequency of 50 kHz and travels at 1500 m/s in water.
Diagram not to scale
(a)
Calculate the wavelength of the ultrasound in water.
(2)
(b)
Explain why a person cannot hear this ultrasound.
(1)
(c)
A pulse of ultrasound reflects from the base of the bath and returns to the surface after 1.6 × 10−4 s. Calculate the depth of water.
(2)
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Notes
Part (a)
λ = v / f
✓ 1
λ = 1500 / 50 000 = 0.030 m
✓ 1
Part (b)
Its frequency is above the upper limit of human hearing (20 000 Hz)
✓ 1
Part (c)
Distance travelled = 1500 × 1.6 × 10−4 = 0.24 m
✓ 1
Depth = 0.24 / 2 = 0.12 m
✓ 1
Exam tip: Convert frequencies in kHz to Hz before using v = fλ.
151IGPC3-3-06r12-14Light, refraction, lenses and dispersionPaper CMedium6 marks
A virtual reality headset uses a converging lens of focal length 8.0 cm as a magnifying glass.
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
An object 5.0 mm tall produces an image 15 mm tall. Calculate the magnification.
(2)
(c)
Explain why the object must be closer to the lens than 8.0 cm when the lens is used as a magnifying glass.
(2)
(d)
State what a diverging lens does to a beam of parallel rays.
(1)
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Notes
Part (a)
The point where rays travelling parallel to the principal axis are brought together (converge) after passing through the lens
✓ 1
Part (b)
m = image height / object height
✓ 1
m = 15 / 5.0 = 3.0
✓ 1
Part (c)
The object must be inside the focal length
✓ 1
so that the lens forms an upright, magnified, virtual image that can be seen through the lens
✓ 1
Part (d)
It spreads the rays out (makes them diverge)
✓ 1
Common error: Magnification has no unit because it is a ratio of two lengths.
152IGPC3-3-06r14-02The electromagnetic spectrumPaper CMedium1 mark
A radio on a rail network uses a signal of frequency 150 MHz. Radio waves travel at 3.0 × 108 m/s in air. What is the wavelength of the signal?
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Notes
λ = v ÷ f = 3.0 × 108 ÷ (150 × 106) = 2.0 m, so B. A (0.50 m) divides f by v. C (2000 m) treats MHz as kHz. D multiplies v by f.
✓ 1
Answer B
Answer: B
Exam tip: Convert MHz to Hz (×10⁶) before using v = fλ.
153IGPC3-3-06r14-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A converging lens has a focal length of 6.0 cm. An object is placed 4.0 cm from the lens. Which description of the image is correct?
Diagram not to scale
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Notes
The object is inside the focal length, so the refracted rays diverge and appear to come from a point behind the object: the image is virtual, upright and magnified (B), as in a magnifying glass. The 'real, inverted' options apply only when the object is beyond the focal length. A converging lens cannot give a diminished image from an object inside the focal length.
✓ 1
Answer B
Answer: B
Exam tip: A real image can be caught on a screen; a virtual image cannot. Objects inside the focal length of a converging lens always give virtual images.
154IGPC3-3-06r14-10Wave properties, reflection, refraction and diffractionPaper CMedium6 marks
A coach observes water waves moving across a swimming pool. Twelve wave crests pass a fixed marker in 8.0 s. The distance between adjacent crests is 0.40 m.
Diagram not to scale
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the speed of the waves. Use the frequency from (a).
(2)
(c)
State whether water waves are transverse or longitudinal and describe how the water surface moves.
(2)
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Notes
Part (a)
frequency = number of waves ÷ time
✓ 1
f = 12 ÷ 8.0 = 1.5 Hz
✓ 1
Part (b)
v = fλ = 1.5 × 0.40
✓ 1
v = 0.60 m/s (ECF from (a))
✓ 1
Part (c)
transverse
✓ 1
the water moves up and down, at right angles to the direction of travel of the wave
✓ 1
Exam tip: The distance between adjacent crests is one wavelength, not the distance across the whole pool.
155IGPC3-3-06r14-14The electromagnetic spectrumPaper CMedium5 marks
A space agency receives a radio signal from a probe orbiting Mars. The probe is 1.8 × 1011 m from the Earth. Radio waves belong to the electromagnetic spectrum.
(a)
State the speed of electromagnetic waves in a vacuum.
(1)
(b)
Calculate the time taken for the radio signal to reach the Earth.
(2)
(c)
Name one type of electromagnetic wave used for imaging bones, and state one danger of exposure to it.
(2)
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Notes
Part (a)
3.0 × 108 m/s
✓ 1
Part (b)
t = distance ÷ speed = 1.8 × 1011 ÷ 3.0 × 108
✓ 1
t = 600 s
✓ 1
Part (c)
X-rays
✓ 1
any one from: damage to cells; mutations; cancer
✓ 1
Exam tip: All electromagnetic waves travel at the same speed in a vacuum, so the type of wave does not change the travel time.
156IGPC3-3-06r14-15Sound and ultrasoundPaper CMedium6 marks
A train driver sounds a horn of frequency 220 Hz. An echo from the wall of a cutting returns after 3.0 s. The speed of sound in air is 330 m/s.
Diagram not to scale
(a)
Calculate the distance from the train to the wall of the cutting.
(2)
(b)
Calculate the wavelength of the sound from the horn.
(2)
(c)
A second horn has the same frequency but a smaller amplitude. State how the sound from the second horn differs from the first in (i) loudness and (ii) pitch.
(2)
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Notes
Part (a)
distance travelled by the sound = 330 × 3.0 = 990 m
✓ 1
distance to wall = 990 ÷ 2 = 495 m
✓ 1
Part (b)
λ = v ÷ f = 330 ÷ 220
✓ 1
λ = 1.5 m
✓ 1
Part (c)
(i) it is quieter (less loud)
✓ 1
(ii) the pitch is unchanged
✓ 1
Common error: Amplitude affects loudness, and frequency affects pitch; do not mix the two up.
157IGPC3-3-06r19-03Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
A ray of light in air strikes a glass window pane in an office building. The ray enters the glass. Which row describes the speed, wavelength and frequency of the light in the glass compared with in the air?
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Notes
Light slows in glass. Its frequency is set by the source and cannot change at a boundary, so since v = fλ the wavelength must decrease too: C is correct. A keeps the wavelength constant and changes frequency, which would break v = fλ. B describes light speeding up, as in a less dense medium. D wrongly says frequency increases.
✓ 1
Answer C
Answer: C
Common error: Saying the frequency changes when a wave enters a new medium; the frequency stays the same, and the speed and wavelength change together.
158IGPC3-3-06r19-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A narrow beam of white light passes through a glass prism and spreads into a spectrum. Which statement is correct?
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Notes
Violet light is slowed most in glass and so bends most towards the normal: D is correct. A reverses the colours. B is wrong because the speed in glass depends on colour, which is why dispersion happens. C wrongly links the longest wavelength with the greatest bending; red light is refracted least.
✓ 1
Answer D
Answer: D
Exam tip: In a prism spectrum, red is deviated least and violet most; the colour with the highest frequency is slowed most.
159IGPC3-3-06r19-07Sound and ultrasoundPaper CMedium1 mark
A dolphin emits ultrasound of frequency 75 kHz. The speed of sound in sea water is 1500 m/s. What is the wavelength of the ultrasound in the water?
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Notes
Wavelength = speed ÷ frequency = 1500 ÷ 75 000 = 0.020 m, so A is correct. B uses 75 ÷ 1500, which treats the frequency as 75 Hz and inverts the equation. C uses 1500 ÷ 75, forgetting to convert kHz to Hz. D uses 75 000 ÷ 1500, which inverts the equation.
✓ 1
Answer A
Answer: A
Common error: Not converting kHz to Hz before using v = fλ; convert to base units first.
160IGPC3-3-06r19-12Sound and ultrasoundPaper CMedium6 marks
A student stands in front of a large wall at a concert hall and claps once. The sound travels to the wall and back.
Diagram not to scale
(a)
Explain why the student hears an echo.
(2)
(b)
The student is 85 m from the wall and hears the echo 0.50 s after the clap. Calculate the speed of sound in air.
(2)
(c)
Sound of frequency 425 Hz travels through the air at 340 m/s. Calculate its wavelength.
(2)
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Notes
Part (a)
the sound wave reflects off the hard wall
✓ 1
the reflected sound reaches the student after the original clap, so it is heard as a separate sound
✓ 1
Part (b)
distance travelled = 2 × 85 = 170 m
✓ 1
speed = 170 ÷ 0.50 = 340 m/s
✓ 1
Part (c)
wavelength = speed ÷ frequency = 340 ÷ 425
✓ 1
0.80 m
✓ 1
Common error: Using the distance to the wall as the distance the sound travels; the sound goes there and back, so the distance is doubled.
161IGPC3-3-06r19-13Light, refraction, lenses and dispersionPaper CMedium6 marks
A marine biologist uses a converging lens of focal length 8.0 cm as a magnifying glass to look at a small shell.
Diagram not to scale
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
The shell is placed 5.0 cm from the lens. State three characteristics of the image seen.
(3)
(c)
The shell is moved to 20 cm from the lens. State two ways in which the image is now different.
(2)
Show mark scheme
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Mark
Notes
Part (a)
the point where rays parallel to the principal axis are brought together (converge) after passing through the lens
✓ 1
Part (b)
virtual
✓ 1
upright
✓ 1
magnified
✓ 1
Part (c)
one valid difference, e.g. it is real, not virtual
✓ 1
a second, different valid difference, e.g. it is inverted, not upright; or it is smaller than the object (diminished), not magnified
✓ 1
Exam tip: The image is virtual and magnified only when the object is closer to the lens than the principal focus; compare the object distance with the focal length first.
162IGPC3-3-08r23-04Sound and ultrasoundPaper CMedium1 mark
A worker in an empty warehouse claps once and hears an echo from a far wall 0.40 s later. The speed of sound in air is 340 m/s. How far away is the wall?
Diagram not to scale
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Notes
The sound travels to the wall and back, so the total distance is 340 × 0.40 = 136 m and the wall is half of this, 68 m away. The value 136 m forgets that the sound makes a round trip. The value 850 m comes from dividing speed by time. The value 1.2 × 10−3 m comes from dividing time by speed.
✓ 1
Answer D
Answer: D
Exam tip: For echoes, the sound covers the distance twice; halve the distance travelled.
163IGPC3-3-08r23-10Sound and ultrasoundPaper CMedium7 marks
A speaker in a retail store emits a sound of frequency 680 Hz. The speed of sound in air is 340 m/s.
Diagram not to scale
(a)
State the range of frequencies that a healthy young person can hear.
(1)
(b)
A microphone is 51 m from the speaker. Calculate the time taken for the sound to reach it.
(2)
(c)
Describe how sound travels through air from the speaker to the microphone.
(2)
(d)
Calculate the wavelength of the sound.
(2)
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Notes
Part (a)
20 Hz to 20 000 Hz
✓ 1
Part (b)
t = d / v
✓ 1
= 51 / 340 = 0.15 s
✓ 1
Part (c)
The speaker vibrates and makes compressions and rarefactions in the air
✓ 1
The air particles vibrate parallel to the direction of travel and pass the energy on
✓ 1
Part (d)
λ = v / f
✓ 1
= 340 / 680 = 0.50 m
✓ 1
Common error: Sound cannot travel through a vacuum because there are no particles to vibrate.
164IGPC3-3-08r23-11The electromagnetic spectrumPaper CMedium6 marks
At an airport, staff use infrared cameras to screen travellers for a raised temperature. Ultraviolet lamps are used to sterilise the area.
(a)
Microwaves, ultraviolet, visible light and X-rays are four parts of the electromagnetic spectrum. Put them in order of increasing frequency.
(2)
(b)
State why infrared cameras can detect travellers with a raised temperature.
(1)
(c)
State the speed of all electromagnetic waves in a vacuum.
(1)
(d)
The ultraviolet radiation has a frequency of 1.2 × 1015 Hz. Calculate its wavelength.
(2)
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Notes
Part (a)
Microwaves lowest and X-rays highest
✓ 1
Visible light before ultraviolet in the middle
✓ 1
Part (b)
Hotter objects emit more infrared radiation
✓ 1
Part (c)
3.0 × 108 m/s
✓ 1
Part (d)
λ = v / f
✓ 1
= 3.0 × 108 / 1.2 × 1015 = 2.5 × 10−7 m
✓ 1
Exam tip: Convert powers of ten carefully; write 3.0 × 10⁸ / 1.2 × 10¹⁵ as 2.5 × 10⁻⁷, not 2.5 × 10⁷.
165IGPC3-3-08r23-15Sound and ultrasoundPaper CMedium5 marks
A hospital uses an ultrasound scanner to check the depth of a tissue boundary. A pulse of ultrasound travels through tissue at 1500 m/s. It reflects from the boundary and returns to the scanner after 8.0 × 10−5 s.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
Calculate the depth of the boundary below the scanner.
(3)
(c)
State one reason why ultrasound is safer than X-rays for scanning a foetus.
(1)
Show mark scheme
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Mark
Notes
Part (a)
Sound with a frequency above 20 000 Hz, which is higher than humans can hear
✓ 1
Part (b)
Distance there and back = speed × time = 1500 × 8.0 × 10−5 = 0.12 m
✓ 1
Depth is half of this
✓ 1
= 0.060 m
✓ 1
Part (c)
Ultrasound is non-ionising, so it does not damage cells
✓ 1
Common error: Remember that an echo travels to the boundary and back, so the depth is half of the total distance.
166IGPC3-3-08r23-16Wave properties, reflection, refraction and diffractionPaper CMedium6 marks
A harbour wall at a wave-power site has a gap in it. Water waves of speed 3.0 m/s and frequency 0.50 Hz approach the gap.
Diagram not to scale
(a)
Calculate the wavelength of the waves.
(2)
(b)
The gap is 4.0 m wide. Describe the pattern of the waves beyond the gap.
(2)
(c)
A different gap is 30 m wide. Explain how the pattern beyond this gap differs from the 4.0 m gap.
(2)
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Notes
Part (a)
λ = v / f
✓ 1
= 3.0 / 0.50 = 6.0 m
✓ 1
Part (b)
The waves spread out into the region beyond the gap, as circular or curved wavefronts
✓ 1
The wavelength stays the same
✓ 1
Part (c)
The gap is much wider than the wavelength, so there is much less diffraction
✓ 1
Most of the waves continue straight on, with only slight bending at the edges
✓ 1
Exam tip: Diffraction is greatest when the gap is about the same size as the wavelength or smaller.
167IGPC3-3-06r26-03Sound and ultrasoundPaper CMedium1 mark
The echo-sounder of a ship sends a pulse of ultrasound towards the sea bed. The echo is detected 0.32 s later. The speed of sound in sea water is 1500 m/s. What is the depth of the water below the ship?
Diagram not to scale
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Notes
The pulse travels down and back, so the one-way distance is half of the total: depth = 1500 × 0.32 ÷ 2 = 240 m. 480 m is the total distance travelled, which forgets to halve. 960 m doubles the total distance instead of halving it. 120 m halves the distance twice.
✓ 1
Answer D
Answer: D
Exam tip: For any echo question, write down that the sound travels twice the distance before you substitute numbers.
168IGPC3-3-06r26-05Light, refraction, lenses and dispersionPaper CMedium1 mark
A volcanologist uses a converging lens of focal length 10 cm as a magnifying glass to look at crystals in a rock. The rock is 4 cm from the lens. Which row describes the image?
Diagram not to scale
Show mark scheme
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Mark
Notes
The object is closer to the lens than the principal focus (4 cm is less than 10 cm), so the lens forms a virtual, upright and magnified image, which is how a magnifying glass works. A real, inverted image forms only when the object is beyond the principal focus, so both real options are wrong. A virtual image from a converging lens is always magnified here, so virtual, upright and diminished is also wrong.
✓ 1
Answer B
Answer: B
Exam tip: Compare the object distance with the focal length first: inside the principal focus gives a virtual image.
169IGPC3-3-06r26-11Sound and ultrasoundPaper CMedium7 marks
The echo-sounder on a ship sends out pulses of ultrasound with a frequency of 50 kHz. The speed of sound in sea water is 1500 m/s.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
Calculate the wavelength of the ultrasound in the sea water.
(2)
(c)
Suggest why ultrasound is used rather than sound that a person can hear.
(1)
(d)
An echo from the sea bed is detected 0.64 s after the pulse is sent. Calculate the depth of the water.
(3)
Show mark scheme
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Mark
Notes
Part (a)
sound with a frequency higher than 20 000 Hz, which is above the range of human hearing
✓ 1
Part (b)
λ = v ÷ f = 1500 ÷ 50 000
✓ 1
0.030 m
✓ 1
Part (c)
any one from: ultrasound has a shorter wavelength so it diffracts less and gives a narrower beam / it is not heard by crew or disturbing to marine animals
✓ 1
Part (d)
distance travelled = 1500 × 0.64
✓ 1
this is the distance down and back, so divide by 2
✓ 1
480 m
✓ 1
Exam tip: Remember that the echo travels to the sea bed and back, so the depth is half of the distance travelled.
170IGPC3-3-06r26-12The electromagnetic spectrumPaper CMedium7 marks
Satellites are used to monitor a volcano. One instrument detects infrared radiation from hot lava. The infrared radiation has a wavelength of 1.2 × 10−5 m. The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
State why infrared radiation is used to detect hot lava.
(1)
(b)
State the speed of electromagnetic waves in a vacuum.
(1)
(c)
Calculate the frequency of the infrared radiation.
(2)
(d)
State one harmful effect on humans of too much exposure to ultraviolet radiation.
(1)
(e)
The satellite is 7.00 × 105 m above the volcano. Calculate the time taken for a radio signal to travel from the satellite to the ground. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
all hot objects emit infrared radiation, and hotter objects emit more
✓ 1
Part (b)
3.0 × 108 m/s
✓ 1
Part (c)
f = v ÷ λ = 3.0 × 108 ÷ 1.2 × 10−5
✓ 1
2.5 × 1013 Hz
✓ 1
Part (d)
any one from: skin cancer / skin burns / damage to the eyes
✓ 1
Part (e)
t = d ÷ v = 7.00 × 105 ÷ 3.0 × 108
✓ 1
2.33 × 10−3 s
✓ 1
Exam tip: Write powers of ten carefully and enter them correctly on your calculator, using the EXP or ×10x key.
171IGPC3-3-06r26-13Wave properties, reflection, refraction and diffractionPaper CMedium7 marks
A theatre lighting technician tests a rectangular glass block that is used as a colour filter. A ray of light in air hits the block with an angle of incidence of 40°. The angle of refraction in the glass is 25°. The speed of light in air is 3.0 × 108 m/s.
Diagram not to scale
(a)
State what is meant by refraction.
(1)
(b)
Calculate the refractive index of the glass. Give your answer to 3 significant figures.
(2)
(c)
State what happens to the frequency and to the wavelength of the light as it enters the glass.
(2)
(d)
Calculate the speed of the light in the glass. Use your unrounded value from the calculation of the refractive index or the value 1.52; either is accepted. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
the change in direction of a wave as it passes from one medium into another (because its speed changes)
✓ 1
Part (b)
n = sin i ÷ sin r = sin 40° ÷ sin 25°
✓ 1
1.52
✓ 1
Part (c)
the frequency stays the same
✓ 1
the wavelength decreases
✓ 1
Part (d)
speed = c ÷ n = 3.0 × 108 ÷ 1.52
✓ 1
1.97 × 108 m/s
✓ 1
Exam tip: Check that your calculator is in degree mode before using sine.
172IGPC3-3-06r26-14Wave properties, reflection, refraction and diffractionPaper CMedium8 marks
Waves are created on the surface of a lake in a volcanic crater. A student counts 12 wave crests passing a fixed point in 20 s. The wavelength of the waves is 0.80 m.
Diagram not to scale
(a)
Calculate the frequency of the waves.
(1)
(b)
Calculate the speed of the waves.
(2)
(c)
Calculate the time period of the waves. Give your answer to 3 significant figures.
(1)
(d)
Near the edge of the lake the water is shallower and the waves slow down. State what happens to the frequency and to the wavelength of the waves.
(2)
(e)
In the shallow water the wavelength is 0.32 m. Calculate the speed of the waves there. Give your answer to 3 significant figures.
(2)
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Part (a)
12 ÷ 20 = 0.60 Hz
✓ 1
Part (b)
v = f × λ = 0.60 × 0.80
✓ 1
0.48 m/s
✓ 1
Part (c)
T = 1 ÷ f = 1 ÷ 0.60 = 1.67 s
✓ 1
Part (d)
the frequency stays the same
✓ 1
the wavelength decreases
✓ 1
Part (e)
v = f × λ = 0.60 × 0.32
✓ 1
0.192 m/s
✓ 1
Common error: The number of crests passing a point per second is the frequency; do not divide by the wavelength.
173IGPC3-3-07r29-03Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
In a food-science laboratory a beam of light passes from air into olive oil. Which row describes the light as it enters the oil?
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Notes
The frequency is set by the source and does not change at a boundary. The speed falls in the denser medium and, since v = fλ with f fixed, the wavelength falls in proportion. The other rows change the frequency, or keep the wavelength or speed fixed, which cannot satisfy v = fλ.
✓ 1
Answer A
Answer: A
Common error: Frequency never changes when a wave crosses a boundary; speed and wavelength change together.
174IGPC3-3-07r29-04Light, refraction, lenses and dispersionPaper CMedium1 mark
A food scientist uses a converging lens of focal length 8 cm as a magnifying glass. A sugar crystal is placed 5 cm from the lens. Which describes the image?
Diagram not to scale
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Notes
The crystal is closer to the lens than the focal length (5 cm < 8 cm), so the refracted rays diverge and appear to come from a larger, upright image on the same side as the crystal; this image is virtual and magnified. Real images only form when the object is beyond the focal length, which is not the case here, and a diminished virtual image is not produced by a converging lens used this way.
✓ 1
Answer C
Answer: C
Exam tip: For a converging lens, an object inside the focal length always gives a virtual, upright, magnified image.
175IGPC3-3-07r29-07Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
A sound of frequency 500 Hz travels in air at 340 m/s and then enters water, where the speed of sound is 1500 m/s. What is the wavelength of the sound in the water?
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Notes
The frequency stays at 500 Hz, so λ = v / f = 1500 / 500 = 3.0 m. 0.33 m comes from dividing frequency by speed. 0.68 m is the wavelength in air, which wrongly assumes the wavelength does not change. 4.4 m is the ratio of the two speeds, not a wavelength.
✓ 1
Answer C
Answer: C
Exam tip: Write v = fλ first, and remember that f is the same in both media.
176IGPC3-3-07r29-11The electromagnetic spectrumPaper CMedium6 marks
A mobile-phone network transmits microwaves of frequency 2.0 GHz (2.0 × 109 Hz). The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
State one use of infrared radiation in telecommunications.
(1)
(b)
Calculate the wavelength of the microwaves.
(2)
(c)
State one hazard of exposure to microwaves.
(1)
(d)
A radio station broadcasts at 100 MHz. State whether its wavelength is longer or shorter than that of the 2.0 GHz microwaves, and give a reason.
(2)
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Notes
Part (a)
Signals sent along optical fibres, or short-range remote controls (any one valid use)
Heating of internal body tissue or cells (any valid hazard)
✓ 1
Part (d)
Longer
✓ 1
Lower frequency and the same speed, so wavelength = speed ÷ frequency is greater
✓ 1
Exam tip: Convert GHz and MHz to Hz before using v = fλ.
177IGPC3-3-07r29-14Sound and ultrasoundPaper CMedium5 marks
In a food-processing plant a pulse of ultrasound with a frequency of 2.5 MHz is used to measure the thickness of a layer of fat on a side of beef. The pulse reflects from the boundary between the fat and the muscle and returns to the detector 52 μs after it was sent. The speed of ultrasound in the fat is 1500 m/s.
Diagram not to scale
(a)
State the range of frequencies that a human can hear.
(1)
(b)
Explain why the pulse is described as ultrasound.
(1)
(c)
Calculate the thickness of the layer of fat. Give your answer in metres.
(3)
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Notes
Part (a)
20 Hz to 20 000 Hz (20 kHz)
✓ 1
Part (b)
Its frequency is higher than 20 000 Hz, the upper limit of human hearing
✓ 1
Part (c)
M1 distance travelled = speed × time = 1500 × 52 × 10−6 = 0.078 m
✓ 1
M1 thickness is half of this distance
✓ 1
A1 0.039 m
✓ 1
Common error: The pulse crosses the layer twice, so halve the distance found from speed × time.
178IGPC3-3-10r32-05Light, refraction, lenses and dispersionPaper CMedium1 mark
A drone camera uses a thin converging lens to focus on a field. The field is further from the lens than twice the focal length. Which describes the image formed on the sensor?
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Notes
For an object beyond twice the focal length, a converging lens forms a real, inverted, diminished image between F and 2F on the other side. Virtual, upright and magnified is the magnifying-glass case (object inside F). Real, inverted and magnified occurs for an object between F and 2F. A real image formed by a single lens is always inverted, so real and upright is not possible.
✓ 1
Answer A
Answer: A
Exam tip: Link the object position (inside F, between F and 2F, beyond 2F) to the image type; a real image from a lens is always inverted.
179IGPC3-3-10r32-06Sound and ultrasoundPaper CMedium1 mark
An ultrasonic sensor at the top of a grain silo sends a pulse downwards. The pulse reflects from the grain surface and returns after 0.025 s. The speed of sound in air is 340 m/s. How far is the sensor above the grain surface?
Diagram not to scale
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Notes
The pulse travels down and back, so 2d = vt = 340 × 0.025 = 8.5 m and d = 4.25 m. 8.5 m forgets that the pulse travels the distance twice. 6800 m divides speed by time instead of multiplying. 13 600 m does the same division and does not halve.
✓ 1
Answer C
Answer: C
Common error: An echo covers the distance twice: halve speed × time to get the distance to the reflector.
180IGPC3-3-10r32-07Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
Radio waves from a transmitter pass through a narrow gap between two hills. The gap width stays the same. Which change to the radio waves makes the diffraction at the gap more pronounced?
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Notes
Diffraction becomes more pronounced as the wavelength gets larger compared with the gap width, so increasing the wavelength is correct. Decreasing the wavelength, or increasing the frequency (which has the same effect), makes the waves spread less. Amplitude affects only the energy of the wave, not how much it spreads.
✓ 1
Answer D
Answer: D
Common error: Higher frequency means shorter wavelength, which means less diffraction, not more.
181IGPC3-3-10r32-11Light, refraction, lenses and dispersionPaper CMedium5 marks
A pilot uses a small plane mirror to check the wing of a light aircraft. A ray of light from the wing strikes the mirror and makes an angle of 35° with the surface of the mirror.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
Calculate the angle of reflection.
(2)
(c)
State two characteristics of the image formed by a plane mirror.
(2)
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Notes
Part (a)
the angle of incidence equals the angle of reflection (both measured from the normal)
✓ 1
Part (b)
M1 angle of incidence is measured from the normal, so 90° − 35°
✓ 1
A1 55°
✓ 1
Part (c)
one correct characteristic (virtual; the same size as the object; the same distance behind the mirror as the object is in front; laterally inverted)
✓ 1
a second, different correct characteristic from the same list
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the surface of the mirror.
182IGPC3-3-10r32-13Light, refraction, lenses and dispersionPaper CMedium5 marks
A seed inspector uses a magnifying glass to examine a seed of length 2.5 mm.
(a)
State the type of lens used in a magnifying glass.
(1)
(b)
The seed is held closer to the lens than the focal length. State two characteristics of the image formed.
(2)
(c)
The image has a length of 7.5 mm. Calculate the magnification.
(2)
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Notes
Part (a)
converging (convex) lens
✓ 1
Part (b)
one correct characteristic (virtual; upright; magnified)
✓ 1
a second, different correct characteristic from the same list
Exam tip: Magnification is a ratio of two lengths, so it has no unit as long as both lengths use the same unit.
183IGPC3-3-10r32-14Sound and ultrasoundPaper CMedium5 marks
A guitarist plays a note of frequency 880 Hz. The speed of sound in air is 340 m/s.
(a)
Describe how the vibrating string produces a sound wave in the air.
(2)
(b)
Calculate the wavelength of the sound in air, to 3 significant figures.
(2)
(c)
The guitarist plays the same note more quietly. State what happens to the amplitude of the sound wave.
(1)
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Notes
Part (a)
the string vibrates and pushes and pulls on the nearby air particles
✓ 1
this creates compressions and rarefactions that travel outwards as a longitudinal wave
✓ 1
Part (b)
M1 λ = v ÷ f = 340 ÷ 880
✓ 1
A1 0.386 m
✓ 1
Part (c)
the amplitude decreases
✓ 1
Common error: Sound is a longitudinal wave: the air particles vibrate parallel to the direction in which the wave travels.
184IGPC3-3-09r34-03Sound and ultrasoundPaper CMedium1 mark
At an excavation site a student claps in front of a rock wall and hears an echo 0.60 s later. The speed of sound in air is 340 m/s. How far away is the wall?
Diagram not to scale
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Notes
The sound travels to the wall and back, so the total distance is 340 × 0.60 = 204 m and the wall is 204 ÷ 2 = 102 m away. 204 m forgets that the path is there and back. 51 m halves the distance twice. 567 m comes from dividing speed by time.
✓ 1
Answer A
Answer: A
Common error: An echo covers the distance twice, so divide speed × time by 2.
185IGPC3-3-09r34-04Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
Light from a cycle lamp passes from air into a plastic lens. Which row describes what happens to the light as it enters the plastic?
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Notes
On entering a denser medium the light slows down. Its frequency is set by the source and stays the same, so since v = fλ the wavelength decreases. The other rows either have the speed rising, change the frequency, or leave the wavelength unchanged despite the lower speed.
✓ 1
Answer D
Answer: D
Exam tip: Frequency never changes at a boundary; speed and wavelength change together.
186IGPC3-3-09r34-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A converging lens has focal length 10 cm. An archaeologist uses it as a magnifying glass to look at a carving. At which object distance from the lens is a magnified, upright image seen?
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Notes
A magnifying glass has the object closer to the lens than the principal focus, which gives a virtual, upright, magnified image; 6 cm is less than 10 cm. At 10 cm the refracted rays are parallel and no image forms. At 15 cm and 25 cm the image is real and inverted.
✓ 1
Answer A
Answer: A
Exam tip: A lens works as a magnifier only when the object is inside the focal length.
187IGPC3-3-09r34-10Light, refraction, lenses and dispersionPaper CMedium6 marks
A cyclist checks traffic behind her using a small plane mirror on the handlebars.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
A ray of light strikes the mirror at an angle of incidence of 35°. State the angle of reflection.
(1)
(c)
Calculate the angle between the incident ray and the reflected ray.
(2)
(d)
State two properties of the image formed in a plane mirror.
(2)
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Notes
Part (a)
The angle of incidence equals the angle of reflection
✓ 1
Part (b)
35°
✓ 1
Part (c)
M1 35° + 35°
✓ 1
A1 70°
✓ 1
Part (d)
Accept any two from: virtual; upright; same size as the object; laterally inverted; same distance behind the mirror as the object is in front.
—
One valid property
✓ 1
A second valid property
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the mirror surface.
188IGPC3-3-09r34-13Light, refraction, lenses and dispersionPaper CMedium5 marks
An archaeologist uses a converging lens as a magnifying glass to study a coin.
(a)
State what is meant by the focal length of a converging lens.
(1)
(b)
The coin is placed closer to the lens than the principal focus. Describe the image.
(2)
(c)
The coin has height 1.5 cm and its image has height 4.5 cm. Calculate the magnification.
(2)
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Notes
Part (a)
The distance from the centre of the lens to the principal focus, the point where rays parallel to the principal axis meet
Exam tip: Magnification has no unit because it is a ratio of two lengths.
189IGPC3-3-09r34-15Sound and ultrasoundPaper CMedium6 marks
Cooling fans in a data centre produce a steady sound of frequency 850 Hz.
(a)
State which property of a sound wave determines its pitch.
(1)
(b)
State which property of a sound wave determines its loudness.
(1)
(c)
Calculate the period of the sound wave. Give your answer to 3 significant figures.
(2)
(d)
Explain why the sound cannot travel through a vacuum.
(2)
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Part (a)
Frequency
✓ 1
Part (b)
Amplitude
✓ 1
Part (c)
M1 T = 1 ÷ f = 1 ÷ 850
✓ 1
A1 1.18 × 10−3 s
✓ 1
Part (d)
Sound is passed on by vibrating particles
✓ 1
A vacuum contains no particles to vibrate and pass on the wave
✓ 1
Exam tip: Period is the time for one cycle, so it is the reciprocal of frequency.
190IGPC3-3-10r34-03Sound and ultrasoundPaper CMedium1 mark
A space probe is 6.0 × 1010 m from Earth. It sends a radio signal to Earth. Radio waves travel at 3.0 × 108 m/s. How long does the signal take to arrive?
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Notes
time = distance ÷ speed = 6.0 × 1010 ÷ 3.0 × 108 = 200 s. 0.0050 s inverts the equation. 100 s halves the distance. 400 s doubles it, as if the signal made a round trip.
✓ 1
Answer C
Answer: C
Common error: Use time = distance ÷ speed and check the power of ten.
191IGPC3-3-10r34-04Light, refraction, lenses and dispersionPaper CMedium1 mark
White light from the glowing chamber of a coffee roaster passes through a glass prism and is dispersed. Which colour is deviated most, and why?
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Notes
Violet light is slowed most in glass, so it is refracted through the largest angle. Red is deviated least despite having the longest wavelength. Green is neither the extreme. Colours travel at the same speed in air but not in glass, which is what causes the dispersion.
✓ 1
Answer B
Answer: B
Exam tip: Red is deviated least and violet most.
192IGPC3-3-10r34-05Sound and ultrasoundPaper CMedium1 mark
An inspector sends an ultrasound pulse into a steel bridge beam. The pulse travels at 5000 m/s and an echo from a crack returns after 40 μs. How far from the surface is the crack?
Diagram not to scale
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Notes
Total distance = 5000 × 40 × 10−6 = 0.20 m, and the crack is half of this, 0.10 m, from the surface. 0.20 m forgets that the pulse goes there and back. 100 m and 200 m come from treating microseconds as milliseconds.
✓ 1
Answer A
Answer: A
Common error: Convert μs to s (×10⁻⁶) and halve the total path.
193IGPC3-3-10r34-08Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
Water waves of wavelength 0.020 m reach a gap between two bridge piers. Which gap width produces the most noticeable diffraction?
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Notes
Diffraction is greatest when the gap is about the same size as the wavelength, which is 0.020 m. Wider gaps give less spreading, so the waves mostly continue straight through.
✓ 1
Answer C
Answer: C
Common error: A wider gap does not give more spreading; diffraction is most noticeable for gaps comparable to the wavelength.
194IGPC3-3-10r34-10The electromagnetic spectrumPaper CMedium6 marks
A coffee roaster uses electromagnetic waves in several ways. The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
Hot coffee beans emit mainly one type of electromagnetic radiation, which a thermal camera detects. Name this radiation.
(1)
(b)
State one harmful effect of ultraviolet radiation on the human body.
(1)
(c)
State one similarity and one difference between gamma rays and radio waves.
(2)
(d)
Orange light from the roaster window has wavelength 6.0 × 10−7 m. Calculate its frequency.
(2)
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Part (a)
Infrared
✓ 1
Part (b)
Any one from: skin cancer; sunburn; eye damage
✓ 1
Part (c)
Similarity: both travel at the same speed in a vacuum, or both are transverse
✓ 1
Difference: gamma rays have a higher frequency or shorter wavelength than radio waves
✓ 1
Part (d)
M1 f = v ÷ λ = 3.0 × 108 ÷ 6.0 × 10−7
✓ 1
A1 5.0 × 1014 Hz
✓ 1
Common error: Gamma rays are not faster than radio waves; they differ in frequency and wavelength, not speed.
195IGPC3-3-10r34-13Sound and ultrasoundPaper CMedium5 marks
An inspector taps one end of a steel girder of length 150 m and a sensor at the other end detects the vibration passing along the steel. The speed of sound in steel is 5000 m/s.
(a)
State which of steel, water and air carries sound fastest.
(1)
(b)
Explain, in terms of particles, why sound travels faster in solids than in gases.
(2)
(c)
Calculate the time taken for the sound to travel along the girder.
(2)
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Notes
Part (a)
Steel
✓ 1
Part (b)
The particles in a solid are much closer together
✓ 1
so vibrations are passed from particle to particle more quickly
✓ 1
Part (c)
M1 t = distance ÷ speed = 150 ÷ 5000
✓ 1
A1 0.030 s
✓ 1
Common error: Sound travels faster in denser, more rigid materials, not slower.
196IGPC3-3-10r34-15Light, refraction, lenses and dispersionPaper CMedium5 marks
A student shines white light from a lamp used to monitor coffee roasting through a glass prism and sees a spectrum on a screen.
(a)
State what is meant by refraction.
(1)
(b)
As light enters the glass from air, state what happens to its frequency and to its wavelength.
(2)
(c)
Describe what the student observes and state which colour is refracted most.
(2)
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Notes
Part (a)
The change in direction of a wave when it changes speed as it passes from one medium to another
✓ 1
Part (b)
The frequency stays the same
✓ 1
The wavelength decreases
✓ 1
Part (c)
White light is split into a spectrum of colours
✓ 1
Violet is refracted the most (and red the least)
✓ 1
Exam tip: This splitting is called dispersion; say that it happens because each colour has a different speed in glass.
197IGPC3-3-09r36-03Sound and ultrasoundPaper CMedium1 mark
A survey boat measures the depth of the sea at a planned wind farm site. A pulse of ultrasound is sent straight down and its echo is detected 0.064 s later. The speed of sound in sea water is 1500 m/s. What is the depth of the sea?
Diagram not to scale
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Notes
The pulse travels to the sea bed and back, so the distance travelled is 1500 × 0.064 = 96 m and the depth is half of this, 48 m. 96 m forgets that the pulse travels down and back. 24 m halves the distance twice. 23 000 m comes from dividing speed by time.
✓ 1
Answer A
Answer: A
Common error: An echo covers the journey there and back, so halve the total distance travelled to find the distance to the reflecting surface.
198IGPC3-3-09r36-04Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
A ray of monochromatic light passes from air into glass. Which row describes what happens to the speed, the wavelength and the frequency of the light as it enters the glass?
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Notes
The frequency is set by the source and does not change at a boundary. Light slows down in glass, and because v = fλ with f constant, the wavelength decreases as well. The option with decreasing frequency wrongly makes the frequency change. The option with increasing wavelength contradicts v = fλ for a slower wave of the same frequency. The option with unchanged speed ignores that light slows in glass and wrongly makes the frequency increase.
✓ 1
Answer C
Answer: C
Exam tip: At a boundary the frequency never changes; use v = fλ to decide how the wavelength changes when the speed changes.
199IGPC3-3-09r36-05Light, refraction, lenses and dispersionPaper CMedium1 mark
An object is placed 30 cm from a converging lens of focal length 12 cm. Which row describes the image?
Diagram not to scale
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Notes
The object is more than twice the focal length (24 cm) from the lens, so the lens forms a real, inverted image that is smaller than the object. The virtual, upright, magnified image is only produced when the object is closer to the lens than the principal focus. A real, inverted, magnified image needs the object between one and two focal lengths from the lens. A real image formed by a converging lens is never upright.
✓ 1
Answer D
Answer: D
Common error: Real images formed by a converging lens are always inverted; an upright image from a converging lens is always virtual.
200IGPC3-3-09r36-06Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
Water waves of wavelength 3.0 cm pass through a gap of width 12 cm in a barrier in a ripple tank. Which change would make the diffraction at the gap more pronounced?
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Notes
Diffraction is most pronounced when the gap is similar in size to the wavelength, so narrowing the gap to 4 cm helps. A wider gap makes the gap even larger compared with the wavelength, so diffraction is less noticeable. A higher frequency in the same water gives a shorter wavelength, which also reduces diffraction. Changing the amplitude does not change how much the waves spread.
✓ 1
Answer B
Answer: B
Exam tip: Compare the gap width with the wavelength: the closer they are in size, the more the waves spread out.
201IGPC3-3-09r36-12Light, refraction, lenses and dispersionPaper CMedium6 marks
A narrow beam of white light is directed at a glass prism and a coloured spectrum is seen on a screen.
(a)
State the name of this effect.
(1)
(b)
State the colour at the end of the spectrum that is deviated the most by the prism.
(1)
(c)
Explain why the different colours of light are separated by the prism.
(2)
(d)
Violet light has a frequency of 7.5 × 1014 Hz in air. Calculate its wavelength in air. The speed of light in air is 3.0 × 108 m/s.
(2)
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Notes
Part (a)
dispersion
✓ 1
Part (b)
violet
✓ 1
Part (c)
different colours travel at different speeds in glass
Exam tip: When asked to explain dispersion, link the different speeds of the colours in glass to the different amounts of refraction.
202IGPC3-3-09r36-14Sound and ultrasoundPaper CMedium6 marks
A student in a library laboratory studies sound from a loudspeaker.
(a)
State what is meant by a longitudinal wave.
(1)
(b)
Describe how sound travels through the air from the loudspeaker to the student's ear.
(2)
(c)
The loudspeaker produces a note of frequency 512 Hz. The speed of sound in air is 340 m/s. Calculate the wavelength of the sound. Give your answer to 3 significant figures.
(2)
(d)
The loudspeaker is then made to produce a louder note with the same frequency. State what happens to the amplitude and to the wavelength of the sound.
(1)
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Mark
Notes
Part (a)
the vibrations (of the particles) are parallel to the direction in which the wave travels (energy is transferred)
✓ 1
Part (b)
the loudspeaker cone vibrates and pushes and pulls on the air, forming compressions and rarefactions
✓ 1
these pass from particle to particle through the air as a longitudinal wave
✓ 1
Part (c)
wavelength = speed / frequency
✓ 1
wavelength = 340 / 512 = 0.664 m
✓ 1
Part (d)
amplitude increases and wavelength stays the same
✓ 1
Common error: Louder does not mean higher: loudness depends on amplitude, while pitch depends on frequency.
203IGPC3-3-09r37-02Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
A beekeeper watches ripples on a pond next to her apiary. A floating leaf moves up and down 24 times in 12 s as the ripples pass. The distance between neighbouring crests is 0.30 m. What is the speed of the ripples?
Diagram not to scale
Show mark scheme
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Mark
Notes
The leaf bobs 24 times in 12 s, so the frequency is 24 ÷ 12 = 2.0 Hz. Speed = frequency × wavelength = 2.0 × 0.30 = 0.60 m/s. 0.15 m/s comes from dividing the wavelength by the frequency. 7.2 m/s comes from multiplying the wavelength by the number of bobs (24) instead of the frequency. 0.025 m/s comes from dividing the wavelength by the time (12 s).
✓ 1
Answer C
Answer: C
Exam tip: Convert a count of oscillations in a time into a frequency (per second) before using speed = frequency × wavelength.
204IGPC3-3-09r37-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A hospital technician uses a converging lens of focal length 10 cm to focus the image of a lamp onto a screen. The lamp is 30 cm from the lens. Which row describes the image?
Diagram not to scale
Show mark scheme
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Mark
Notes
The lamp is more than twice the focal length (20 cm) from the lens, so the lens forms a real image that is inverted and smaller than the object; it can be caught on a screen. A virtual, upright, magnified image is formed only when the object is closer than the focal length. A real, inverted, magnified image needs the object between one and two focal lengths from the lens. A virtual image cannot be shown on a screen.
✓ 1
Answer A
Answer: A
Exam tip: A screen image is always real and inverted; compare the object distance with the focal length and twice the focal length to decide its size.
205IGPC3-3-09r37-07Sound and ultrasoundPaper CMedium1 mark
A track marshal at a Formula 1 circuit hears the echo of a passing car from a large wall 0.80 s after the sound of the car reaches him directly. Assume the car is very close to him. The speed of sound in air is 340 m/s. How far away is the wall?
Show mark scheme
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Mark
Notes
The sound travels to the wall and back in 0.80 s, so the total distance is 340 × 0.80 = 272 m and the wall is half of that, 136 m, away. 272 m forgets that the sound travels there and back. 68 m halves the distance twice. 425 m comes from dividing the speed by the time instead of multiplying.
✓ 1
Answer D
Answer: D
Common error: For an echo, the time includes the journey to the wall and back, so halve the total distance.
206IGPC3-3-09r37-11The electromagnetic spectrumPaper CMedium5 marks
In a hospital, gamma rays are used to sterilise surgical instruments. The gamma rays have a frequency of 2.0 × 1020 Hz. The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
State one hazard to human health of exposure to gamma rays.
(1)
(b)
Calculate the wavelength of these gamma rays.
(2)
(c)
A forehead thermometer in the hospital detects infrared radiation. State whether the frequency of infrared radiation is higher or lower than the frequency of gamma rays.
(1)
(d)
State one property that gamma rays and infrared radiation have in common.
(1)
Show mark scheme
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Mark
Notes
Part (a)
damage to cells / cancer / mutation of DNA / radiation burns (any one)
both travel at the same speed in a vacuum / both are transverse waves / both are electromagnetic waves (any one)
✓ 1
Exam tip: Rewrite frequencies and wavelengths in standard form before dividing; subtract the powers of ten carefully.
207IGPC3-3-09r37-13Light, refraction, lenses and dispersionPaper CMedium6 marks
A surgeon uses a converging lens of focal length 12 cm as a magnifying glass to inspect a small object.
Diagram not to scale
(a)
State what is meant by the focal length of a converging lens.
(1)
(b)
The object is 5.0 cm from the lens. State three properties of the image formed.
(3)
(c)
The surgeon moves the object so that it is 20 cm from the lens. State whether the image is now real or virtual, and whether it is upright or inverted.
(2)
Show mark scheme
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Mark
Notes
Part (a)
the distance from the centre of the lens to the principal focus
✓ 1
Part (b)
virtual
✓ 1
upright
✓ 1
magnified
✓ 1
Part (c)
real
✓ 1
inverted
✓ 1
Common error: A magnifying glass only works when the object is closer to the lens than the focal length; beyond that the image is real and inverted.
208IGPC3-3-09r37-14Sound and ultrasoundPaper CMedium6 marks
A hospital scanner sends a short pulse of ultrasound into a patient. The pulse reflects from a boundary inside the body and the echo returns to the probe 80 μs after the pulse was emitted. The speed of ultrasound in body tissue is 1500 m/s.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
Calculate the depth of the boundary below the probe. Give your answer in cm.
(3)
(c)
Suggest why ultrasound is used instead of X-rays to scan an unborn baby.
(2)
Show mark scheme
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Mark
Notes
Part (a)
sound with a frequency higher than 20 000 Hz / above the upper limit of human hearing
✓ 1
Part (b)
distance travelled = speed × time = 1500 × 80 × 10−6 = 0.12 m
✓ 1
the pulse travels to the boundary and back, so divide by 2
✓ 1
6.0 cm
✓ 1
Part (c)
ultrasound is non-ionising
✓ 1
so it does not damage cells or DNA / X-rays are ionising and could harm the baby
✓ 1
Common error: The time of an echo covers the trip to the boundary and back, so the depth is half of speed × time.
209IGPC3-3-09r37-16Wave properties, reflection, refraction and diffractionPaper CMedium7 marks
During a Formula 1 race, the pit crew speak to a driver using radio waves of frequency 450 MHz. Radio waves travel at 3.00 × 108 m/s in air.
(a)
State whether radio waves are transverse or longitudinal.
(1)
(b)
Calculate the wavelength of the radio waves. Give your answer to 3 significant figures.
(2)
(c)
A message from the pit wall reaches a car that is 2.4 km away. Calculate the time taken by the signal.
(2)
(d)
The radio waves pass through a doorway of width 0.70 m in a garage wall. State whether the waves are noticeably diffracted and give a reason.
the width of the gap is similar to the wavelength of the waves (ECF from (b))
✓ 1
Common error: Convert prefixes before calculating: 450 MHz is 450 × 10⁶ Hz and 2.4 km is 2400 m.
210IGPC3-3-08r40-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A security camera in a library uses a converging lens to focus light from a person standing far beyond twice the focal length of the lens. Which statement describes the image formed by the lens?
Show mark scheme
Marking point
Mark
Notes
When an object is beyond twice the focal length of a converging lens, the rays from it actually meet on the far side of the lens, so the image is real, inverted and diminished. A virtual image forms only when the object is closer to the lens than its focal length. A real image from a single converging lens is always inverted, so an upright real image is not possible.
✓ 1
Answer A
Answer: A
Exam tip: A real image can be caught on a screen and is inverted; a virtual image cannot be caught on a screen and is upright.
211IGPC3-3-08r40-07Sound and ultrasoundPaper CMedium1 mark
Ultrasound is used to test the thickness of a wind-turbine blade. A pulse is sent in from the outer surface and its echo from the inner surface returns after 4.0 × 10−5 s. The speed of ultrasound in the blade material is 2800 m/s. What is the thickness of the blade?
Diagram not to scale
Show mark scheme
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Notes
The pulse crosses the blade twice, so thickness = 2800 × 4.0 × 10−5 ÷ 2 = 0.056 m = 5.6 cm. 11 cm is the there-and-back distance (not halved). 2.8 cm comes from halving twice. 56 cm comes from a slip of a power of ten when converting the time.
✓ 1
Answer C
Answer: C
Common error: Writing 4.0 × 10⁻⁵ as 4.0 × 10⁻⁴ multiplies the answer by ten; check the power of ten on your calculator.
212IGPC3-3-08r40-11Sound and ultrasoundPaper CMedium7 marks
A signal generator in a school library produces a sound of frequency 200 Hz. The speed of sound in air is 340 m/s.
(a)
Calculate the wavelength of the sound.
(2)
(b)
The librarian turns the amplitude control up and then turns the frequency control up. State what change to the sound is heard for each.
(2)
(c)
A dog whistle produces a sound of frequency 28 kHz. Explain why a person cannot hear it.
(2)
(d)
Explain why the sound from the generator would not be heard if it was placed in an evacuated jar.
(1)
Show mark scheme
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Mark
Notes
Part (a)
wavelength = speed ÷ frequency = 340 ÷ 200
✓ 1
1.7 m
✓ 1
Part (b)
larger amplitude: louder sound
✓ 1
higher frequency: higher pitch
✓ 1
Part (c)
the range of human hearing is about 20 Hz to 20 kHz
✓ 1
28 kHz is above the upper limit of this range
✓ 1
Part (d)
sound needs particles (a medium) to pass vibrations on, and there are none in a vacuum
✓ 1
Exam tip: 'Louder' links to amplitude and 'higher pitch' links to frequency; do not swap them.
213IGPC3-3-08r40-16Wave properties, reflection, refraction and diffractionPaper CMedium6 marks
Water waves approach an offshore wind farm. The waves have a speed of 4.0 m/s and a frequency of 0.50 Hz.
(a)
Calculate the wavelength of the waves.
(2)
(b)
A wave strikes a smooth flat sea wall. State the law of reflection.
(1)
(c)
The waves pass through a gap of width 8.0 m in a harbour barrier. Describe what happens to the waves after the gap.
(2)
(d)
State how the diffraction would change if the gap were much wider than 8.0 m.
(1)
Show mark scheme
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Mark
Notes
Part (a)
wavelength = speed ÷ frequency = 4.0 ÷ 0.50
✓ 1
8.0 m
✓ 1
Part (b)
the angle of incidence equals the angle of reflection
✓ 1
Part (c)
the waves spread out into the region behind the barrier (diffraction)
✓ 1
the wavelength and frequency stay the same
✓ 1
Part (d)
there is much less spreading / the waves mostly continue straight on
✓ 1
Exam tip: Diffraction is greatest when the gap is about the same size as the wavelength.
214IGPC3-3-09r42-03Light, refraction, lenses and dispersionPaper CMedium1 mark
In a gaming headset, a ray of light travels from air into an acrylic lens of refractive index 1.50. The angle of incidence is 40°. What is the angle of refraction?
Diagram not to scale
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Mark
Notes
Using n = sin i ÷ sin r, sin r = sin 40° ÷ 1.50 = 0.4285, so r = 25.4°, which is 25° to the nearest degree. 27° comes from dividing the angle itself by 1.50 instead of using sines. 60° comes from multiplying the angle by 1.50. 65° is the angle between the refracted ray and the surface, not the normal.
✓ 1
Answer C
Answer: C
Common error: Apply the refractive index to the sines of the angles, not to the angles themselves.
215IGPC3-3-09r42-04Sound and ultrasoundPaper CMedium1 mark
An ultrasonic pulse is sent vertically downwards from the top of a water storage tank. The echo from the bottom returns after 0.080 s. The speed of sound in water is 1500 m/s. What is the depth of the water?
Diagram not to scale
Show mark scheme
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Notes
The pulse travels down and back, so the distance travelled is 1500 × 0.080 = 120 m and the depth is half of this, 60 m. 120 m forgets that the pulse makes a two-way trip. 30 m halves the distance twice. 19 000 m comes from dividing the speed by the time instead of multiplying.
✓ 1
Answer D
Answer: D
Exam tip: In echo problems, the pulse covers the distance twice, so halve the total distance.
216IGPC3-3-09r42-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A pharmacist reads small print on a medicine label using a converging lens of focal length 10 cm. The label is held 4.0 cm from the lens. Which row describes the image?
Diagram not to scale
Show mark scheme
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Mark
Notes
The object is closer to the lens than the principal focus (4.0 cm is less than 10 cm), so the refracted rays diverge and appear to come from a point behind the object. The image is virtual, upright and magnified, as in a magnifying glass. The two real-image rows describe objects placed beyond the focal length, and a virtual image of a nearby object cannot be smaller than the object.
✓ 1
Answer D
Answer: D
Exam tip: Compare the object distance with the focal length first: inside the focal length always gives a virtual, upright, magnified image.
217IGPC3-3-09r42-12Light, refraction, lenses and dispersionPaper CMedium6 marks
A pharmacist uses a converging lens as a magnifying glass to read the batch number on a small tablet pack.
(a)
State where the pack must be placed relative to the lens for it to act as a magnifying glass.
(1)
(b)
State two properties of the image formed.
(2)
(c)
The batch number has height 1.5 mm. Its image has height 7.5 mm. Calculate the magnification.
(2)
(d)
Explain why this image cannot be projected onto a screen.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
between the lens and its principal focus / closer to the lens than the focal length
✓ 1
Part (b)
any two from: virtual / upright / magnified
✓ 2
Part (c)
magnification = image height ÷ object height
✓ 1
= 7.5 ÷ 1.5 = 5.0
✓ 1
Part (d)
image is virtual, so light rays do not actually meet at the image / rays only appear to come from it
✓ 1
Common error: Magnification has no unit because it is a ratio of two lengths.
218IGPC3-3-09r42-13Sound and ultrasoundPaper CMedium7 marks
A survey boat uses ultrasound to measure the depth of water in a reservoir. A pulse sent from the boat returns after 0.44 s. The speed of sound in water is 1500 m/s.
Diagram not to scale
(a)
Calculate the depth of the water.
(2)
(b)
State what is meant by ultrasound.
(1)
(c)
The ultrasound has a frequency of 50 kHz. Calculate its wavelength in water. Give your answer to 2 significant figures.
(2)
(d)
Explain why sound cannot travel through a vacuum.
(2)
Show mark scheme
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Notes
Part (a)
distance travelled = 1500 × 0.44 = 660 m
✓ 1
depth = 660 ÷ 2 = 330 m
✓ 1
Part (b)
sound with a frequency above 20 000 Hz / above the range of human hearing
✓ 1
Part (c)
λ = v ÷ f
✓ 1
= 1500 ÷ 50 000 = 0.030 m
✓ 1
Part (d)
sound is a longitudinal wave passed on by particles vibrating (compressions and rarefactions)
✓ 1
a vacuum contains no particles to pass on the vibration
✓ 1
Exam tip: Convert kilohertz to hertz before substituting into v = f × λ.
219IGPC3-3-08r43-03Sound and ultrasoundPaper CMedium1 mark
A librarian taps a bell and the sound travels through the air of the reading room. Which statement about this sound wave is correct?
Show mark scheme
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Mark
Notes
Sound is a longitudinal wave: the air particles vibrate back and forth along the direction of travel, forming compressions and rarefactions. Vibration at right angles describes a transverse wave. Sound needs a medium, so it cannot travel through a vacuum. Sound travels faster in solids such as steel than in air.
✓ 1
Answer D
Answer: D
Common error: Sound is longitudinal and needs a medium; do not describe it as transverse.
220IGPC3-3-08r43-04Sound and ultrasoundPaper CMedium1 mark
A depth sensor fixed at the surface of a swimming pool sends out a pulse of ultrasound. The echo from the floor returns 0.0040 s later. The speed of sound in water is 1500 m/s. What is the depth of the water?
Diagram not to scale
Show mark scheme
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Mark
Notes
The pulse travels to the floor and back, so the total distance is 1500 × 0.0040 = 6.0 m and the depth is half of this, 3.0 m. The value 6.0 m forgets to halve. The value 0.66 m uses the speed of sound in air (330 m/s) instead of water. The value 12 m doubles the distance instead of halving it.
✓ 1
Answer B
Answer: B
Common error: An echo covers the distance twice; halve the total distance travelled to find how far away the surface is.
221IGPC3-3-08r43-06Light, refraction, lenses and dispersionPaper CMedium1 mark
A librarian uses a converging lens as a magnifying glass to read small print. Where must the book be placed relative to the lens?
Show mark scheme
Marking point
Mark
Notes
A converging lens gives an upright, magnified, virtual image only when the object is closer to the lens than the principal focus. Exactly twice the focal length gives a real, inverted image the same size as the object. Beyond twice the focal length gives a real, inverted, smaller image. Between one and two focal lengths gives a real, inverted, magnified image which is not what a magnifying glass produces.
✓ 1
Answer C
Answer: C
Exam tip: A magnifying glass works with the object inside the focal length, giving a virtual, upright, enlarged image.
222IGPC3-3-08r43-10Sound and ultrasoundPaper CMedium7 marks
A library fits ultrasonic sensors above its doors to count visitors. Each sensor sends out sound of frequency 40 kHz. Use 340 m/s for the speed of sound in air.
Diagram not to scale
(a)
State the lowest frequency of sound that counts as ultrasound.
(1)
(b)
Calculate the wavelength of the sound from the sensor.
(2)
(c)
Suggest why ultrasound is better than audible sound for this sensor.
(1)
(d)
A visitor stands 1.7 m directly below a sensor. Calculate the time between a pulse leaving the sensor and its echo returning.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
20 000 Hz (20 kHz)
✓ 1
Part (b)
M1 wavelength = speed ÷ frequency = 340 ÷ 40 000
✓ 1
A1 0.0085 m
✓ 1
Part (c)
it cannot be heard by people, so readers are not disturbed
✓ 1
Part (d)
M1 total distance = 2 × 1.7 = 3.4 m
✓ 1
M1 time = distance ÷ speed = 3.4 ÷ 340
✓ 1
A1 0.010 s
✓ 1
Common error: Remember to convert kHz to Hz, and that an echo travels to the object and back.
223IGPC3-3-08r43-11Light, refraction, lenses and dispersionPaper CMedium6 marks
A librarian uses a converging lens as a magnifying glass. The lens has a focal length of 10 cm and the book is held 6 cm from the lens.
Diagram not to scale
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
Describe the image formed.
(3)
(c)
Explain why this image cannot be projected onto a screen.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the point where rays parallel to the principal axis meet (converge) after passing through the lens
✓ 1
Part (b)
virtual
✓ 1
upright
✓ 1
magnified / larger than the object
✓ 1
Part (c)
the rays leaving the lens spread out (diverge) and do not meet
✓ 1
a virtual image is formed only where the rays appear to come from, so no light actually reaches that position
✓ 1
Exam tip: Use the words real or virtual, upright or inverted, and magnified or diminished when describing any image.
224IGPC3-3-08r43-14The electromagnetic spectrumPaper CMedium5 marks
The control centre of a wind farm receives data from turbines using radio waves of frequency 150 MHz. Use 3.0 × 108 m/s for the speed of electromagnetic waves in air.
(a)
A technician finds an overheating bearing with a thermal imaging camera. State the type of radiation the camera detects.
(1)
(b)
Calculate the wavelength of the radio waves.
(2)
(c)
State one harmful effect on people of too much exposure to ultraviolet radiation.
(1)
(d)
Radio waves and gamma rays travel at the same speed in a vacuum. State one way in which gamma rays differ from radio waves.
gamma rays have a shorter wavelength OR higher frequency OR more energy / are ionising
✓ 1
Common error: Convert MHz to Hz (1 MHz = 10⁶ Hz) before using v = fλ.
225IGPC3-3-10r44-03Sound and ultrasoundPaper CMedium1 mark
A cyclist stops in a valley and shouts towards a cliff. She hears the echo 1.2 s later. The speed of sound in air is 340 m/s. How far away is the cliff?
Diagram not to scale
Show mark scheme
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Mark
Notes
The sound travels to the cliff and back, so the total distance is 340 × 1.2 = 408 m and the cliff is half of this, 204 m, away. 408 m forgets that the sound makes a round trip. 283 m comes from dividing speed by time. 102 m halves the distance twice.
✓ 1
Answer D
Answer: D
Common error: An echo covers the distance twice. Halve the total distance travelled by the sound.
226IGPC3-3-10r44-04Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
A light ray passes from air into a glass paperweight found at an excavation site. Which row describes the changes to the light as it enters the glass?
Show mark scheme
Marking point
Mark
Notes
Light slows down in glass. The frequency is set by the source and does not change at a boundary, so from v = fλ a smaller speed means a smaller wavelength. The option with frequency decreasing or increasing is wrong because frequency stays the same. The option with speed and wavelength increasing describes light leaving glass for air.
✓ 1
Answer B
Answer: B
Exam tip: At a boundary, frequency stays constant while speed and wavelength change together.
227IGPC3-3-10r44-11Sound and ultrasoundPaper CMedium7 marks
A team of archaeologists surveys a sunken harbour with sonar on a boat. A pulse of ultrasound is sent to the sea bed and the echo returns after 0.080 s. The speed of sound in sea water is 1500 m/s.
Diagram not to scale
(a)
Calculate the depth of the sea bed below the boat.
(3)
(b)
State what is meant by ultrasound and suggest one reason why it is used for the survey.
(2)
(c)
Describe how a sound wave travels through water.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total distance = speed × time = 1500 × 0.080 = 120 m
✓ 1
M1 depth is half of the total distance
✓ 1
A1 60 m
✓ 1
Part (b)
B1 sound of frequency above 20 000 Hz (above the range of human hearing)
✓ 1
B1 reason, for example it cannot be heard by divers, or it gives a narrow beam and shorter wavelength for more detail
✓ 1
Part (c)
B1 particles of the water vibrate back and forth parallel to the direction in which the wave travels (longitudinal wave)
✓ 1
B1 forming a series of compressions and rarefactions
✓ 1
Common error: Remember the pulse travels to the sea bed and back, so halve the distance found from speed × time.
228IGPC3-3-10r44-12Light, refraction, lenses and dispersionPaper CMedium6 marks
An archaeologist photographs a mosaic using a camera with a converging lens. The mosaic is more than twice the focal length from the lens. Later she uses a glass prism to demonstrate the behaviour of white light.
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
Describe the image of the mosaic formed on the camera sensor.
(2)
(c)
Describe and explain what happens when a narrow beam of white light passes through the prism.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 the point where rays parallel to the principal axis meet after passing through the lens
✓ 1
Part (b)
B1 real and inverted
✓ 1
B1 smaller than the object (diminished)
✓ 1
Part (c)
B1 the white light is split into a spectrum of colours (dispersion)
✓ 1
B1 red is refracted (deviated) least and violet the most
✓ 1
B1 because the different colours travel at different speeds in glass
✓ 1
Exam tip: Describe a lens image using three words: real or virtual, upright or inverted, and magnified or diminished.
229IGPC3-3-08r45-02Sound and ultrasoundPaper CMedium1 mark
A librarian claps once in a large empty hall. She hears the echo from the far wall 0.20 s later. The speed of sound in air is 340 m/s. How far away is the far wall?
Diagram not to scale
Show mark scheme
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Mark
Notes
The sound travels to the wall and back, so the distance to the wall is 340 × 0.20 ÷ 2 = 34 m. 68 m forgets that the journey is there and back, 17 m halves twice, and 1700 m divides the speed by the time.
✓ 1
Answer D
Answer: D
Common error: For an echo, the sound covers twice the distance to the reflecting surface.
230IGPC3-3-08r45-10Light, refraction, lenses and dispersionPaper CMedium7 marks
A reader stands in front of a plane mirror on a library wall.
Diagram not to scale
(a)
A ray of light hits the mirror with an angle of incidence of 35°. State the angle of reflection.
(1)
(b)
Calculate the angle between the incident ray and the reflected ray.
(2)
(c)
State two characteristics of the image formed by a plane mirror.
(2)
(d)
The reader stands 2.5 m from the mirror. Calculate the distance between the reader and her image.
(2)
Show mark scheme
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Mark
Notes
Part (a)
35°
✓ 1
Part (b)
35 + 35, or 2 × 35
✓ 1
70°
✓ 1
Part (c)
Any two from: virtual; same size as the object; upright; laterally inverted; same distance behind the mirror as the object is in front
✓ 2
Part (d)
image is 2.5 m behind the mirror
✓ 1
2 × 2.5 = 5.0 m
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the mirror surface.
231IGPC3-3-08r45-11The electromagnetic spectrumPaper CMedium6 marks
A wind farm control centre sends radio signals to the turbines. Radio waves are part of the electromagnetic spectrum.
(a)
State the speed of electromagnetic waves in a vacuum.
(1)
(b)
State one hazard of exposure to X-rays.
(1)
(c)
Put these four types of wave in order of increasing frequency: X-rays, microwaves, visible light, infrared.
(1)
(d)
A radio signal has a frequency of 150 MHz. Calculate its wavelength.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
3.0 × 108 m/s
✓ 1
Part (b)
damage to cells / cancer / mutation of cells
✓ 1
Part (c)
microwaves, infrared, visible light, X-rays
✓ 1
Part (d)
wavelength = speed ÷ frequency
✓ 1
150 MHz = 1.5 × 108 Hz
✓ 1
3.0 × 108 ÷ 1.5 × 108 = 2.0 m
✓ 1
Common error: Convert prefixes such as MHz to hertz before you substitute into v = f × λ.
232IGPC3-3-08r45-12Sound and ultrasoundPaper CMedium6 marks
At a swimming gala, the starter fires a starting pistol at one end of the pool.
(a)
State the range of frequencies of sound that a normal human ear can detect.
(1)
(b)
Describe how sound travels through the air from the pistol to a listener.
(3)
(c)
A timekeeper stands 85 m from the starter. The speed of sound in air is 340 m/s. Calculate the time for the sound to reach the timekeeper.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
20 Hz to 20 000 Hz (accept 20 kHz)
✓ 1
Part (b)
air particles vibrate backwards and forwards along the direction of travel of the wave
✓ 1
this produces compressions and rarefactions
✓ 1
energy is passed on from particle to particle, with no overall movement of the air
✓ 1
Part (c)
time = distance ÷ speed
✓ 1
85 ÷ 340 = 0.25 s
✓ 1
Common error: Sound is a longitudinal wave: the particles vibrate parallel to the direction of travel, not across it.
233IGPC3-3-08r45-16Wave properties, reflection, refraction and diffractionPaper CMedium7 marks
A student uses a ripple tank. Plane water waves of frequency 5.0 Hz travel at 0.20 m/s towards a gap in a barrier.
(a)
Calculate the wavelength of the waves.
(2)
(b)
State what is meant by diffraction.
(1)
(c)
The gap is first 0.040 m wide and then 0.20 m wide. Describe the waves after the barrier in each case.
(2)
(d)
Houses near a wind farm can receive radio waves of wavelength 300 m from behind a hill, but not waves of wavelength 0.03 m. Explain why.
(2)
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Notes
Part (a)
λ = v ÷ f
✓ 1
0.20 ÷ 5.0 = 0.040 m
✓ 1
Part (b)
spreading out of waves as they pass through a gap or around an obstacle
✓ 1
Part (c)
0.040 m gap: waves spread out strongly into the region behind the barrier
✓ 1
0.20 m gap: waves mostly continue straight on, with only slight spreading at the edges
✓ 1
Part (d)
diffraction is greater when the wavelength is similar to or larger than the size of the obstacle
✓ 1
the 300 m waves bend round the hill, but the 0.03 m waves are much smaller than the hill and do not
✓ 1
Exam tip: Describe diffraction by comparing the wavelength with the gap size; the effect is largest when they are similar.
234IGPC3-3-08r46-04Light, refraction, lenses and dispersionPaper CMedium1 mark
A converging lens of focal length 12 cm is used to form an image of a small model stadium placed 30 cm from the lens. Which row describes the image?
Diagram not to scale
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Notes
The object is more than twice the focal length (24 cm) from the lens, so the image is real, inverted and smaller than the object. A larger real image needs the object between one and two focal lengths away. A virtual upright magnified image needs the object closer than one focal length, as in a magnifying glass. A real image formed by a converging lens is never upright.
✓ 1
Answer D
Answer: D
Exam tip: Compare the object distance with the focal length and twice the focal length before choosing the image type.
235IGPC3-3-08r46-05Sound and ultrasoundPaper CMedium1 mark
A sprinter at the edge of a stadium claps once and hears an echo from a distant stand 0.50 s later. The speed of sound in air is 340 m/s. How far away is the stand?
Diagram not to scale
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Notes
The sound travels to the stand and back, a total distance of 340 × 0.50 = 170 m, so the stand is 170 ÷ 2 = 85 m away. Giving 170 m forgets that the sound travels there and back. 680 m results from dividing the speed by the time. 43 m halves the distance twice.
✓ 1
Answer C
Answer: C
Common error: An echo covers the distance twice, so halve the total distance travelled.
236IGPC3-3-08r46-08Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
Water waves of wavelength 2.0 m approach a harbour wall. There is a gap in the wall, and the waves spread out noticeably after passing through it. Which gap width produces the greatest diffraction?
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Notes
Diffraction is most pronounced when the gap width is about the same as the wavelength, so 2.0 m gives the greatest spreading. Wider gaps, 8.0 m, 20 m and 50 m, allow the waves to pass almost straight through with little spreading at the edges. Wider gaps do not produce more spreading just because more water passes through.
✓ 1
Answer B
Answer: B
Exam tip: Compare the gap width with the wavelength: the closer they are in size, the more the wave spreads.
237IGPC3-3-08r46-12Light, refraction, lenses and dispersionPaper CMedium8 marks
A student uses a converging lens in a camera model to photograph a toy train.
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
The student uses the lens as a magnifying glass. Describe the image when the object is placed closer to the lens than the principal focus.
(2)
(c)
The toy train is 9.6 cm tall and its image on the screen is 2.4 cm tall. Calculate the magnification.
(2)
(d)
Describe the image formed on the screen in (c).
(2)
(e)
State what happens to the size of the image on the screen as the object is moved closer to the principal focus.
(1)
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Notes
Part (a)
the point where rays parallel to the principal axis meet (converge) after passing through the lens
Common error: Powers of ten must be entered carefully on a calculator; use the exponent key rather than typing × 10.
239IGPC3-3-09r46-03Sound and ultrasoundPaper CMedium1 mark
A sound pulse from a horn at a wind farm reflects from the tower of a distant turbine. The tower is 255 m from the horn and the echo is heard 1.5 s after the pulse is sent. What is the speed of sound in the air?
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Notes
The pulse travels to the tower and back, so the distance is 2 × 255 = 510 m and speed = 510 ÷ 1.5 = 340 m/s. 170 m/s uses only one way. 380 m/s multiplies 255 by 1.5. 510 m/s forgets to divide by the time.
✓ 1
Answer B
Answer: B
Common error: In echo questions the sound covers the distance twice, so use twice the separation as the distance travelled.
240IGPC3-3-09r46-08Sound and ultrasoundPaper CMedium1 mark
An ultrasound pulse is sent down from a boat beside an offshore wind turbine. The pulse returns from the seabed after 0.040 s. The speed of ultrasound in sea water is 1500 m/s. What is the depth of the water?
Diagram not to scale
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Notes
The pulse travels down and back, so depth = (1500 × 0.040) ÷ 2 = 30 m. 60 m forgets to halve. 37 500 m divides speed by time. 2.7 × 10−5 m divides time by speed.
✓ 1
Answer A
Answer: A
Common error: Halve the total distance for any reflection question: the signal travels to the surface and back.
241IGPC3-3-09r46-10Sound and ultrasoundPaper CMedium7 marks
A sensor in the ceiling of a library counts visitors. It sends a pulse of ultrasound of frequency 40 kHz downwards and detects the pulse reflected from the head of a child. The sensor is 2.60 m above the floor. The speed of sound in air is 340 m/s.
Diagram not to scale
(a)
Explain why a sound of frequency 40 kHz is described as ultrasound.
(1)
(b)
Calculate the wavelength of the ultrasound.
(2)
(c)
The pulse returns to the sensor 9.0 ms after it was sent. Calculate the distance from the sensor to the child's head.
(3)
(d)
Calculate the height of the child's head above the floor.
(1)
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Notes
Part (a)
B1 its frequency is above 20 000 Hz, the upper limit of human hearing
✓ 1
Part (b)
M1 wavelength = speed ÷ frequency
✓ 1
A1 340 ÷ 40 000 = 0.0085 m
✓ 1
Part (c)
M1 total distance = 340 × 0.0090 = 3.06 m
✓ 1
M1 halve for one-way distance
✓ 1
A1 1.53 m
✓ 1
Part (d)
A1 2.60 − 1.53 = 1.07 m (ECF from (c))
✓ 1
Common error: Convert milliseconds to seconds (divide by 1000) before using distance = speed × time.
242IGPC3-3-09r46-14Light, refraction, lenses and dispersionPaper CMedium7 marks
A narrow beam of white light passes through a glass prism and a coloured band of light forms on a screen.
(a)
Name the effect that separates the white light into colours.
(1)
(b)
Describe the order of the colours in the band and state which colour is bent the most.
(2)
(c)
Explain why the different colours are bent by different amounts.
(2)
(d)
Green light has frequency 5.0 × 1014 Hz. Calculate its wavelength in air. The speed of light in air is 3.0 × 108 m/s.
(2)
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Notes
Part (a)
B1 dispersion
✓ 1
Part (b)
B1 red at one end through to violet at the other end
✓ 1
B1 violet is deviated the most
✓ 1
Part (c)
B1 each colour has a different frequency/wavelength and so travels at a different speed in glass
✓ 1
B1 so each colour is refracted by a different amount
✓ 1
Part (d)
M1 wavelength = speed ÷ frequency
✓ 1
A1 6.0 × 10−7 m
✓ 1
Common error: Do not say the colours 'have different refractive indices' without linking this to a different speed in the glass.
243IGPC3-3-09r48-02Wave properties, reflection, refraction and diffractionPaper CMedium1 mark
Water waves of wavelength 2.0 cm travel towards a barrier that has a gap in it. Which gap width makes the waves spread out the most as they pass through?
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Notes
Diffraction is most noticeable when the gap is about the same size as, or smaller than, the wavelength, so a gap of 1.5 cm gives strong spreading. The 4.0 cm, 10 cm and 40 cm gaps are progressively wider than the wavelength, so more of the wavefront passes straight through and the spreading is less. A wider gap does not produce more diffraction.
✓ 1
Answer C
Answer: C
Common error: Diffraction depends on gap width compared with wavelength, not on the gap width alone.
244IGPC3-3-09r48-04Sound and ultrasoundPaper CMedium1 mark
A research ship sends an ultrasound pulse vertically downwards. The echo from the seabed returns after 0.80 s. The speed of sound in sea water is 1500 m/s. What is the depth of the sea at this point?
Diagram not to scale
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Notes
The pulse travels down and back, so the one-way time is 0.40 s and depth = 1500 × 0.40 = 600 m. 1200 m forgets that the time covers a round trip. 2400 m doubles the distance instead of halving it. 1875 m comes from dividing speed by time.
✓ 1
Answer D
Answer: D
Common error: An echo time always covers the journey there and back, so halve the distance travelled by the pulse.
245IGPC3-3-09r48-06Light, refraction, lenses and dispersionPaper CMedium1 mark
An object is placed 30 cm in front of a converging lens of focal length 10 cm. Which description of the image is correct?
Diagram not to scale
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Notes
The object is more than twice the focal length from the lens, so the image is real, inverted and smaller than the object. Virtual, upright and magnified is the result when the object is closer than the focal length. A real image formed by a converging lens is always inverted, so real and upright is impossible. Real, inverted and magnified occurs when the object is between one and two focal lengths from the lens.
✓ 1
Answer B
Answer: B
Exam tip: Compare the object distance with f and 2f before deciding what kind of image forms.
246IGPC3-3-09r48-12Sound and ultrasoundPaper CMedium7 marks
A dolphin finds fish by emitting very short clicks of sound and listening for the echoes. The clicks have a frequency of 1.0 × 105 Hz. The speed of sound in sea water is 1500 m/s.
Diagram not to scale
(a)
State the range of frequencies that a healthy young human can hear and explain why a human cannot hear the dolphin's clicks.
(2)
(b)
An echo from a fish returns to the dolphin 0.040 s after the click. Calculate the distance from the dolphin to the fish.
(3)
(c)
Calculate the wavelength of the clicks in sea water.
(2)
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Notes
Part (a)
20 Hz to 20 000 Hz
✓ 1
the clicks (100 000 Hz) have a frequency above the upper limit of human hearing / are ultrasound
✓ 1
Part (b)
M1 distance = speed × time
✓ 1
M1 divide by 2 because the sound travels to the fish and back
✓ 1
A1 1500 × 0.040 ÷ 2 = 30 m
✓ 1
Part (c)
M1 wavelength = speed ÷ frequency
✓ 1
A1 1500 ÷ 1.0 × 105 = 0.015 m
✓ 1
Common error: Convert frequency to hertz before using v = f × λ; 100 kHz is 1.0 × 10⁵ Hz.
247IGPC3-3-09r48-14Wave properties, reflection, refraction and diffractionPaper CMedium6 marks
A harbour has a breakwater wall with a gap in it, which is the harbour entrance. Water waves of wavelength 25 m and speed 5.0 m/s approach the wall. The gap is 20 m wide.
Diagram not to scale
(a)
State what is meant by diffraction.
(1)
(b)
Describe what happens to the waves after they pass through the gap.
(2)
(c)
The gap is made narrower. State the effect on the amount of diffraction.
(1)
(d)
Calculate the frequency of the waves.
(2)
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Notes
Part (a)
the spreading out of waves as they pass through a gap or around an edge
✓ 1
Part (b)
waves spread out into the region behind the wall
✓ 1
wavelength (and speed) of the waves is unchanged
✓ 1
Part (c)
more diffraction / waves spread out more
✓ 1
Part (d)
M1 frequency = speed ÷ wavelength
✓ 1
A1 5.0 ÷ 25 = 0.20 Hz
✓ 1
Exam tip: In diffraction the wavelength and frequency do not change; only the direction in which the waves travel changes.
248IGPC-037-03-10Sound and ultrasoundPaper CHard1 mark
A signal generator and loudspeaker produce a sound in a room. The frequency is then doubled, and the temperature of the air stays the same. What happens to the speed and the wavelength of the sound?
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Notes
The speed of sound depends on the medium, and the air has not changed, so the speed is constant. From v = fλ, if v is constant and f doubles, then λ must halve.
✓ 1
Answer B
Answer: B
Common error: Changing the source does not change the wave speed. The medium sets the speed, so f and λ change in opposite directions.
249IGPC-037-02-10The electromagnetic spectrumPaper CHard1 mark
Two electromagnetic waves, P and Q, travel through a vacuum. The frequency of P is twice the frequency of Q. Which statement is correct?
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Notes
All electromagnetic waves have the same speed in a vacuum. Since v = fλ with v constant, doubling f halves λ. Each wrong option either changes the speed or makes wavelength increase with frequency.
✓ 1
Answer B
Answer: B
Common error: In a vacuum the speed is fixed, so frequency and wavelength are inversely proportional. If one doubles, the other halves.
250IGPC3-3-03c-16Sound and ultrasoundPaper CHard9 marks
A student stands facing a tall flat wall 102 m away and claps repeatedly. She adjusts her clapping until each clap coincides with the echo of the previous clap. She times 25 intervals between claps and finds that they take 15.0 s in total.
Diagram not to scale
(a)
Calculate the time between successive claps.
(2)
(b)
Calculate the speed of sound in air from these results.
(3)
(c)
Suggest why the student times 25 intervals rather than one interval.
(2)
(d)
The experiment is repeated on a much hotter day. State and explain the effect on the time between claps.
(2)
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Part (a)
time = 15.0 / 25
✓ 1
time = 0.60 s
✓ 1
Part (b)
distance travelled by the sound between claps = 2 × 102 = 204 m
✓ 1
speed = distance / time = 204 / 0.60, ECF from (a)
✓ 1
speed = 340 m/s
✓ 1
Part (c)
the delay in starting and stopping the stopwatch is the same whichever number of intervals is timed
✓ 1
so it is a much smaller fraction of the longer time, giving a smaller percentage uncertainty
✓ 1
Part (d)
the time between claps decreases
✓ 1
sound travels faster in warmer air so the echo returns sooner
✓ 1
Common error: Sound travels to the wall and back between claps, so the distance used in speed = distance / time is twice the wall distance.
251IGPC3-3-06r14-08Wave properties, reflection, refraction and diffractionPaper CHard1 mark
Water waves of speed 0.80 m/s and frequency 2.0 Hz pass through a gap of width 0.40 m in a harbour wall. The frequency of the waves is now changed to 4.0 Hz and the speed stays the same. How does the diffraction of the waves at the gap change?
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Notes
Wavelength = v ÷ f, so it falls from 0.40 m to 0.20 m. The gap is now twice the wavelength instead of equal to it, so diffraction is less pronounced (B). A is wrong because energy does not determine the amount of spreading. C ignores that the wavelength has changed. D is wrong because diffraction still happens but is less noticeable.
✓ 1
Answer B
Answer: B
Common error: The amount of diffraction depends on the gap width compared with the wavelength: longer wavelength or narrower gap gives more spreading.
252IGPC3-3-06r19-15Wave properties, reflection, refraction and diffractionPaper CHard8 marks
The entrance to a harbour is a gap in a sea wall 25 m wide. Water waves of period 8.0 s and wavelength 20 m move towards the gap in deep water.
Diagram not to scale
(a)
Describe what happens to the waves as they pass through the gap.
(2)
(b)
Calculate the speed of the waves in deep water.
(2)
(c)
In shallow water near the shore the waves slow to 1.6 m/s and the frequency does not change. Calculate the wavelength of the waves in the shallow water.
(2)
(d)
The gap is widened to 100 m. State and explain how the diffraction of the 20 m waves changes.
(2)
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Notes
Part (a)
the waves spread out
✓ 1
into the region behind the wall / beyond the gap (diffraction)
✓ 1
Part (b)
frequency = 1 ÷ 8.0 = 0.125 Hz
✓ 1
speed = 0.125 × 20 = 2.5 m/s
✓ 1
Part (c)
wavelength = speed ÷ frequency = 1.6 ÷ 0.125
✓ 1
12.8 m
✓ 1
Part (d)
there is less diffraction / the waves spread out less
✓ 1
because the gap is now much wider than the wavelength (diffraction is greatest when the gap is about one wavelength)
✓ 1
Common error: Saying the wavelength stays the same when a wave slows down; the frequency stays the same, so the wavelength decreases with the speed.
253IGPC3-3-08r23-07Wave properties, reflection, refraction and diffractionPaper CHard1 mark
A ripple tank models a wave-power site. Water waves move from deep to shallow water. In the deep water the speed is 0.60 m/s and the wavelength is 0.30 m. In the shallow water the speed is 0.40 m/s. What is the wavelength in the shallow water?
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Notes
The frequency stays the same when waves cross a boundary. f = 0.60 ÷ 0.30 = 2.0 Hz, so λ = 0.40 ÷ 2.0 = 0.20 m. The value 0.45 m comes from scaling the wavelength by the speed ratio the wrong way round. The value 0.12 m comes from multiplying 0.30 by 0.40. The value 0.30 m assumes the wavelength does not change.
✓ 1
Answer C
Answer: C
Common error: When a wave changes speed at a boundary, frequency stays constant and the wavelength changes in proportion to the speed.
254IGPC3-3-08r40-14Sound and ultrasoundPaper CHard7 marks
Ultrasound pulses are used to check a wind-turbine blade for internal cracks. The speed of ultrasound in the blade material is 2400 m/s. A pulse is sent in from the outer surface. An echo from a crack returns after 3.2 × 10−5 s and an echo from the inner surface of the blade returns after 5.6 × 10−5 s.
Diagram not to scale
(a)
Calculate the depth of the crack below the outer surface.
(2)
(b)
Calculate the thickness of the blade.
(2)
(c)
Calculate the distance between the crack and the inner surface. Use the unrounded values from (a) and (b).
(1)
(d)
Explain why ultrasound rather than audible sound is used for this test.
(1)
(e)
The ultrasound has a frequency of 2.0 MHz (2.0 × 106 Hz). Calculate its wavelength in the blade material.
ultrasound has a shorter wavelength so it diffracts less and forms a narrow beam that can find small cracks / it cannot be heard by workers
✓ 1
Part (e)
wavelength = 2400 ÷ 2.0 × 106 = 1.2 × 10−3 m
✓ 1
Common error: The echo time covers the journey to the reflector and back, so the depth is half of speed × time.
255IGPC3-3-10r44-14Wave properties, reflection, refraction and diffractionPaper CHard8 marks
Water waves travel from a deep region into a shallow region of a tank. In the deep region the speed is 0.60 m/s and the wavelength is 0.12 m. In the shallow region the speed is 0.40 m/s.
(a)
Calculate the frequency of the waves in the deep region.
(2)
(b)
State the frequency of the waves in the shallow region and give a reason.
(2)
(c)
Calculate the wavelength in the shallow region.
(2)
(d)
The wavefronts meet the boundary at an angle. Explain why their direction changes.
(2)
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Notes
Part (a)
M1 f = v ÷ λ = 0.60 ÷ 0.12
✓ 1
A1 5.0 Hz
✓ 1
Part (b)
B1 5.0 Hz (ECF from (a))
✓ 1
B1 the frequency is set by the source, so the number of waves arriving each second does not change
✓ 1
Part (c)
M1 λ = v ÷ f = 0.40 ÷ 5.0 (ECF from (b))
✓ 1
A1 0.080 m
✓ 1
Part (d)
B1 the part of each wavefront that reaches the shallow region first slows down first, while the rest is still moving faster
✓ 1
B1 so the wavefront pivots and changes direction
✓ 1
Common error: At a boundary, frequency stays the same. It is the speed and wavelength that change.
256IGPC3-3-10r44-16Wave properties, reflection, refraction and diffractionPaper CHard8 marks
Archaeologists use ground-penetrating radar. A pulse of radio waves of frequency 100 MHz travels in air at 3.0 × 108 m/s, then enters soil where its speed is 1.0 × 108 m/s. The pulse is reflected by a buried wall and the echo returns to the surface 40 ns after the pulse was sent. 1 ns = 1.0 × 10−9 s.
(a)
Calculate the wavelength of the waves in air.
(2)
(b)
State the frequency of the waves in the soil.
(1)
(c)
Calculate the wavelength of the waves in the soil. Use your answer to (b).
(2)
(d)
Calculate the depth of the buried wall. Assume all of the travel time is in the soil.
(3)
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Notes
Part (a)
M1 λ = v ÷ f = 3.0 × 108 ÷ 1.0 × 108
✓ 1
A1 3.0 m
✓ 1
Part (b)
B1 100 MHz (unchanged)
✓ 1
Part (c)
M1 λ = v ÷ f = 1.0 × 108 ÷ 1.0 × 108
✓ 1
A1 1.0 m
✓ 1
Part (d)
M1 total distance = speed × time = 1.0 × 108 × 40 × 10−9 = 4.0 m
✓ 1
M1 depth is half of the total distance
✓ 1
A1 2.0 m
✓ 1
Exam tip: Convert ns and MHz to seconds and hertz first, and halve the distance for a reflected pulse.
257IGPC3-3-09r46-06Wave properties, reflection, refraction and diffractionPaper CHard1 mark
Water waves of frequency 2.0 Hz travel at 0.80 m/s in the deep end of a swimming pool and slow to 0.50 m/s over the shallow end. What is the wavelength of the waves in the shallow end?
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Notes
The frequency is fixed by the source and does not change, so wavelength = speed ÷ frequency = 0.50 ÷ 2.0 = 0.25 m. 0.40 m is the deep-end wavelength, wrongly assumed to be unchanged. 0.80 m copies the deep-end speed. 4.0 m divides frequency by speed.
✓ 1
Answer D
Answer: D
Common error: When waves cross a boundary the frequency stays the same; speed and wavelength change together.
258IGPC3-3-09r46-16Wave properties, reflection, refraction and diffractionPaper CHard8 marks
Water waves of wavelength 0.30 m and frequency 1.5 Hz travel along a swimming pool towards a barrier with a narrow gap.
(a)
State what is meant by diffraction.
(1)
(b)
Describe how the gap width affects the amount of diffraction.
(2)
(c)
Calculate the speed of the waves.
(2)
(d)
The waves then move into shallower water where their speed is 0.30 m/s. State what happens to the frequency and calculate the new wavelength.
(3)
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Notes
Part (a)
B1 spreading out of waves as they pass through a gap or past an edge
✓ 1
Part (b)
B1 diffraction is greatest when the gap is about the same size as the wavelength (or narrower)
✓ 1
B1 little diffraction when the gap is much wider than the wavelength
✓ 1
Part (c)
M1 speed = frequency × wavelength
✓ 1
A1 1.5 × 0.30 = 0.45 m/s
✓ 1
Part (d)
B1 frequency is unchanged at 1.5 Hz
✓ 1
M1 wavelength = speed ÷ frequency
✓ 1
A1 0.30 ÷ 1.5 = 0.20 m
✓ 1
Common error: Frequency stays the same when waves change speed; it is the wavelength that changes.
259IGPC3-3-09r48-16Sound and ultrasoundPaper CHard8 marks
A reversing sensor on a delivery truck sends out a short pulse of ultrasound of frequency 40 kHz and detects the echo from an obstacle. Take the speed of sound in air as 340 m/s.
Diagram not to scale
(a)
Suggest why ultrasound rather than audible sound is used by the sensor.
(1)
(b)
The echo returns after 14.0 ms. Calculate the distance from the sensor to the obstacle.
(3)
(c)
On a hot day the speed of sound in air is greater than 340 m/s, but the sensor still uses 340 m/s in its calculation. State whether the sensor will over-estimate or under-estimate the distance and explain your answer.
(2)
(d)
Calculate the wavelength of the ultrasound in air at 340 m/s.
(2)
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Notes
Part (a)
it cannot be heard by people (it is above 20 000 Hz), so it does not disturb them
✓ 1
Part (b)
M1 convert 14.0 ms to 0.0140 s
✓ 1
M1 distance = speed × time ÷ 2
✓ 1
A1 340 × 0.0140 ÷ 2 = 2.38 m
✓ 1
Part (c)
under-estimates the distance
✓ 1
the sound actually travels faster so goes further in the measured time than the sensor assumes
✓ 1
Part (d)
M1 wavelength = speed ÷ frequency, with frequency 40 000 Hz
✓ 1
A1 340 ÷ 40 000 = 0.0085 m
✓ 1
Exam tip: Convert kilohertz and milliseconds to hertz and seconds before substituting into any equation.
260IGPC-037-03-13Sound and ultrasoundPaper EEasy5 marks
Hospitals use ultrasound scanners to produce images of structures inside the body.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
State one medical use of ultrasound scanning.
(1)
(c)
Explain how ultrasound pulses can be used to find how deep a boundary between two tissues lies below the skin.
(2)
(d)
Suggest why ultrasound, rather than X-rays, is used to scan an unborn baby.
(1)
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Notes
Part (a)
Sound with a frequency higher than 20 kHz (20 000 Hz), OR sound with a frequency above the upper limit of human hearing .
✓ 1
Part (b)
Any one: checking a developing foetus during pregnancy; imaging soft tissue or organs such as the heart, liver or kidneys .
✓ 1
Part (c)
Pulses are sent into the body and are partly reflected at the boundary between tissues . The time for the echo to return is measured and, with the speed of sound in the tissue, gives the depth using depth = speed × time ÷ 2 .
✓ 2
Part (d)
Ultrasound is not ionising, so it does not damage cells or harm the baby .
✓ 1
Exam tip: For any ultrasound use, the same idea earns the marks: a pulse is sent, it reflects at a boundary, and the echo time gives a distance. Remember to halve the total time, because the pulse travels there and back.
261IGPC-037-03-06Sound and ultrasoundPaper EMedium1 mark
The same sound travels through air, water and steel. Which list puts the speed of sound in order from fastest to slowest?
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In general sound travels fastest in solids, slower in liquids and slowest in gases. Particles are closer together and more strongly linked in solids, so vibrations are passed on more quickly. The order is steel, then water, then air.
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Answer C
Answer: C
Common error: Do not think that denser materials 'block' sound and slow it down. Sound is faster in solids than in liquids, and faster in liquids than in gases.
262IGPC-037-03-07Sound and ultrasoundPaper EMedium1 mark
A sound wave travels through air. Which statement describes a rarefaction?
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In a longitudinal sound wave, the particles vibrate backwards and forwards along the direction of travel. In a rarefaction the particles are spread out, so the pressure is lower than normal. The second option describes a compression, and the third describes the wavelength. The particles do not travel with the wave.
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Answer A
Answer: A
Exam tip: Compression means particles are close together (higher pressure). Rarefaction means particles are spread out (lower pressure). The particles only vibrate about fixed positions.
263IGPC-037-03-08Sound and ultrasoundPaper EMedium1 mark
Which statement about ultrasound is correct?
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Ultrasound is sound with a frequency above the upper limit of human hearing, 20 kHz. It is a longitudinal mechanical wave, not an electromagnetic wave. Sound below 20 Hz is infrasound, and loudness depends on amplitude, not frequency.
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Answer D
Answer: D
Common error: Ultrasound is defined by its frequency (above 20 kHz), not by its loudness. It is still a sound wave and still needs a medium.
264IGPC-037-01-04Light, refraction, lenses and dispersionPaper EMedium1 mark
A ray of light passes from air into a transparent plastic. The angle of incidence is 50° and the angle of refraction is 30°. What is the refractive index of the plastic?
Diagram not to scale
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n = sin i / sin r = sin 50° / sin 30° = 0.766 / 0.500 = 1.53. The value 1.67 comes from dividing the angles themselves (50 ÷ 30) instead of their sines. The value 0.65 is sin r / sin i, which is upside down. The value 1.35 uses cosines instead of sines. A refractive index from air into a solid is always greater than 1.
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Answer C
Answer: C
Exam tip: Check that your calculator is in degree mode. Then check that your refractive index is greater than 1; if it is not, you have probably inverted the ratio.
265IGPC-037-01-05Light, refraction, lenses and dispersionPaper EMedium1 mark
A type of glass has a refractive index of 1.60. What is the critical angle for light travelling from this glass into air?
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sin c = 1/n = 1/1.60 = 0.625, so c = sin−1(0.625) = 38.7° ≈ 39°. 51° is the complement (90° − 39°). 32° comes from tan−1(1/1.60), which uses the wrong trig function. 58° comes from tan−1(1.60).
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Answer B
Answer: B
Exam tip: Use n = 1/sin c, so c = sin⁻¹(1/n). A denser material has a larger n and therefore a smaller critical angle.
266IGPC-037-01-09Light, refraction, lenses and dispersionPaper EMedium1 mark
A person is short-sighted and cannot see distant objects clearly. Where does the eye focus light from a distant object, and which lens corrects this?
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In a short-sighted eye, light from a distant object is focused in front of the retina because the eye converges the light too strongly. A diverging lens spreads the rays out a little before they enter the eye, which moves the focus back onto the retina. Focusing behind the retina describes long sight, which is corrected by a converging lens.
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Answer B
Answer: B
Common error: Short sight is corrected with a diverging lens and long sight with a converging lens. Learn where each eye defect focuses the light, not just the lens type.
267IGPC-037-02-06The electromagnetic spectrumPaper EMedium1 mark
A radio station broadcasts at a frequency of 95 MHz. What is the wavelength of the radio waves in air?
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Radio waves travel at 3.0 × 108 m/s in air, and 95 MHz = 95 × 106 Hz. λ = v/f = 3.0 × 108 / 95 × 106 = 3.2 m. 0.32 m and 32 m come from mistakes converting MHz into Hz. 2.9 × 1016 m comes from multiplying v by f instead of dividing.
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Answer B
Answer: B
Common error: Mega means 10⁶. Write 95 MHz as 95 × 10⁶ Hz before substituting into λ = v/f.
268IGPC-037-02-07The electromagnetic spectrumPaper EMedium1 mark
Mobile phones communicate with nearby masts using microwaves. Why are microwaves suitable?
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Microwaves penetrate building walls well enough to reach a phone indoors. Their short wavelength means only a short aerial is needed for sending and receiving. Strong diffraction around hills is a property of long-wavelength radio waves. Microwaves are not ionising. All electromagnetic waves travel at the same speed.
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Answer B
Answer: B
Exam tip: For mobile phones and wireless internet, give two reasons: microwaves pass through walls, and only a short aerial is needed.
269IGPC-037-02-08The electromagnetic spectrumPaper EMedium1 mark
High-speed broadband uses glass optical fibres carrying visible light or infrared. Why are these regions of the spectrum used?
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Glass absorbs very little visible light and infrared, so signals travel long distances along the fibre. Their high frequency allows a very high rate of data transfer. Light is guided round bends by total internal reflection, not diffraction. Light travels more slowly in glass than in air. Some absorption always occurs, so long links still need repeaters.
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Answer B
Answer: B
Common error: Light follows an optical fibre by total internal reflection at the glass surface. Diffraction plays no part in guiding it.
270IGPC-037-00-06Wave properties, reflection, refraction and diffractionPaper EMedium1 mark
Plane water waves of wavelength 2.0 cm travel towards a barrier that has a gap in it. Which gap width produces the greatest spreading of the waves?
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Diffraction is greatest when the gap width is similar to the wavelength. A 2.0 cm gap matches the 2.0 cm wavelength, so the waves spread out the most. Wider gaps let more of the wave through, but the wave spreads much less, mostly at the edges.
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Answer A
Answer: A
Common error: A wider gap does not give more diffraction. What matters is how the gap width compares with the wavelength.
271IGPC-037-00-16Wave properties, reflection, refraction and diffractionPaper EMedium6 marks
Sea waves of wavelength 12 m approach a harbour. The harbour wall has a single entrance gap 15 m wide.
Diagram not to scale
(a)
State what is meant by diffraction.
(1)
(b)
Describe the waves inside the harbour, just after they pass through the 15 m gap.
(2)
(c)
The entrance is widened to 120 m. Describe and explain how the waves inside the harbour change.
(2)
(d)
Long-wave radio signals can be received in a valley behind a hill, but light from the transmitter cannot reach it. Suggest why.
(1)
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Part (a)
The spreading out of a wave as it passes through a gap or around the edge of an obstacle.
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Part (b)
The waves spread out a lot / become curved and reach most of the harbour.
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The wavelength stays the same (12 m).
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Part (c)
There is much less spreading; the waves continue mostly straight ahead and spread only at the edges.
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This is because the gap is now much larger than the wavelength.
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Part (d)
Radio waves have a wavelength comparable to the size of the hill, so they diffract around it; the wavelength of light is far too small for significant diffraction.
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Exam tip: Say that the wavelength is unchanged by diffraction. Link the amount of spreading to how the gap size compares with the wavelength.
272IGPC-037-03-15Sound and ultrasoundPaper EMedium6 marks
Engineers use ultrasound to check steel bars for hidden cracks. A probe held against one end of a bar sends a short pulse along the bar and detects any echoes. The speed of ultrasound in steel is 5900 m/s.
Diagram not to scale
(a)
For an undamaged bar, the only echo comes from the far end of the bar. It arrives 0.12 ms after the pulse is sent. Calculate the length of the bar.
(3)
(b)
In a second bar, an extra echo arrives 0.040 ms after the pulse is sent. Calculate the distance from the probe to the crack.
(2)
(c)
Suggest one advantage of testing the bars with ultrasound rather than cutting them open.
(1)
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Part (a)
distance travelled = 5900 × 0.12 × 10−3 = 0.708 m
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The pulse travels the length of the bar twice, so divide by 2
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length = 0.354 m ≈ 0.35 m
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Part (b)
distance = 5900 × 0.040 × 10−3 ÷ 2
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= 0.118 m ≈ 0.12 m
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Part (c)
Any one valid advantage , e.g.:
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• the bar is not damaged, so it can still be used;
—
• cracks inside the bar can be found without destroying it;
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• testing is quicker or cheaper;
—
• every bar can be tested, not just samples, or bars can be tested in place.
—
Exam tip: Convert milliseconds to seconds (×10⁻³) before calculating. Remember that an echo travels to the reflector and back, so halve the total distance.
273IGPC-037-01-12Light, refraction, lenses and dispersionPaper EMedium6 marks
A ray of light travels from air into a swimming pool. At the water surface, the angle of incidence is 40° and the angle of refraction is 29°.
Diagram not to scale
(a)
Calculate the refractive index of the water.
(2)
(b)
Calculate the critical angle for light travelling from the water into air.
(2)
(c)
A lamp at the bottom of the pool sends a ray up towards the water surface. The ray strikes the surface from below at an angle of incidence of 55°. Explain why this ray does not leave the water.
(2)
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Part (a)
n = sin i / sin r = sin 40° / sin 29°
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n = 0.643 / 0.485 = 1.33
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Part (b)
sin c = 1/n = 1/1.326 = 0.754 (accept 1/1.33 = 0.752)
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c = 49°
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Part (c)
The ray is travelling from the denser medium (water) and 55° is greater than the critical angle of 49°
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so the ray is totally internally reflected back into the water
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Exam tip: Keep the unrounded value of n (1.326) in your calculator for part (b). Using a rounded value can change the final digit of an answer that is given to more significant figures.
274IGPC-037-00-14Wave properties, reflection, refraction and diffractionPaper EMedium6 marks
A ray of light in air enters a rectangular block of transparent plastic. The angle of incidence is 55° and the angle of refraction is 33°. The speed of light in air is 3.0 × 108 m/s.
Diagram not to scale
(a)
Explain why the ray bends towards the normal as it enters the block.
(1)
(b)
Calculate the refractive index of the plastic.
(2)
(c)
Refractive index can also be defined as a ratio of speeds. Use your answer to the previous part to calculate the speed of light in the plastic.
(2)
(d)
The ray leaves the block through the opposite parallel face. State the direction of the emerging ray compared with the original incident ray.
(1)
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Part (a)
Light travels more slowly in the plastic than in air (accept: the plastic is optically denser, so the light slows down).
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Part (b)
n = sin i ÷ sin r = sin 55° ÷ sin 33°
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n = 1.50 (accept 1.5)
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Part (c)
speed in plastic = speed in air ÷ n = 3.0 × 108 ÷ 1.50
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= 2.0 × 108 m/s
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Part (d)
Parallel to the incident ray (but displaced sideways).
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Exam tip: Keep your calculator in degree mode. Carry the unrounded value of n into the next part.
275IGPC3-3-03c-05Light, refraction, lenses and dispersionPaper EMedium1 mark
A watch repairer holds a converging lens of focal length 10 cm at a distance of 4 cm from a tiny gear. Which description applies to the image seen through the lens?
Diagram not to scale
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An object closer to a converging lens than its principal focus gives a virtual, upright, magnified image on the same side as the object (D). A is the image for an object between f and 2f. B: a diverging lens gives virtual diminished images but this is a converging lens. C does not occur for a single lens.
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Answer D
Answer: D
Exam tip: Compare the object distance with the focal length first; inside f the lens acts as a magnifying glass.
276IGPC3-3-03c-11Wave properties, reflection, refraction and diffractionPaper EMedium8 marks
A ray of light in air strikes a glass block of refractive index 1.52 at an angle of incidence of 40°.
Diagram not to scale
(a)
Calculate the angle of refraction in the glass.
(3)
(b)
The speed of light in air is 3.0 × 108 m/s. Calculate the speed of light in the glass.
(2)
(c)
Calculate the critical angle for light travelling from this glass into air.
(3)
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Part (a)
n = sin i / sin r
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sin r = sin 40° / 1.52 = 0.423
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r = 25.0°
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Part (b)
n = c / v
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v = 3.0 × 108 / 1.52 = 2.0 × 108 m/s
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Part (c)
sin c = 1 / n
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sin c = 1 / 1.52 = 0.658
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c = 41.1°
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Exam tip: Keep the unrounded value of sin r on the calculator before taking the inverse sine.
277IGPC3-3-03c-15Light, refraction, lenses and dispersionPaper EMedium6 marks
A student uses a converging lens of focal length 12 cm.
Diagram not to scale
(a)
Light from the Sun passes through the lens parallel to its principal axis. Name the point where the light converges.
(1)
(b)
State the distance between the lens and the point named in (a).
(1)
(c)
A candle is placed 30 cm from the lens. Describe the image formed.
(3)
(d)
State one device that uses a lens to form an image of the type described in (c).
(1)
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Part (a)
principal focus (focal point)
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Part (b)
12 cm
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Part (c)
real
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inverted
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diminished (smaller than the object)
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Part (d)
camera (or eye)
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Exam tip: An object further than twice the focal length always gives a real, inverted, smaller image.
278IGPC3-3-05r9-05Light, refraction, lenses and dispersionPaper EMedium1 mark
A ray of light in air hits the glass lens of a cycle lamp at an angle of incidence of 50° and is refracted at an angle of refraction of 30°. What is the refractive index of the glass?
Diagram not to scale
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Correct: D. n = sin i / sin r = sin 50° ÷ sin 30° = 0.766 ÷ 0.50 = 1.53, which is 1.53. A inverts the ratio (sin 30° ÷ sin 50°). B divides the angles (50 ÷ 30 = 1.67) instead of their sines. C (1.19) uses sin 40° as the denominator, wrongly using the angle between the refracted ray and the surface.
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Answer D
Answer: D
Common error: Refractive index uses the sines of the angles measured from the normal, not the angles themselves.
279IGPC3-3-05r9-14Light, refraction, lenses and dispersionPaper EMedium6 marks
Light travels along the glass core of an optical fibre in a telecoms cable. The refractive index of the core is 1.48.
(a)
State the two conditions needed for total internal reflection at the boundary between the core and the cladding.
(2)
(b)
Calculate the critical angle of the core. Give your answer to 3 significant figures.
(3)
(c)
A ray in the core meets the boundary at an angle of incidence of 38°. State what happens to the ray. Use your answer to (b); ECF from (b).
(1)
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Part (a)
light is travelling from the more dense material (core) towards the less dense material (cladding)
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angle of incidence is greater than the critical angle
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Part (b)
sin c = 1 ÷ n
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sin c = 1 ÷ 1.48 = 0.676
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c = 42.5°
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Part (c)
38° is less than the critical angle, so the ray is (partly) refracted out of the core into the cladding
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Common error: Use sin c = 1/n and then take the inverse sine; do not divide the angle by n.
280IGPC3-3-06r12-05Light, refraction, lenses and dispersionPaper EMedium1 mark
Light travels in a glass fibre of refractive index 1.60 used in an inspection tool for tablet coatings. What is the critical angle for the boundary between this glass and air?
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sin c = 1/n = 1/1.60 = 0.625, so c = 39°. 51° is 90° − 39°, from using cosine. 58° comes from tan c = 1.60. 56° comes from 90°/1.60, which is not a valid relationship.
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Answer C
Answer: C
Exam tip: Use sin c = 1/n and take the inverse sine of the answer.
281IGPC3-3-06r12-07Sound and ultrasoundPaper EMedium1 mark
An ultrasound pulse used in a pharmaceutical test passes from air into water. Which row describes what happens?
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The frequency is set by the source and does not change at a boundary. Sound travels faster in water than in air, and λ = v/f, so the wavelength increases: A. B wrongly changes the frequency. C has the speed decreasing, which is wrong for water. D wrongly changes the frequency.
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Answer A
Answer: A
Common error: Frequency never changes when a wave crosses a boundary; speed and wavelength do.
282IGPC3-3-06r12-13Light, refraction, lenses and dispersionPaper EMedium7 marks
A beam of light travels from air into a clear liquid medicine in a tank. The angle of incidence is 50° and the angle of refraction is 34°. The speed of light in air is 3.0 × 108 m/s.
Diagram not to scale
(a)
Calculate the refractive index of the liquid. Give your answer to 3 significant figures.
(3)
(b)
Calculate the speed of light in the liquid. Give your answer to 2 significant figures.
(2)
(c)
State what happens to the speed and to the direction of the light as it enters the liquid.
(2)
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Part (a)
n = sin i / sin r
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n = sin 50° / sin 34°
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n = 1.37
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Part (b)
n = c / v
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v = 3.0 × 108 / 1.37 = 2.2 × 108 m/s (ECF from (a))
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Part (c)
The light slows down
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It bends towards the normal
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Exam tip: Set your calculator to degrees before using sin i / sin r.
283IGPC3-3-06r14-05Light, refraction, lenses and dispersionPaper EMedium1 mark
Light travels inside a block of glass of refractive index 1.50 and reaches the boundary with air. Which angle of incidence at the boundary will cause total internal reflection?
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The critical angle is given by sin c = 1/1.50, so c = 41.8°. Total internal reflection needs the angle of incidence to be greater than c, so only 50° (D) qualifies. 25°, 32° and 40° are all below 41.8°, so some light is refracted out; 40° is close but still below the critical angle.
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Answer D
Answer: D
Exam tip: Total internal reflection needs two things: light going from a denser to a less dense medium and an angle of incidence greater than the critical angle.
284IGPC3-3-06r14-07Sound and ultrasoundPaper EMedium1 mark
A rail inspector sends a pulse of ultrasound into a steel rail. The speed of sound in the steel is 5000 m/s. An echo from a crack returns to the probe after 40 μs. How far below the probe is the crack?
Diagram not to scale
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The pulse travels to the crack and back, so distance to crack = v × t ÷ 2 = 5000 × 40 × 10−6 ÷ 2 = 0.10 m = 10 cm (C). D (20 cm) forgets to halve. B (5.0 cm) halves twice. A (1.0 cm) is a power-of-ten error converting microseconds.
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Answer C
Answer: C
Common error: An echo covers the distance twice, so halve the total distance travelled by the pulse.
285IGPC3-3-06r14-11Light, refraction, lenses and dispersionPaper EMedium5 marks
A signal lamp on a railway has a glass cover. In a test, a ray of light travels from air into the glass. The angle of incidence is 52° and the angle of refraction is 31°.
Diagram not to scale
(a)
Calculate the refractive index of the glass.
(2)
(b)
A second ray enters the glass with an angle of incidence of 30°. Use n = 1.53 to calculate its angle of refraction.
(2)
(c)
State what happens to the speed of light as it enters the glass from air.
(1)
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Part (a)
n = sin i ÷ sin r
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n = sin 52° ÷ sin 31° = 1.53
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Part (b)
sin r = sin 30° ÷ 1.53
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r = 19.1°
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Part (c)
it decreases
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Common error: Make sure your calculator is in degree mode, and remember that the ray bends towards the normal when it enters a denser medium.
286IGPC3-3-06r19-05Wave properties, reflection, refraction and diffractionPaper EMedium1 mark
Light travels at 3.0 × 108 m/s in air. An architect specifies a transparent roof panel in which the light speed is 2.0 × 108 m/s. What is the refractive index of the panel material?
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Refractive index n = speed in air ÷ speed in material = 3.0 × 108 ÷ 2.0 × 108 = 1.5, so C is correct. A inverts the ratio (2.0 ÷ 3.0). B subtracts the speeds (3.0 − 2.0). D adds them (3.0 + 2.0).
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Answer C
Answer: C
Exam tip: The refractive index of any transparent material is greater than 1, because light is slower there than in air; use this to check the ratio is the right way up.
287IGPC3-3-06r19-11Light, refraction, lenses and dispersionPaper EMedium6 marks
An architect is choosing glass for a window. In a test, a ray of light in air strikes a flat sheet of glass at an angle of incidence of 40°. The angle of refraction in the glass is 25°.
Diagram not to scale
(a)
Calculate the refractive index of the glass. Give your answer to 3 significant figures.
(2)
(b)
The speed of light in air is 3.0 × 108 m/s. Calculate the speed of light in the glass. Use n = 1.52 and give your answer to 3 significant figures.
(2)
(c)
A second ray strikes the glass at an angle of incidence of 60°. Calculate the angle of refraction. Use n = 1.52 and give your answer to 3 significant figures.
(2)
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Part (a)
n = sin i ÷ sin r = sin 40° ÷ sin 25°
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1.52
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Part (b)
speed in glass = c ÷ n = 3.0 × 108 ÷ 1.52
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1.97 × 108 m/s
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Part (c)
sin r = sin 60° ÷ 1.52 = 0.570
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r = 34.7°
✓ 1
Exam tip: Check that the refracted ray is bent towards the normal when light enters a denser medium: your angle of refraction must be smaller than the angle of incidence.
288IGPC3-3-06r19-14Light, refraction, lenses and dispersionPaper EMedium7 marks
Optical fibres are used with cameras to inspect pipes on a marine research vessel. The glass of the fibre core has a refractive index of 1.50, and a ray travels inside the core towards the boundary with air.
Diagram not to scale
(a)
Calculate the critical angle for the glass-air boundary. Give your answer to 3 significant figures.
(2)
(b)
A ray in the fibre hits the boundary at an angle of incidence of 60°. State and explain what happens to the ray.
(2)
(c)
The fibre is 2.4 km long. The speed of light in air is 3.0 × 108 m/s. Calculate the time taken by a pulse of light to travel along the fibre in a straight line.
(3)
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Part (a)
sin c = 1 ÷ n = 1 ÷ 1.50
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c = 41.8°
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Part (b)
it is totally internally reflected
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because the angle of incidence (60°) is greater than the critical angle
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Part (c)
speed in glass = 3.0 × 108 ÷ 1.50 = 2.0 × 108 m/s
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time = distance ÷ speed = 2400 ÷ (2.0 × 108)
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1.2 × 10−5 s
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Exam tip: Total internal reflection needs both conditions: light travelling from the denser to the less dense material, and an angle of incidence greater than the critical angle.
289IGPC3-3-08r23-03Light, refraction, lenses and dispersionPaper EMedium1 mark
A ray of light in air strikes a glass display block in a shop at an angle of incidence of 50°. The angle of refraction in the glass is 30°. What is the refractive index of the glass?
Diagram not to scale
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Notes
n = sin i ÷ sin r = sin 50° ÷ sin 30° = 0.766 ÷ 0.500 = 1.53. The value 1.67 comes from dividing the angles instead of their sines. The value 0.65 comes from inverting the ratio of the sines. The value 0.60 comes from inverting the ratio of the angles.
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Answer C
Answer: C
Common error: Use the sines of the angles, not the angles themselves, and measure both angles from the normal.
290IGPC3-3-08r23-05Light, refraction, lenses and dispersionPaper EMedium1 mark
Light is trapped inside a transparent block of refractive index 1.60 in a solar concentrator. What is the critical angle for light travelling from the block into air?
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Notes
sin c = 1 ÷ n = 1 ÷ 1.60 = 0.625, so c = 38.7°. The value 51.3° is 90° − 38.7°, which is the angle to the surface instead of to the normal. The value 56° comes from dividing 90° by 1.60. The value 32.0° comes from using tan instead of sin.
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Answer A
Answer: A
Common error: Critical angles are measured from the normal, and the sine of the critical angle is 1 ÷ n, not n.
291IGPC3-3-08r23-06Light, refraction, lenses and dispersionPaper EMedium1 mark
A small shop projector uses a thin converging lens to form an image of a slide on a screen. The slide is placed between one and two focal lengths from the lens. Which row describes the image on the screen?
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Notes
An object between f and 2f gives a real image beyond 2f that is inverted and magnified, and a real image can be shown on a screen. A virtual image cannot be projected on a screen and is formed only when the object is closer than f. A diminished image is formed only when the object is beyond 2f.
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Answer B
Answer: B
Exam tip: Remember the order: object beyond 2f gives a small real image; between f and 2f a large real image; inside f a virtual image.
292IGPC3-3-06r26-04Light, refraction, lenses and dispersionPaper EMedium1 mark
Light travels from glass of refractive index 1.60 towards the boundary with air. What is the critical angle for this glass-air boundary?
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Notes
sin c = 1 ÷ n = 1 ÷ 1.60 = 0.625, so c = 38.7°. 51.3° is 90° minus the correct angle, which comes from using cosine or the complementary angle. 32.0° uses tangent instead of sine. 45.0° assumes that the critical angle is always 45° for glass, but it depends on the refractive index.
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Answer B
Answer: B
Common error: Use sin c = 1/n for the critical angle; do not use cos or tan, and do not assume a fixed value.
293IGPC3-3-06r26-15Light, refraction, lenses and dispersionPaper EMedium8 marks
A glass prism with refractive index 1.52 is used in a signalling lamp on a ship.
Diagram not to scale
(a)
Calculate the critical angle for the glass-air boundary. Give your answer to 3 significant figures.
(2)
(b)
A ray of light inside the glass hits the glass-air boundary at an angle of incidence of 45°. Explain what happens to the ray.
(2)
(c)
State the other condition that must be met for total internal reflection to take place.
(1)
(d)
Another ray inside the glass hits the glass-air boundary at an angle of incidence of 30°. Calculate the angle of refraction in the air. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
sin c = 1 ÷ n = 1 ÷ 1.52
✓ 1
41.1°
✓ 1
Part (b)
45° is greater than the critical angle
✓ 1
total internal reflection occurs, so all of the light is reflected back into the glass
✓ 1
Part (c)
the light must be travelling from the more dense medium to the less dense medium
✓ 1
Part (d)
n = sin r ÷ sin i for light leaving the glass, so sin r = 1.52 × sin 30°
✓ 1
sin r = 0.76
✓ 1
r = 49.5°
✓ 1
Common error: When light leaves the glass, the refractive index multiplies the sine of the angle in the glass; do not divide.
294IGPC3-3-07r29-05Light, refraction, lenses and dispersionPaper EMedium1 mark
Glass used in a telecoms cable has a refractive index of 1.5. What is the critical angle for light travelling in this glass towards a boundary with air?
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Mark
Notes
At the critical angle sin c = 1/n = 1/1.5, so c = 41.8°. 48.2° results from using cosine instead of sine. 60.0° comes from dividing 90° by the refractive index. 33.7° results from using tangent instead of sine.
✓ 1
Answer D
Answer: D
Common error: Use sin c = 1/n; check that the angle is below 90° and that your calculator is in degree mode.
295IGPC3-3-07r29-10Wave properties, reflection, refraction and diffractionPaper EMedium7 marks
A food scientist checks the sugar content of a syrup by measuring its refractive index. A ray of light in air strikes the flat surface of the syrup at an angle of incidence of 52°. The angle of refraction is 36°. The speed of light in air is 3.00 × 108 m/s.
Diagram not to scale
(a)
Calculate the refractive index of the syrup. Give your answer to 3 significant figures.
(3)
(b)
Calculate the speed of light in the syrup. Use n = 1.34 (the unrounded value gives the same answer to 3 significant figures).
(2)
(c)
Explain why the wavelength of the light in the syrup is shorter than in air.
(2)
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Notes
Part (a)
M1 n = sin i ÷ sin r
✓ 1
M1 n = sin 52° ÷ sin 36°
✓ 1
A1 1.34
✓ 1
Part (b)
M1 speed = c ÷ n = 3.00 × 108 ÷ 1.34
✓ 1
A1 2.24 × 108 m/s (ECF from (a))
✓ 1
Part (c)
The frequency of the light does not change when it enters the syrup
✓ 1
The speed decreases and v = fλ, so the wavelength decreases in proportion
✓ 1
Exam tip: Make sure sin i is the angle in the less dense medium when calculating n = sin i ÷ sin r.
296IGPC3-3-10r32-04Light, refraction, lenses and dispersionPaper EMedium1 mark
A glass prism in an aircraft cockpit instrument has refractive index 1.60. What is the critical angle for light travelling in the glass towards a boundary with air?
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Notes
sin c = 1 ÷ n = 1 ÷ 1.60 = 0.625, so c = 38.7°. 51.3° is 90° − c, the angle to the surface rather than to the normal. 56.3° comes from 90° ÷ 1.60. 0.625° forgets to take the inverse sine.
✓ 1
Answer D
Answer: D
Common error: After finding sin c = 1/n, remember to take the inverse sine to get the angle.
297IGPC3-3-10r32-12Light, refraction, lenses and dispersionPaper EMedium6 marks
An optical fibre in a farm's crop-monitoring network is made of glass. A ray of light in air strikes a flat piece of the same glass at an angle of incidence of 50.0° and is refracted at an angle of 30.0°.
Diagram not to scale
(a)
Calculate the refractive index of the glass.
(2)
(b)
Calculate the critical angle of the glass. Use n = 1.53.
(2)
(c)
Explain why light can travel along the fibre by total internal reflection.
(2)
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Notes
Part (a)
M1 n = sin i ÷ sin r = sin 50.0° ÷ sin 30.0°
✓ 1
A1 1.53
✓ 1
Part (b)
M1 sin c = 1 ÷ n = 1 ÷ 1.53
✓ 1
A1 40.8°
✓ 1
Part (c)
the light meets the surface of the glass at an angle of incidence greater than the critical angle
✓ 1
the light is travelling in the glass (higher refractive index) towards a material of lower refractive index (such as air or cladding), so all of it is reflected
✓ 1
Common error: Total internal reflection needs both conditions: light moving towards a less dense medium and an angle of incidence above the critical angle.
298IGPC3-3-09r34-05Light, refraction, lenses and dispersionPaper EMedium1 mark
A bare strand of glass in a data centre has refractive index 1.60 and is surrounded by air. What is the critical angle for light travelling inside the strand towards the glass–air surface?
Show mark scheme
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Mark
Notes
sin c = 1/n = 1/1.60 = 0.625, so c = 38.7°. 51.3° is the angle from the surface rather than from the normal. 32.0° results from using tan instead of sin. 58.0° is 90° minus 32.0°.
✓ 1
Answer C
Answer: C
Common error: Critical angle is measured from the normal, and the formula uses sin c = 1/n.
299IGPC3-3-09r34-11Light, refraction, lenses and dispersionPaper EMedium8 marks
A thin block of transparent resin is used to preserve a small artefact. A ray of light in air strikes the resin at an angle of incidence of 50° and is refracted at an angle of 30°. The speed of light in air is 3.00 × 108 m/s.
Diagram not to scale
(a)
Calculate the refractive index of the resin. Give your answer to 3 significant figures.
(2)
(b)
Calculate the speed of light in the resin. Use n = 1.53.
(2)
(c)
Calculate the critical angle for the resin. Use n = 1.53 and give your answer to 3 significant figures.
(2)
(d)
A ray inside the resin meets the resin–air surface at an angle of incidence of 45°. State what happens to the ray and give a reason.
(2)
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Notes
Part (a)
M1 n = sin i ÷ sin r = sin 50° ÷ sin 30°
✓ 1
A1 1.53
✓ 1
Part (b)
M1 v = c ÷ n = 3.00 × 108 ÷ 1.53
✓ 1
A1 1.96 × 108 m/s
✓ 1
Part (c)
M1 sin c = 1 ÷ n = 1 ÷ 1.53
✓ 1
A1 40.8°
✓ 1
Part (d)
The ray is totally internally reflected
✓ 1
The angle of incidence (45°) is greater than the critical angle (40.8°)
✓ 1
Exam tip: For total internal reflection, compare the angle of incidence with the critical angle, and state both conditions: dense to less dense and angle greater than critical.
300IGPC3-3-10r34-11Light, refraction, lenses and dispersionPaper EMedium6 marks
A camera with a converging lens of focal length 50 mm is used to photograph a distant bridge. The speed of light in air is 3.0 × 108 m/s.
(a)
The bridge is so far away that rays from it are almost parallel. State the distance from the lens to the sensor for a sharp image.
(1)
(b)
Describe the image formed on the sensor.
(2)
(c)
State how the lens must be moved to photograph a nearer object sharply.
(1)
(d)
The speed of light in the glass of the lens is 2.0 × 108 m/s. Calculate the refractive index of the glass.
(2)
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Notes
Part (a)
50 mm (the focal length)
✓ 1
Part (b)
The image is real and inverted
✓ 1
The image is smaller than the object, that is diminished
✓ 1
Part (c)
The lens is moved further from the sensor
✓ 1
Part (d)
M1 n = c ÷ v = 3.0 × 108 ÷ 2.0 × 108
✓ 1
A1 1.5
✓ 1
Exam tip: Refractive index has no unit; it is the ratio of two speeds.
301IGPC3-3-09r36-13Light, refraction, lenses and dispersionPaper EMedium6 marks
A ray of light in air strikes the surface of the water in a swimming pool at an angle of incidence of 40°. The angle of refraction in the water is 28°.
Diagram not to scale
(a)
Calculate the refractive index of the water. Give your answer to 3 significant figures.
(2)
(b)
Take the refractive index of the water as 1.37. The speed of light in air is 3.0 × 108 m/s. Calculate the speed of light in the water. Give your answer to 3 significant figures.
(2)
(c)
A lamp at the bottom of the pool shines light upwards at the water surface. Calculate the critical angle for light travelling from the water into air. Use n = 1.37 and give your answer to 1 decimal place.
(2)
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Notes
Part (a)
n = sin i / sin r
✓ 1
n = sin 40° / sin 28° = 1.37
✓ 1
Part (b)
n = speed in air / speed in water
✓ 1
speed = 3.0 × 108 / 1.37 = 2.19 × 108 m/s
✓ 1
Part (c)
sin c = 1 / n
✓ 1
c = sin−1(1 / 1.37) = 46.9°
✓ 1
Exam tip: For the critical angle use sin c = 1/n; the angle must be measured in the denser material, from the normal.
302IGPC3-3-09r36-15Light, refraction, lenses and dispersionPaper EMedium5 marks
A librarian uses a converging lens of focal length 10 cm as a magnifying glass to read small print on an old catalogue card.
Diagram not to scale
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
The print is held 6 cm from the lens. State two characteristics of the image seen.
(2)
(c)
State the range of distances between the print and the lens for which the lens acts as a magnifying glass.
(1)
(d)
Explain why the image seen in (b) cannot be projected onto a screen.
(1)
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Notes
Part (a)
the point where rays that are parallel to the principal axis meet (converge) after passing through the lens
✓ 1
Part (b)
any two from: virtual; upright; magnified
✓ 2
Part (c)
less than 10 cm, or between the lens and the principal focus
✓ 1
Part (d)
the image is virtual because the rays only appear to come from it; the rays do not actually meet there
✓ 1
Common error: A magnifying glass needs the object inside the focal length; beyond the focal length the image is real and inverted.
303IGPC3-3-09r37-04Wave properties, reflection, refraction and diffractionPaper EMedium1 mark
Water waves of frequency 4.0 Hz travel across a hospital hydrotherapy pool. They pass from a deep region, where the wavelength is 3.0 cm, into a shallow region, where the wavelength is 2.0 cm. What is the speed of the waves in the shallow region?
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Notes
The frequency is set by the source and stays at 4.0 Hz when the waves enter the shallow region, so speed = 4.0 × 2.0 = 8.0 cm/s. 12 cm/s is the speed in the deep region (4.0 × 3.0), which wrongly assumes the speed does not change. 5.3 cm/s assumes that the frequency falls in proportion to the wavelength (2.7 Hz) and then multiplies by 2.0 cm. 2.0 cm/s just reuses the number for the wavelength.
✓ 1
Answer D
Answer: D
Exam tip: When waves cross a boundary, the frequency stays the same; the speed and wavelength change together.
304IGPC3-3-09r37-12Light, refraction, lenses and dispersionPaper EMedium7 marks
A beekeeper tests a sample of honey. Light travels through the honey at 2.00 × 108 m/s. The speed of light in air is 3.00 × 108 m/s.
Diagram not to scale
(a)
Calculate the refractive index of the honey.
(2)
(b)
A ray of light in air strikes the surface of the honey with an angle of incidence of 45°. Use a refractive index of 1.50 for the honey. Calculate the angle of refraction. Give your answer to 3 significant figures.
(3)
(c)
Explain why the ray bends towards the normal as it enters the honey.
(2)
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Notes
Part (a)
refractive index = speed in air ÷ speed in honey = 3.00 × 108 ÷ 2.00 × 108
✓ 1
1.50
✓ 1
Part (b)
n = sin i ÷ sin r, so sin r = sin 45° ÷ 1.50
✓ 1
sin r = 0.471
✓ 1
r = 28.1°
✓ 1
Part (c)
the light slows down in the honey
✓ 1
the part of the wavefront that enters first slows first, so the wavefront and the ray change direction
✓ 1
Exam tip: Rearrange the refraction equation for sin r before using the inverse sine key, and check your calculator is in degree mode.
305IGPC3-3-08r40-04Light, refraction, lenses and dispersionPaper EMedium1 mark
A ray of light in air strikes the surface of a swimming pool at an angle of incidence of 60° to the normal. The refractive index of the water is 1.33. What is the angle of refraction in the water?
Diagram not to scale
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Mark
Notes
Using n = sin i ÷ sin r, sin r = sin 60° ÷ 1.33 = 0.651, so r = 41°. 45° comes from dividing the angle itself by 1.33. 80° comes from multiplying the angle by 1.33. 60° would mean the ray is not refracted at all, but light bends towards the normal when it enters a denser medium.
✓ 1
Answer D
Answer: D
Common error: Apply the refractive index to the sines of the angles, not to the angles themselves.
306IGPC3-3-08r40-05Light, refraction, lenses and dispersionPaper EMedium1 mark
An optical fibre carrying data around a library has a core of refractive index 1.52. What is the critical angle for light travelling from the core into the surrounding air?
Show mark scheme
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Mark
Notes
At the critical angle sin c = 1 ÷ n = 1 ÷ 1.52 = 0.658, so c = 41°. 33° comes from using tan instead of sin. 49° is 90° minus the critical angle, which is the angle measured from the surface rather than from the normal. 57° comes from using tan with n instead of the inverse sine of 1 ÷ n.
✓ 1
Answer B
Answer: B
Exam tip: Critical angle and total internal reflection are measured from the normal, and only occur when light travels from a denser to a less dense medium.
307IGPC3-3-08r40-12Light, refraction, lenses and dispersionPaper EMedium7 marks
A ray of light in air is incident on the surface of a swimming pool at an angle of incidence of 55°. The angle of refraction in the water is 38°.
Diagram not to scale
(a)
Calculate the refractive index of the water. Give your answer to 3 significant figures.
(2)
(b)
A lamp under the water shines light towards the surface. Calculate the critical angle for water-to-air, using n = 1.33. Give your answer to 2 significant figures.
(2)
(c)
The lamp directs a ray at the surface with an angle of incidence of 60°. Explain what happens to the ray at the surface.
(3)
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Notes
Part (a)
n = sin i ÷ sin r = sin 55° ÷ sin 38°
✓ 1
1.33
✓ 1
Part (b)
sin c = 1 ÷ n = 1 ÷ 1.33
✓ 1
49°
✓ 1
Part (c)
the angle of incidence (60°) is greater than the critical angle (49°)
✓ 1
the light is travelling from water (denser) to air (less dense)
✓ 1
total internal reflection occurs, so the ray is reflected back into the water and none emerges
✓ 1
Common error: Total internal reflection needs both conditions: light going into a less dense medium, and an angle of incidence greater than the critical angle.
308IGPC3-3-09r42-08Light, refraction, lenses and dispersionPaper EMedium1 mark
In glass, red light is compared with violet light. Which statement is correct?
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Notes
Glass has a smaller refractive index for red light than for violet, so red light travels faster in glass and is deviated less. Red light has a lower frequency than violet and a longer wavelength in a vacuum, and its refractive index in glass is smaller, not greater.
✓ 1
Answer B
Answer: B
Exam tip: Violet is deviated most by a prism, so it has the largest refractive index and the slowest speed in glass.
309IGPC3-3-09r42-11Light, refraction, lenses and dispersionPaper EMedium7 marks
An endoscope used in a hospital carries light along a glass fibre of refractive index 1.52. The speed of light in a vacuum is 3.00 × 108 m/s.
(a)
Calculate the critical angle for the glass–air boundary.
(3)
(b)
State two conditions needed for total internal reflection at a boundary.
(2)
(c)
Calculate the speed of light in the glass fibre.
(2)
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Notes
Part (a)
sin c = 1 ÷ n
✓ 1
sin c = 1 ÷ 1.52 = 0.658
✓ 1
c = 41.1°
✓ 1
Part (b)
light is travelling in the denser medium (higher refractive index) towards the less dense medium
✓ 1
angle of incidence is greater than the critical angle
✓ 1
Part (c)
v = c ÷ n
✓ 1
= 3.00 × 108 ÷ 1.52 = 1.97 × 108 m/s
✓ 1
Exam tip: Check that the light travels from the denser to the less dense medium before you use the critical angle idea.
310IGPC3-3-08r43-05Light, refraction, lenses and dispersionPaper EMedium1 mark
A ray of light in air strikes the glass of a library display case at an angle of incidence of 40°. The refractive index of the glass is 1.50. What is the angle of refraction in the glass?
Diagram not to scale
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Notes
Use n = sin i / sin r, so sin r = sin 40° ÷ 1.50 = 0.4285 and r = 25.4°. The value 26.7° comes from dividing the angle (40°) by 1.50 instead of dividing the sine. The value 40.0° assumes the ray does not bend. The value 74.6° comes from multiplying the sine by 1.50 instead of dividing.
✓ 1
Answer A
Answer: A
Common error: Divide the sine of the angle by n, not the angle itself, then use the inverse sine.
311IGPC3-3-08r43-13Wave properties, reflection, refraction and diffractionPaper EMedium6 marks
A ray of light in air strikes the surface of the water in a swimming pool at an angle of incidence of 55°. The refractive index of water is 1.33.
Diagram not to scale
(a)
Calculate the angle of refraction.
(3)
(b)
State and explain what happens to the speed, the wavelength and the frequency of the light as it enters the water.
(3)
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Mark
Notes
Part (a)
M1 n = sin i ÷ sin r, so sin r = sin 55° ÷ 1.33
✓ 1
M1 sin r = 0.616, r = sin−1(0.616)
✓ 1
A1 38.0°
✓ 1
Part (b)
speed decreases
✓ 1
frequency stays the same
✓ 1
wavelength decreases, since speed = frequency × wavelength and the frequency is constant
✓ 1
Common error: Frequency is set by the source and does not change at a boundary; it is the wavelength and speed that change.
312IGPC3-3-10r44-05Light, refraction, lenses and dispersionPaper EMedium1 mark
A conservator shines a ray of light from air into a clear plastic sample. The angle of incidence is 50° and the angle of refraction is 30°. What is the refractive index of the plastic?
Diagram not to scale
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Notes
n = sin i ÷ sin r = sin 50° ÷ sin 30° = 0.766 ÷ 0.500 = 1.53. 1.67 divides the angles themselves instead of their sines. 0.65 inverts the ratio. 2.06 uses tangents instead of sines.
✓ 1
Answer C
Answer: C
Common error: Use the sines of the angles, with the angle in air on top, and check that n for a material is greater than 1.
313IGPC3-3-10r44-07Light, refraction, lenses and dispersionPaper EMedium1 mark
An archaeologist holds a converging lens of focal length 10 cm 6.0 cm above a small inscription on a stone. Which row describes the image seen through the lens?
Diagram not to scale
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Notes
The object is closer to the lens than the principal focus, so the refracted rays diverge and appear to come from a virtual image on the same side as the object. This image is upright and magnified, as in a magnifying glass. A real, inverted image needs the object beyond the principal focus. A diminished image is formed only when the object is more than twice the focal length away.
✓ 1
Answer A
Answer: A
Exam tip: Compare the object distance with the focal length first: inside the focal length gives a virtual, upright, magnified image.
314IGPC3-3-10r44-13Light, refraction, lenses and dispersionPaper EMedium7 marks
A glass optical fibre in a data centre has a refractive index of 1.50. Assume the fibre is surrounded by air.
Diagram not to scale
(a)
Calculate the critical angle for the glass.
(2)
(b)
State two conditions needed for total internal reflection to occur at a boundary.
(2)
(c)
Explain why signals can travel a long distance along the fibre.
(1)
(d)
A ray in air enters the end of the fibre with an angle of incidence of 30°. Calculate the angle of refraction. Give your answer to 1 decimal place.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 sin c = 1 ÷ n = 1 ÷ 1.50
✓ 1
A1 41.8°
✓ 1
Part (b)
B1 light must travel in the denser medium towards the less dense medium (for example, from glass to air)
✓ 1
B1 the angle of incidence must be greater than the critical angle
✓ 1
Part (c)
B1 the light is totally internally reflected at each boundary so almost no light escapes through the sides
✓ 1
Part (d)
M1 sin r = sin i ÷ n = sin 30° ÷ 1.50
✓ 1
A1 19.5°
✓ 1
Common error: In n = sin i ÷ sin r, i is the angle in air. For a critical angle, use sin c = 1 ÷ n.
315IGPC3-3-08r45-06Light, refraction, lenses and dispersionPaper EMedium1 mark
A student holds a converging lens as a magnifying glass in a library. The object is closer to the lens than the principal focus. Which description of the image is correct?
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Notes
With the object between the lens and the principal focus the refracted rays diverge, so they only appear to come from a point behind the object. The image is virtual, upright and magnified. The real, inverted images occur when the object is beyond the principal focus, and a diminished virtual image is formed by a diverging lens.
✓ 1
Answer C
Answer: C
Exam tip: A magnifying glass always has the object inside the focal length; only then can you see an upright enlarged image.
316IGPC3-3-08r45-14Light, refraction, lenses and dispersionPaper EMedium6 marks
A narrow beam of white light passes through a glass prism.
(a)
Describe what happens to the white light as it passes through the prism.
(3)
(b)
State which colour of visible light has the highest frequency.
(1)
(c)
Red light has a wavelength of 7.0 × 10−7 m in a vacuum. Calculate its frequency. Give your answer to 2 significant figures.
(2)
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Notes
Part (a)
the light is refracted (bent) when it enters and leaves the prism
✓ 1
different colours (frequencies) are refracted by different amounts
✓ 1
a spectrum is formed, with violet bent most and red bent least
✓ 1
Part (b)
violet
✓ 1
Part (c)
f = v ÷ λ, with v = 3.0 × 108 m/s
✓ 1
4.3 × 1014 Hz
✓ 1
Exam tip: Name the property that differs: each colour has its own frequency, so each is refracted by a different amount.
317IGPC3-3-08r45-15Sound and ultrasoundPaper EMedium7 marks
Ultrasound is used to check wind turbine blades for hidden cracks. A probe on the surface of a blade sends a short pulse into the blade and detects the echo from a crack.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
The echo from a crack returns to the probe 6.0 × 10−5 s after the pulse is sent. The speed of ultrasound in the blade material is 2800 m/s. Calculate the depth of the crack below the surface.
(3)
(c)
Explain why part of the pulse is reflected at the crack.
(1)
(d)
The ultrasound has a frequency of 2.5 MHz. Calculate its wavelength in the blade material. Give your answer to 2 significant figures.
(2)
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Notes
Part (a)
sound with a frequency above 20 000 Hz, higher than humans can hear
✓ 1
Part (b)
distance = speed × time
✓ 1
pulse travels to the crack and back, so use half the total distance, 2800 × 6.0 × 10−5 ÷ 2
✓ 1
0.084 m
✓ 1
Part (c)
the crack contains air, so the pulse meets a boundary between two different materials, where part of it is reflected
✓ 1
Part (d)
λ = v ÷ f
✓ 1
2800 ÷ 2.5 × 106 = 1.1 × 10−3 m
✓ 1
Exam tip: In echo questions, divide the total path length by two to find the distance to the reflector.
318IGPC3-3-08r46-07Wave properties, reflection, refraction and diffractionPaper EMedium1 mark
A coach uses an underwater camera to study a diver. A ray in air strikes the water surface with an angle of incidence of 55° and the angle of refraction is 38°. What is the refractive index of the water?
Diagram not to scale
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Notes
n = sin i ÷ sin r = sin 55° ÷ sin 38° = 0.819 ÷ 0.616 = 1.33. The value 1.45 comes from dividing the angles instead of their sines. 0.75 results from inverting the ratio, sin r ÷ sin i. 0.62 is just sin 38° and ignores sin 55°.
✓ 1
Answer D
Answer: D
Common error: Use the sines of the angles in n = sin i / sin r, not the angles themselves, and keep air as the incident side.
319IGPC3-3-08r46-13Light, refraction, lenses and dispersionPaper EMedium8 marks
Light travels through a glass fibre used for railway signalling. The refractive index of the glass is 1.48. The speed of light in a vacuum is 3.0 × 108 m/s.
Diagram not to scale
(a)
State the two conditions needed for total internal reflection at a boundary.
(2)
(b)
Calculate the critical angle for the glass-air boundary.
(2)
(c)
A ray inside the glass strikes the boundary at 38° to the normal. Explain whether the ray undergoes total internal reflection.
(2)
(d)
Calculate the speed of light in the glass. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
light travels from a more dense (higher refractive index) medium to a less dense medium
✓ 1
the angle of incidence is greater than the critical angle
✓ 1
Part (b)
sin c = 1/n = 1/1.48
✓ 1
c = 42.5°
✓ 1
Part (c)
38° is less than the critical angle (42.5°)
✓ 1
so total internal reflection does not occur and the ray partly refracts out of the glass
✓ 1
Part (d)
speed = 3.0 × 108 ÷ 1.48
✓ 1
2.03 × 108 m/s
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Exam tip: Check that the angle in your critical-angle comparison is measured from the normal.
320IGPC3-3-09r46-04Light, refraction, lenses and dispersionPaper EMedium1 mark
A librarian uses a converging lens of focal length 8 cm as a magnifying glass. A page is held 5 cm from the lens. Which row describes the image seen?
Diagram not to scale
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The object is closer to the lens than the principal focus, so the refracted rays diverge and appear to come from a point behind the page: the image is virtual, upright and larger. Real inverted images need the object beyond the focal length. A diminished upright image is not formed by a converging lens in this case.
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Answer D
Answer: D
Exam tip: If the object is inside the focal length of a converging lens, expect a virtual, upright, magnified image.
321IGPC3-3-09r46-12Light, refraction, lenses and dispersionPaper EMedium6 marks
A window in the side of a swimming pool is made of acrylic of refractive index 1.50. A ray of light in air strikes the outer surface of the window at an angle of incidence of 40°. The speed of light in air is 3.0 × 108 m/s.
Diagram not to scale
(a)
State what is meant by refraction.
(1)
(b)
Calculate the angle of refraction in the acrylic. Give your answer to 3 significant figures.
(3)
(c)
Calculate the speed of light in the acrylic.
(2)
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Part (a)
B1 change of direction of a wave when it passes from one medium to another because its speed changes
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Part (b)
M1 n = sin i ÷ sin r
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M1 sin r = sin 40° ÷ 1.50 = 0.4285
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A1 r = 25.4°
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Part (c)
M1 n = speed in air ÷ speed in acrylic, so speed = (3.0 × 108) ÷ 1.50
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A1 2.0 × 108 m/s
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Exam tip: Set your calculator to degrees before using sin or sin⁻¹.
322IGPC3-3-09r46-15Light, refraction, lenses and dispersionPaper EMedium7 marks
A pupil uses a converging lens as a magnifying glass to read small print in a library book. The letters in the print are 2.5 mm tall.
(a)
State what is meant by the principal focus of a converging lens.
(2)
(b)
The print is held closer to the lens than the principal focus. State three features of the image.
(3)
(c)
The image of a letter is 7.5 mm tall. Calculate the magnification.
(2)
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Notes
Part (a)
B1 point on the axis
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B1 where rays parallel to the axis meet / converge after passing through the lens
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Part (b)
B1 virtual
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B1 upright
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B1 magnified / larger than the object
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Part (c)
M1 magnification = image height ÷ object height
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A1 7.5 ÷ 2.5 = 3.0
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Exam tip: Magnification has no unit because it is a ratio of two lengths.
323IGPC3-3-09r48-10Light, refraction, lenses and dispersionPaper EMedium9 marks
An architect designs a glass atrium. A ray of light in air strikes a flat glass panel at an angle of incidence of 45.0°. The ray is refracted and travels through the glass at an angle of refraction of 28.1° to the normal.
Diagram not to scale
(a)
Calculate the refractive index of the glass. Give your answer to 3 significant figures.
(3)
(b)
State what happens to the speed of the light as it enters the glass.
(1)
(c)
Light travelling inside the glass strikes the glass–air surface at an angle of incidence of 50°. Use n = 1.50 to calculate the critical angle of the glass.
(3)
(d)
Use your answer to (c) to state and explain what happens to the light that strikes the glass–air surface at 50°.
(2)
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Part (a)
M1 n = sin i ÷ sin r
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M1 n = sin 45.0° ÷ sin 28.1°
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A1 1.50
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Part (b)
speed decreases
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Part (c)
M1 sin c = 1 ÷ n
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M1 c = sin−1(1 ÷ 1.50)
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A1 41.8°
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Part (d)
B1 total internal reflection occurs
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B1 because the angle of incidence (50°) is greater than the critical angle (accept ECF from (c))
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Common error: Use sin i ÷ sin r with both angles measured from the normal, and put the larger refractive index in the denominator when finding the critical angle.
324IGPC3-3-09r48-13Light, refraction, lenses and dispersionPaper EMedium7 marks
A marine biologist uses a converging lens of focal length 6.0 cm as a magnifying glass to examine a small plankton sample.
Diagram not to scale
(a)
State what is meant by the principal focus of a converging lens.
(1)
(b)
The sample is placed 4.0 cm from the lens. State three characteristics of the image.
(3)
(c)
The sample is 0.80 cm long and its image is 2.4 cm long. Calculate the magnification.
(2)
(d)
Explain why this image cannot be shown on a screen.
(1)
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Notes
Part (a)
the point on the principal axis where rays that were parallel to the axis meet (converge) after passing through the lens
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Part (b)
virtual
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upright
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magnified
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Part (c)
M1 magnification = image size ÷ object size
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A1 2.4 ÷ 0.80 = 3.0
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Part (d)
it is a virtual image because the rays only appear to come from it; they do not actually meet
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Exam tip: A magnifying glass works only when the object is closer to the lens than its principal focus.
325IGPC-037-03-09Sound and ultrasoundPaper EHard1 mark
A ship sends a pulse of ultrasound straight down to the sea bed. The echo is detected 0.084 s after the pulse is sent. The speed of sound in sea water is 1500 m/s. What is the depth of the sea below the ship?
Diagram not to scale
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Distance travelled = 1500 × 0.084 = 126 m. This is down and back up, so depth = 126 ÷ 2 = 63 m. 126 m forgets to halve, 31.5 m halves twice, and 13.9 m uses the speed of sound in air (330 m/s) instead of in water.
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Answer C
Answer: C
Exam tip: Use the speed of sound for the medium the wave actually travels through. Here that is sea water, which is much faster than air.
326IGPC-037-03-16Sound and ultrasoundPaper EHard7 marks
A long steel safety railing is 1020 m long. A worker strikes one end of it once with a hammer. A second worker, at the other end, puts her ear close to the railing and hears two separate bangs. The speed of sound in air is 340 m/s and the speed of sound in steel is 5100 m/s.
Diagram not to scale
(a)
Explain why the second worker hears two bangs.
(2)
(b)
Calculate the time between the two bangs.
(3)
(c)
Describe the motion of the air particles as the sound wave passes through the air, and state what is meant by a compression.
(2)
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Part (a)
The sound travels to her through two different media, the steel and the air . Sound is faster in steel than in air, so the sound through the steel arrives first .
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Part (b)
time in air = 1020 ÷ 340 = 3.00 s
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time in steel = 1020 ÷ 5100 = 0.200 s
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time between bangs = 3.00 − 0.200 = 2.80 s
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Part (c)
The particles vibrate backwards and forwards parallel to the direction the wave travels, so the wave is longitudinal . A compression is a region where the particles are close together and the pressure is higher .
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Common error: Do not divide the distance by the difference in speeds. Work out each travel time separately, then subtract the two times.
327IGPC-037-01-10Light, refraction, lenses and dispersionPaper EHard1 mark
Inside a glass block of refractive index 1.50, a ray of light strikes the glass–air boundary at an angle of incidence of 45°. What happens to the ray at the boundary?
Diagram not to scale
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Notes
The critical angle is c = sin−1(1/1.50) = 41.8°. The angle of incidence, 45°, is greater than the critical angle, so the ray is totally internally reflected. You can check this another way: the angle of refraction would need sin r = 1.50 × sin 45° = 1.06, which is impossible because a sine cannot exceed 1. The 28° option divides by n instead of multiplying, which would bend the ray towards the normal. A ray travels along the boundary only when the angle of incidence equals the critical angle exactly.
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Answer D
Answer: D
Exam tip: For light leaving a dense medium, find the critical angle first. If the angle of incidence is larger than the critical angle, the ray is totally internally reflected.
328IGPC-037-01-14Light, refraction, lenses and dispersionPaper EHard8 marks
An optical fibre has a glass core with a refractive index of 1.52 and is surrounded by air. A ray of light in air strikes the flat end face of the fibre at an angle of incidence of 30°. It refracts into the core and then travels on to strike the side wall of the fibre. The end face is at 90° to the side wall.
Diagram not to scale
(a)
Calculate the critical angle for the glass–air boundary.
(2)
(b)
Calculate the angle of refraction at the end face.
(2)
(c)
Calculate the angle of incidence at the side wall. State whether the ray escapes from the fibre, giving a reason.
(3)
(d)
State one use of optical fibres.
(1)
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Part (a)
sin c = 1/1.52 = 0.658
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c = 41°
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Part (b)
sin r = sin 30° / 1.52 = 0.329
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r = 19°
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Part (c)
The normal to the side wall is at 90° to the normal to the end face, so angle of incidence = 90° − 19.2°
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= 71°
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71° is greater than the critical angle of 41°, so the ray is totally internally reflected and does not escape
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Part (d)
Any one from: telecommunications or broadband data transmission; an endoscope for viewing inside the body
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Exam tip: When two faces meet at 90°, their normals are also at 90°. The angle of incidence at the second face is therefore 90° minus the angle of refraction at the first face.
329IGPC-037-02-09The electromagnetic spectrumPaper EHard1 mark
An electromagnetic wave has a frequency of 1.5 × 1018 Hz. What is its wavelength, and in which region of the spectrum is it?
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λ = v/f = 3.0 × 108 / 1.5 × 1018 = 2.0 × 10−10 m. A wavelength this short, with a very high frequency, places the wave in the X-ray region. Radio waves have long wavelengths and low frequencies. 4.5 × 1026 m comes from multiplying v by f. 5.0 × 109 m comes from dividing f by v.
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Answer A
Answer: A
Exam tip: Check that an answer makes sense. A frequency around 10¹⁸ Hz must give a tiny wavelength, roughly the size of an atom.
330IGPC-037-02-16The electromagnetic spectrumPaper EHard8 marks
A town receives radio broadcasts from two stations. Station A transmits at a frequency of 1.2 MHz. Station B transmits at 100 MHz, which gives a wavelength of 3.0 m.
(a)
Calculate the wavelength of the waves from station A.
(2)
(b)
A house is in a valley behind a range of hills from both transmitters. Suggest, with a reason, which station is received better at the house.
(2)
(c)
X-rays are also electromagnetic waves. Explain why X-rays are hazardous to people but these radio waves are not.
(2)
(d)
An X-ray machine produces X-rays of wavelength 1.0 × 10−10 m. Calculate the frequency of these X-rays.
(2)
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Part (a)
λ = v/f = 3.0 × 108 / 1.2 × 106 ; λ = 250 m
✓ 2
Part (b)
station A ; its much longer wavelength is diffracted more around the hills
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Part (c)
X-rays have a much higher frequency / energy and are ionising ; they can damage cells and cause mutation / cancer, whereas radio waves cannot
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Part (d)
f = v/λ = 3.0 × 108 / 1.0 × 10−10 ; f = 3.0 × 1018 Hz
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Exam tip: Diffraction is greatest when the wavelength is comparable to the size of the obstacle or gap. Long-wavelength radio waves bend around hills; short-wavelength waves do not.
331IGPC-037-00-08Wave properties, reflection, refraction and diffractionPaper EHard1 mark
The critical angle for light travelling from a certain glass into air is 38°. What is the refractive index of the glass?
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Use n = 1 ÷ sin c = 1 ÷ sin 38° = 1 ÷ 0.616 = 1.62. The value 0.62 is sin c, which forgets to take the reciprocal. The value 0.79 is cos 38°. The value 1.27 is 1 ÷ cos 38°, which wrongly uses the angle to the surface instead of the angle to the normal.
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Answer D
Answer: D
Exam tip: Check that your answer is greater than 1. A refractive index below 1 cannot be right.
332IGPC-037-00-10Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Light of wavelength 600 nm in air has a frequency of 5.0 × 1014 Hz. It enters glass of refractive index 1.5. What are the frequency and the wavelength of the light inside the glass?
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The frequency is set by the source, so it stays at 5.0 × 1014 Hz. The speed in the glass is the speed in air divided by 1.5. Since v = fλ and f is unchanged, the wavelength also falls by a factor of 1.5: 600 ÷ 1.5 = 400 nm. The 900 nm option multiplies by n instead of dividing.
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Answer A
Answer: A
Exam tip: n = speed in air ÷ speed in the medium. Because the frequency is constant, the wavelength changes by the same factor as the speed.
333IGPC-037-01-16Light, refraction, lenses and dispersionPaper EHard8 marks
A ray of light in air strikes the top face of a rectangular glass block at an angle of incidence of 60°. The glass has a refractive index of 1.50, and the block is 4.0 cm thick. The top and bottom faces are parallel.
Diagram not to scale
(a)
Calculate the angle of refraction at the top face.
(2)
(b)
State the angle at which the ray leaves the bottom face, measured from the normal. Explain your answer.
(2)
(c)
Calculate the distance the ray travels inside the block.
(2)
(d)
Explain why, whatever the angle of incidence at the top face, the ray can never be totally internally reflected at the bottom face.
(2)
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Part (a)
sin r = sin 60° / 1.50 = 0.577
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r = 35.3° (accept 35°)
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Part (b)
60°
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The faces are parallel, so the ray meets the bottom face at the angle of refraction from (a); the refraction is reversed, so the emerging ray is parallel to the incident ray
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Part (c)
distance = thickness / cos r = 4.0 / cos 35.3°
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= 4.9 cm
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Part (d)
sin r = sin i / 1.50 and the largest possible value of sin i is 1 (i = 90°), so sin r can never be more than 1/1.50, and r can never be more than 41.8°
✓ 1
The critical angle is given by sin c = 1/1.50, so c = 41.8°. The ray meets the bottom face at angle r, which is never greater than c (it equals c only at grazing incidence), so total internal reflection cannot occur
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Exam tip: In part (c), the ray inside the glass is the hypotenuse of a right-angled triangle. The block thickness is the side next to angle r.
334IGPC-037-00-15Wave properties, reflection, refraction and diffractionPaper EHard6 marks
In a classroom demonstration, a bare glass fibre with no cladding is surrounded by air. The glass has a refractive index of 1.52. Light travels along the fibre by repeatedly striking the boundary between the glass and the air.
Diagram not to scale
(a)
State the two conditions needed for total internal reflection.
(2)
(b)
Calculate the critical angle for light passing from the glass into air.
(2)
(c)
A ray inside the fibre strikes the glass–air boundary with an angle of incidence of 48°. State and explain what happens to the ray.
(2)
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Part (a)
Light must travel from an optically denser medium towards a less dense medium, that is from higher to lower refractive index.
✓ 1
The angle of incidence must be greater than the critical angle.
✓ 1
Part (b)
sin c = 1 ÷ n = 1 ÷ 1.52 = 0.658
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c = 41° (41.1°)
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Part (c)
The ray is totally internally reflected, so it stays inside the fibre.
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This is because 48° is greater than the critical angle of 41°.
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Common error: Total internal reflection needs the angle of incidence to be greater than the critical angle. At exactly the critical angle, the refracted ray travels along the boundary. Below it, most of the light refracts out of the glass.
335IGPC3-3-03c-07Light, refraction, lenses and dispersionPaper EHard1 mark
A ray inside a block of glass of refractive index 1.60 reaches the glass–air surface at an angle of incidence of 42°. What happens to the ray?
Diagram not to scale
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sin c = 1 / 1.60 = 0.625, so c = 38.7°. Since 42° is greater than c, the ray is totally internally reflected (C). A: light going into a less dense medium bends away from the normal, and here it does not leave. B uses 90° − 38.7° = 51.3°, an angle from the surface. D: the ray emerges along the surface only if the angle equals 38.7°.
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Answer C
Answer: C
Common error: The critical angle is measured from the normal, not from the surface, and comes from sin c = 1/n.
336IGPC3-3-03c-14Wave properties, reflection, refraction and diffractionPaper EHard10 marks
In a ripple tank a vibrating bar produces straight water waves of frequency 8.0 Hz. The waves travel from a deep region into a shallow region. In the deep region the wavelength is 2.5 cm. In the shallow region the wavelength is 1.5 cm.
Diagram not to scale
(a)
Calculate the speed of the waves in the deep region, in m/s.
(2)
(b)
Determine the speed of the waves in the shallow region.
(3)
(c)
The wavefronts meet the boundary at an angle. State and explain how the direction of travel changes as the waves cross the boundary.
(2)
(d)
The direction of travel in the deep region makes an angle of 40° with the normal to the boundary. For these waves, sin i / sin r = v1 / v2. Calculate the angle between the direction of travel in the shallow region and the normal.
(3)
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Part (a)
v = f λ = 8.0 × 0.025
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v = 0.20 m/s
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Part (b)
frequency is unchanged at 8.0 Hz
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v = 8.0 × 0.015
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v = 0.12 m/s
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Part (c)
the waves bend towards the normal
✓ 1
because their speed decreases in the shallow region
✓ 1
Part (d)
sin r = sin i × v2 / v1
✓ 1
sin r = sin 40° × 0.12 / 0.20 = 0.386, ECF from (b)
✓ 1
r = 22.7°
✓ 1
Common error: Frequency is set by the source and stays the same across a boundary; it is the speed and wavelength that change.
337IGPC3-3-05r9-06Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Water waves in a harbour travel from deep water into shallow water. In the deep water the wavelength is 6.0 cm and the speed is 12 cm/s. The speed in the shallow water is 8.0 cm/s. What is the wavelength in the shallow water?
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Correct: B. The frequency stays the same: f = 12 ÷ 6.0 = 2.0 Hz. In shallow water λ = v ÷ f = 8.0 ÷ 2.0 = 4.0 cm. A inverts the speed ratio (6.0 × 12 ÷ 8.0). C assumes the wavelength is unchanged. D multiplies speed by frequency (8.0 × 2.0).
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Answer B
Answer: B
Common error: On refraction the frequency of a wave stays constant. The speed and wavelength change together in the same proportion.
338IGPC3-3-05r9-15Wave properties, reflection, refraction and diffractionPaper EHard8 marks
An archaeologist examines a glass paperweight of refractive index 1.52. The speed of light in air is 3.0 × 108 m/s. Light of wavelength 608 nm in air hits the glass at an angle of incidence of 35°. Use n = 1.52 directly in each part rather than a rounded answer from an earlier part.
Diagram not to scale
(a)
Calculate the speed of the light in the glass. Give your answer to 3 significant figures.
(2)
(b)
Calculate the angle of refraction. Give your answer to 3 significant figures.
(3)
(c)
The frequency of the light does not change as it enters the glass. Calculate the wavelength of the light in the glass.
(3)
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Part (a)
speed in glass = c ÷ n = 3.0 × 108 ÷ 1.52
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1.97 × 108 m/s
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Part (b)
sin r = sin 35° ÷ 1.52 (n = sin i ÷ sin r)
✓ 1
sin r = 0.377
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r = 22.2°
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Part (c)
frequency is the same, so wavelength is proportional to speed
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wavelength in glass = 608 ÷ 1.52
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400 nm
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Common error: When light enters glass, its frequency stays the same; the speed and wavelength both decrease by the same factor n.
339IGPC3-3-05r9-16Wave properties, reflection, refraction and diffractionPaper EHard8 marks
A research boat on a flooded quarry uses a loudspeaker to send sound of frequency 1700 Hz through the air. The speed of sound in air is 340 m/s. The sound then enters the water of the quarry, where its speed is 1500 m/s.
(a)
Calculate the wavelength of the sound in air.
(2)
(b)
State which of the speed, frequency and wavelength change when the sound enters the water.
(2)
(c)
Calculate the wavelength of the sound in water. Give your answer to 2 significant figures.
(2)
(d)
The sound passes through a gap of width 0.50 m in a wall, once in air and once in water. Explain in which case the diffraction is greater.
(2)
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Part (a)
wavelength = speed ÷ frequency = 340 ÷ 1700
✓ 1
0.20 m
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Part (b)
speed and wavelength change (both increase)
✓ 1
frequency stays the same
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Part (c)
wavelength = 1500 ÷ 1700
✓ 1
0.88 m
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Part (d)
diffraction is greater when the wavelength is closer to, or larger than, the gap width
✓ 1
in water the wavelength (0.88 m) is larger than the gap whereas in air (0.20 m) it is smaller, so there is more diffraction in water
✓ 1
Exam tip: Diffraction depends on the wavelength compared with the gap width; always find the wavelength in each medium before comparing.
340IGPC3-3-06r12-08Light, refraction, lenses and dispersionPaper EHard1 mark
In a water treatment plant, a ray of light in a glass light pipe of refractive index 1.50 hits the glass–air boundary at an angle of incidence of 40°. What happens to the ray?
Diagram not to scale
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Notes
The critical angle is sin−1(1/1.50) = 41.8°. The angle of incidence, 40°, is smaller, so there is no total internal reflection and the ray emerges, bending away from the normal: sin r = 1.50 × sin 40° = 0.964, so r is about 75°. A is wrong because 40° is below the critical angle. C uses sin r = sin 40°/1.50, which bends the ray towards the normal as if it entered glass. D ignores refraction.
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Answer B
Answer: B
Exam tip: Always compare the angle of incidence with the critical angle before deciding between refraction and total internal reflection.
341IGPC3-3-06r12-15Wave properties, reflection, refraction and diffractionPaper EHard8 marks
Water waves travel across a reservoir from a deep region into a shallow region. In the deep region the waves have frequency 4.0 Hz and wavelength 2.5 cm. In the shallow region the wavelength is 1.5 cm.
Diagram not to scale
(a)
Calculate the speed of the waves in the deep region.
(2)
(b)
State the frequency of the waves in the shallow region and calculate their speed there.
(2)
(c)
Calculate the percentage decrease in the wave speed.
(2)
(d)
Waves meet the boundary at an angle. Explain why the wavefronts change direction.
(2)
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Part (a)
v = f λ
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v = 4.0 × 0.025 = 0.10 m/s
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Part (b)
Frequency is unchanged at 4.0 Hz
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v = 4.0 × 0.015 = 0.060 m/s
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Part (c)
(0.10 − 0.060) / 0.10
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40% (ECF from (a) and (b))
✓ 1
Part (d)
The part of each wavefront that reaches the shallow region first slows down
✓ 1
while the rest still travels faster in the deep region, so the wavefront pivots and changes direction
✓ 1
Common error: When waves slow down at a boundary, the frequency stays the same and the wavelength shortens.
342IGPC3-3-06r12-16Sound and ultrasoundPaper EHard8 marks
At a gaming event, a loudspeaker is 85 m from a gamer. A wireless headset worn by the gamer receives radio signals from a console also 85 m away.
Diagram not to scale
(a)
A clap through the loudspeaker is heard 0.25 s after it is made. Calculate the speed of sound in air.
(2)
(b)
Calculate the time taken for a radio signal to travel 85 m. The speed of radio waves is 3.0 × 108 m/s. Give your answer to 2 significant figures.
(2)
(c)
State one way in which radio waves differ from sound waves, and one way they are the same.
(2)
(d)
Calculate how many times greater the speed of radio waves is than the speed of sound. Use 340 m/s for sound and give your answer in standard form to 2 significant figures.
(2)
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Part (a)
v = d / t
✓ 1
v = 85 / 0.25 = 340 m/s
✓ 1
Part (b)
t = d / v
✓ 1
t = 85 / 3.0 × 108 = 2.8 × 10−7 s
✓ 1
Part (c)
Radio waves are transverse and can travel through a vacuum, whereas sound is longitudinal and needs a medium
✓ 1
Both transfer energy and can be reflected, refracted and diffracted
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Part (d)
ratio = 3.0 × 108 / 340
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= 8.8 × 105
✓ 1
Exam tip: Compare speeds as a ratio by dividing the larger by the smaller.
343IGPC3-3-06r14-12Light, refraction, lenses and dispersionPaper EHard8 marks
A communications cable alongside a railway contains an optical fibre. The glass core of the fibre has a refractive index of 1.48. Light signals travel along the fibre by total internal reflection at the boundary between the core and the outside.
Diagram not to scale
(a)
State two conditions needed for total internal reflection.
(2)
(b)
Calculate the critical angle for the glass-air boundary.
(3)
(c)
A ray meets the boundary at an angle of incidence of 38°. State whether the ray stays in the fibre. Explain your answer.
(2)
(d)
Another ray meets the boundary at an angle of incidence of 50°. Calculate the angle between the reflected ray and the boundary surface.
(1)
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Notes
Part (a)
light travels from a denser medium towards a less dense medium
✓ 1
the angle of incidence is greater than the critical angle
✓ 1
Part (b)
sin c = 1 ÷ n
✓ 1
c = asin(1 ÷ 1.48)
✓ 1
c = 42.5°
✓ 1
Part (c)
no, it does not stay in the fibre (ECF from (b))
✓ 1
because 38° is less than the critical angle, so some light is refracted out
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Part (d)
the angle of reflection is 50° from the normal, so the angle to the surface is 90 − 50 = 40°
✓ 1
Common error: Angles of incidence and reflection are measured from the normal, not from the surface.
344IGPC3-3-06r14-13Sound and ultrasoundPaper EHard7 marks
A physiotherapist sends a short pulse of ultrasound into a patient's thigh. The echo from the top surface of the thigh muscle returns to the probe after 40 μs. The echo from the surface of the bone below the muscle returns after 100 μs. Assume the speed of the ultrasound is 1500 m/s in all of the tissue.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
Calculate the depth of the top surface of the muscle below the probe.
(2)
(c)
Calculate the thickness of the muscle layer, in cm.
(3)
(d)
Gel is spread on the skin before the probe is used. Suggest why.
(1)
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Notes
Part (a)
sound with a frequency higher than the upper limit of human hearing (above 20 000 Hz)
✓ 1
Part (b)
d = vt ÷ 2 = 1500 × 40 × 10−6 ÷ 2
✓ 1
d = 0.030 m (3.0 cm)
✓ 1
Part (c)
the extra time taken to cross the muscle and back is 100 − 40 = 60 μs (or the depth of the bone is 7.5 cm)
✓ 1
thickness = 1500 × 60 × 10−6 ÷ 2 (or 7.5 − 3.0)
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thickness = 4.5 cm (0.045 m)
✓ 1
Part (d)
it removes air between the probe and the skin, because air would reflect most of the ultrasound (so more enters the body)
✓ 1
Exam tip: Remember that the time given is for the journey there and back; only half of it applies to the depth.
345IGPC3-3-06r19-08Light, refraction, lenses and dispersionPaper EHard1 mark
A ray of light in air hits a flat glass panel in a building facade at an angle of incidence of 45°. The refractive index of the glass is 1.50. What is the angle of refraction in the glass?
Diagram not to scale
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Notes
Using n = sin i ÷ sin r: sin r = sin 45° ÷ 1.50 = 0.471, so r = 28.1° (B). A divides the angle by n (45 ÷ 1.5). C is the angle between the refracted ray and the surface (90° − 28.1°). D assumes the ray is not deviated.
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Answer B
Answer: B
Common error: Dividing the angle of incidence by the refractive index; the refractive index divides the sines of the angles, not the angles themselves.
346IGPC3-3-06r19-16Light, refraction, lenses and dispersionPaper EHard8 marks
A marine biologist uses an underwater lamp to study fish at night. The lamp is in sea water of refractive index 1.34 and shines light upward at the surface. The speed of light in air is 3.0 × 108 m/s.
Diagram not to scale
(a)
Calculate the speed of light in the sea water. Give your answer to 3 significant figures.
(2)
(b)
A ray in the water reaches the surface at an angle of incidence of 30°. Calculate the angle of refraction in the air. Give your answer to 3 significant figures.
(2)
(c)
Calculate the critical angle for the water-air boundary. Give your answer to 3 significant figures.
(2)
(d)
A ray in the water reaches the surface at an angle of incidence of 55°. State what happens and explain your answer.
(2)
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Part (a)
speed = c ÷ n = 3.0 × 108 ÷ 1.34
✓ 1
2.24 × 108 m/s
✓ 1
Part (b)
sin r = 1.34 × sin 30° = 0.670
✓ 1
r = 42.1°
✓ 1
Part (c)
sin c = 1 ÷ 1.34
✓ 1
c = 48.3°
✓ 1
Part (d)
it is totally internally reflected
✓ 1
because 55° is greater than the critical angle of 48.3°
✓ 1
Exam tip: When light goes from water into air, use n = sin(angle in water) ÷ sin(angle in air) with the water value on the top line (or invert it in your calculation), and check the ray bends away from the normal.
347IGPC3-3-08r23-13Wave properties, reflection, refraction and diffractionPaper EHard8 marks
A solar-energy company tests a transparent plastic lens. A ray of light in air strikes the plastic at an angle of incidence of 42° and refracts at an angle of 27°. Use the refractive index to 3 significant figures in later parts.
Diagram not to scale
(a)
Calculate the refractive index of the plastic.
(2)
(b)
The speed of light in air is 3.0 × 108 m/s. Calculate the speed of light in the plastic. Give your answer to 3 significant figures.
(2)
(c)
Calculate the critical angle for the plastic.
(2)
(d)
A ray inside the plastic strikes the plastic-air surface at 50°. Describe what happens to the ray.
(2)
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Part (a)
n = sin i / sin r
✓ 1
= sin 42° / sin 27° = 1.47
✓ 1
Part (b)
v = c / n
✓ 1
= 3.0 × 108 / 1.47 = 2.04 × 108 m/s
✓ 1
Part (c)
sin c = 1 / n
✓ 1
c = 42.9°
✓ 1
Part (d)
50° is greater than the critical angle
✓ 1
The ray undergoes total internal reflection, so none leaves the plastic
✓ 1
Common error: Total internal reflection only occurs when light travels from a more dense medium towards a less dense one.
348IGPC3-3-08r23-14Light, refraction, lenses and dispersionPaper EHard9 marks
A retailer uses a thin converging lens in a slide projector. The slide is 2.0 cm tall and the image on the screen is 15 cm tall.
(a)
Describe what happens to a ray of light that travels parallel to the principal axis towards the lens.
(2)
(b)
A magnifying glass is a thin converging lens. State three features of the image seen when the object is closer to the lens than the principal focus.
(3)
(c)
Calculate the magnification of the projector.
(2)
(d)
The magnification also equals image distance divided by object distance. The screen is 3.0 m from the lens. Calculate the distance from the slide to the lens.
(2)
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Notes
Part (a)
The ray is refracted by the lens
✓ 1
It passes through the principal focus on the other side
✓ 1
Part (b)
Virtual
✓ 1
Upright
✓ 1
Magnified
✓ 1
Part (c)
magnification = image height / object height
✓ 1
= 15 / 2.0 = 7.5
✓ 1
Part (d)
object distance = image distance / magnification
✓ 1
= 3.0 / 7.5 = 0.40 m
✓ 1
Exam tip: A magnification has no unit; make sure both heights are in the same unit before dividing.
349IGPC3-3-06r26-07Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Water waves travel at 6.0 m/s in a deep channel and at 4.0 m/s in a shallower region. A wave crosses the boundary between the two regions. In the deep channel the angle between the direction of travel and the normal to the boundary is 48°. What is the angle between the direction of travel and the normal in the shallow region?
Diagram not to scale
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Notes
For waves, sin i ÷ sin r = v₁ ÷ v₂. So sin r = sin 48° × 4.0 ÷ 6.0 = 0.495, giving r = 29.7°. 32.0° assumes the angle is proportional to the speed (48° × 4 ÷ 6). 72.0° multiplies by the inverted speed ratio and ignores the sines. 19.3° uses the speed ratio squared.
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Answer A
Answer: A
Common error: The angles are not in proportion to the speeds: the sines of the angles are.
350IGPC3-3-06r26-08Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Light of frequency 5.0 × 1014 Hz passes from air into a glass filter in a stage lamp. The speed of light in air is 3.0 × 108 m/s and in the glass it is 2.0 × 108 m/s. Which row gives the frequency and wavelength of the light in the glass?
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Notes
The frequency is set by the source and does not change at the boundary. The wavelength in glass = speed ÷ frequency = 2.0 × 108 ÷ 5.0 × 1014 = 4.0 × 10−7 m. The first alternative has the frequency falling and the wavelength unchanged, which is the wrong quantity changing. The second alternative gives the wavelength in air, so it ignores the slower speed. The last alternative has a frequency that increases, which does not happen when light is refracted.
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Answer C
Answer: C
Exam tip: On crossing a boundary, frequency stays the same; speed and wavelength change together.
351IGPC3-3-06r26-16Wave properties, reflection, refraction and diffractionPaper EHard9 marks
Water waves with a frequency of 0.40 Hz travel at 5.0 m/s across deep water towards a harbour. The waves then enter shallower water, where their speed is 3.2 m/s. In the deep water, the direction of travel of the waves makes an angle of 52° with the normal to the boundary between the deep and the shallow water.
Diagram not to scale
(a)
Calculate the wavelength of the waves in the deep water.
(2)
(b)
Calculate the wavelength of the waves in the shallow water.
(2)
(c)
Explain why the wavelength decreases when the frequency stays the same.
(2)
(d)
Calculate the angle between the direction of travel of the waves in the shallow water and the normal. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
λ = v ÷ f = 5.0 ÷ 0.40
✓ 1
12.5 m
✓ 1
Part (b)
frequency is unchanged, so λ = 3.2 ÷ 0.40
✓ 1
8.0 m
✓ 1
Part (c)
the frequency is determined by the source, so the number of waves arriving each second does not change
✓ 1
v = fλ, so when the speed decreases at constant frequency the wavelength decreases
✓ 1
Part (d)
sin i ÷ sin r = v₁ ÷ v₂, so sin r = sin 52° × 3.2 ÷ 5.0
✓ 1
sin r = 0.504
✓ 1
r = 30.3°
✓ 1
Common error: The waves bend towards the normal when they slow down; check that your angle in the slower region is smaller.
352IGPC3-3-07r29-08Wave properties, reflection, refraction and diffractionPaper EHard1 mark
A telecoms engineer tests a 2.4 GHz signal and a 5.0 GHz signal passing through a doorway 0.90 m wide. Both travel at 3.0 × 108 m/s. Which statement about diffraction at the doorway is correct?
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Notes
The wavelengths are 3.0 × 108 / 2.4 × 109 = 0.125 m and 3.0 × 108 / 5.0 × 109 = 0.060 m. Diffraction is greatest when the wavelength is comparable to or larger than the gap, so the longer wavelength spreads more. Energy does not control the amount of diffraction, the speed being equal does not make the wavelengths equal, and shorter wavelengths diffract less, not more.
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Answer A
Answer: A
Common error: More diffraction needs a longer wavelength or a narrower gap, so compare wavelength with gap width.
353IGPC3-3-07r29-13Light, refraction, lenses and dispersionPaper EHard9 marks
An optical fibre used in a telecommunications cable has a glass core of refractive index 1.52. The core is surrounded by cladding of lower refractive index.
Diagram not to scale
(a)
State two conditions needed for total internal reflection to occur at a boundary.
(2)
(b)
Calculate the critical angle for a boundary between the glass core and air.
(3)
(c)
A ray inside a bare section of the core meets the glass–air boundary at an angle of incidence of 38°. State what happens to most of the light and explain your answer.
(2)
(d)
Explain why the core is covered with cladding of lower refractive index.
(2)
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Notes
Part (a)
Light travels from a more dense medium to a less dense medium
✓ 1
The angle of incidence is greater than the critical angle
✓ 1
Part (b)
M1 sin c = 1 ÷ n
✓ 1
M1 sin c = 1 ÷ 1.52
✓ 1
A1 41.1°
✓ 1
Part (c)
Most of the light refracts out into the air (accept: total internal reflection does not occur)
✓ 1
The angle of incidence (38°) is less than the critical angle (ECF from (b))
✓ 1
Part (d)
So that total internal reflection can occur at the core–cladding boundary
✓ 1
Any one of: protects the surface from scratches; stops light leaking between neighbouring fibres
✓ 1
Common error: Total internal reflection needs both conditions; a large angle alone is not enough if light is going from less dense to more dense.
354IGPC3-3-10r32-08Light, refraction, lenses and dispersionPaper EHard1 mark
Light travelling in glass of refractive index 1.50 strikes the boundary with air at an angle of incidence of 30°. What is the angle of refraction in the air?
Diagram not to scale
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Notes
Light moving from glass to air obeys n sin i = sin r, so sin r = 1.50 × sin 30° = 0.75 and r = 48.6°. 19.5° divides by n, which is the calculation for light entering the glass. 45° multiplies the angle by 1.50 instead of using sines. 30° assumes no bending occurs at the boundary.
✓ 1
Answer B
Answer: B
Exam tip: Going from the denser to the less dense medium the ray bends away from the normal, so the angle in air must be larger than the angle in glass.
355IGPC3-3-10r32-15Sound and ultrasoundPaper EHard7 marks
An ultrasound scanner is used on a farm to measure the thickness of the layer of fat on a pig. The ultrasound has a frequency of 3.0 MHz and travels in body tissue at 1540 m/s. Echoes from the top and bottom surfaces of the fat layer return to the scanner after 3.9 × 10−5 s and 6.5 × 10−5 s respectively.
Diagram not to scale
(a)
Calculate the wavelength of the ultrasound in the tissue, to 2 significant figures.
(2)
(b)
Calculate the thickness of the layer of fat.
(3)
(c)
Suggest two reasons why ultrasound is more suitable than audible sound for this scan.
(2)
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Notes
Part (a)
M1 λ = v ÷ f = 1540 ÷ (3.0 × 106)
✓ 1
A1 5.1 × 10−4 m
✓ 1
Part (b)
M1 time difference between the echoes = 6.5 × 10−5 − 3.9 × 10−5 = 2.6 × 10−5 s
✓ 1
M1 thickness = speed × time difference ÷ 2, since the pulse crosses the layer twice
✓ 1
A1 0.0200 m (2.00 cm)
✓ 1
Part (c)
ultrasound has a short wavelength so it diffracts less and can reveal small features / thin layers
✓ 1
ultrasound has a frequency above 20 kHz, so it cannot be heard and does not disturb the animal or the operator
✓ 1
Exam tip: With two echoes, use the difference in arrival times; the time to reach the first surface does not belong to the layer.
356IGPC3-3-10r32-16Light, refraction, lenses and dispersionPaper EHard8 marks
A ray of white light in an aircraft's cockpit display strikes a parallel-sided glass block at an angle of incidence of 55°. For one colour in the ray the refractive index of the glass is 1.52. The speed of light in a vacuum is 3.00 × 108 m/s.
Diagram not to scale
(a)
Calculate the angle of refraction in the glass for this colour.
(2)
(b)
State the angle at which the ray emerges into the air from the opposite parallel face, and give a reason.
(2)
(c)
Explain why white light passing through a glass prism is dispersed into colours.
(2)
(d)
Calculate the speed of light in the glass for this colour.
(2)
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Notes
Part (a)
M1 sin r = sin 55° ÷ 1.52
✓ 1
A1 32.6°
✓ 1
Part (b)
55°
✓ 1
the faces are parallel, so the angle of incidence at the second face equals the angle of refraction at the first, and the ray bends back by the same amount
✓ 1
Part (c)
different colours (wavelengths) travel at different speeds in glass, so the glass has a different refractive index for each colour
✓ 1
so each colour is refracted by a different amount, with violet bent most and red least
✓ 1
Part (d)
M1 n = c ÷ v, so v = c ÷ n = 3.00 × 108 ÷ 1.52
✓ 1
A1 1.97 × 108 m/s
✓ 1
Common error: Refractive index n = c ÷ v, so light is slower in glass than in air or a vacuum, never faster.
357IGPC3-3-09r34-07Sound and ultrasoundPaper EHard1 mark
A survey ship sends a sound wave of frequency 500 Hz from air into water. The speed of sound is 340 m/s in air and 1500 m/s in water. What is the wavelength of the sound in water?
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Notes
The frequency stays at 500 Hz, so wavelength = 1500 ÷ 500 = 3.0 m. 0.68 m is the wavelength in air. 0.15 m inverts the speed ratio. 4.4 m is the ratio of the speeds, not a wavelength.
✓ 1
Answer D
Answer: D
Common error: Keep the frequency fixed and divide the new speed by it; do not scale the wavelength the wrong way round.
358IGPC3-3-09r34-14Sound and ultrasoundPaper EHard8 marks
An ultrasonic scanner is used to measure the thickness of a stone wall. The scanner sends a short pulse of ultrasound of frequency 250 kHz into one face of the wall. The speed of sound in the stone is 2800 m/s.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
Calculate the wavelength of the ultrasound in the stone. Give your answer to 3 significant figures.
(2)
(c)
An echo from the opposite face of the wall returns after 90 μs. Calculate the thickness of the wall.
(3)
(d)
A second wall is 0.40 m thick. Calculate the time between the pulse being sent and its echo returning. Give your answer in seconds to 3 significant figures.
(2)
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Notes
Part (a)
Sound with a frequency higher than 20 000 Hz, the upper limit of human hearing
✓ 1
Part (b)
M1 λ = v ÷ f = 2800 ÷ 250 000
✓ 1
A1 0.0112 m
✓ 1
Part (c)
M1 distance travelled = speed × time = 2800 × 90 × 10−6 (= 0.252 m)
✓ 1
M1 divide by 2 because the pulse crosses the wall twice
✓ 1
A1 0.126 m
✓ 1
Part (d)
M1 t = 2 × 0.40 ÷ 2800
✓ 1
A1 2.86 × 10−4 s
✓ 1
Common error: Remember that the pulse travels twice the thickness; convert microseconds with ×10⁻⁶.
359IGPC3-3-09r34-16Wave properties, reflection, refraction and diffractionPaper EHard7 marks
In a ripple tank model of an ancient harbour, water waves of wavelength 0.060 m travel at 0.30 m/s in the deep region. They then pass into a shallow region where their speed is 0.18 m/s.
Diagram not to scale
(a)
Calculate the frequency of the waves in the deep region.
(2)
(b)
Calculate the wavelength of the waves in the shallow region.
(2)
(c)
State why the frequency is the same in the shallow region.
(1)
(d)
The wavefronts meet the boundary at an angle. Describe and explain what happens to their direction.
(2)
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Notes
Part (a)
M1 f = v ÷ λ = 0.30 ÷ 0.060
✓ 1
A1 5.0 Hz
✓ 1
Part (b)
M1 λ = v ÷ f = 0.18 ÷ 5.0
✓ 1
A1 0.036 m, ECF from (a)
✓ 1
Part (c)
The frequency is set by the source, so the same number of waves per second leave the deep region as arrive in the shallow region
✓ 1
Part (d)
The wavefronts change direction, bending towards the normal
✓ 1
Because the waves slow down in the shallow region
✓ 1
Common error: When speed changes, frequency stays fixed and the wavelength changes.
360IGPC3-3-10r34-07Light, refraction, lenses and dispersionPaper EHard1 mark
Light travels inside a glass fibre used to inspect a bridge. The refractive index of the glass is 1.45. The light meets the glass–air surface at an angle of incidence of 42°. What happens to the light?
Diagram not to scale
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Notes
The critical angle is sin−1(1/1.45) = 43.6°. The angle of incidence, 42°, is less than this, so there is no total internal reflection and most light refracts out, bending away from the normal as it enters the less dense medium. Travelling along the surface would need exactly 43.6°. Bending towards the normal would happen only going into a denser medium.
✓ 1
Answer B
Answer: B
Common error: Always calculate the critical angle before deciding; an angle close to it does not guarantee total internal reflection.
361IGPC3-3-10r34-14Sound and ultrasoundPaper EHard8 marks
An ultrasonic sensor above a hopper of coffee beans measures how far down the beans are. The sensor sends pulses of frequency 40 kHz. The speed of sound in air is 340 m/s.
Diagram not to scale
(a)
Calculate the wavelength of the ultrasound in air.
(2)
(b)
An echo from the beans returns 5.0 ms after the pulse is sent. Calculate the distance from the sensor to the beans.
(2)
(c)
More beans are added until the surface is 0.60 m from the sensor. Calculate the new time for the echo to return. Give your answer in seconds to 3 significant figures.
(2)
(d)
Suggest two reasons why ultrasound is used rather than sound that people can hear.
(2)
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Notes
Part (a)
M1 λ = v ÷ f = 340 ÷ 40 000
✓ 1
A1 8.5 × 10−3 m
✓ 1
Part (b)
M1 distance = 340 × 5.0 × 10−3 ÷ 2
✓ 1
A1 0.85 m
✓ 1
Part (c)
M1 t = 2 × 0.60 ÷ 340
✓ 1
A1 3.53 × 10−3 s
✓ 1
Part (d)
It cannot be heard, so it does not disturb workers
✓ 1
Its short wavelength means less diffraction, giving a narrower, more precise beam
✓ 1
Common error: Use the total path (there and back) when converting between distance and echo time.
362IGPC3-3-10r34-16Wave properties, reflection, refraction and diffractionPaper EHard7 marks
Water waves in a harbour pass through a gap in a breakwater beside a bridge. The waves travel at 1.6 m/s and have frequency 0.80 Hz.
(a)
Calculate the wavelength of the waves.
(2)
(b)
The gap is 0.50 m wide. Describe and explain what happens to the waves as they pass through it.
(2)
(c)
A different gap is 10 m wide. State what happens to the waves.
(1)
(d)
State what happens to the wavelength and the speed of the waves after they pass through a gap.
(2)
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Notes
Part (a)
M1 λ = v ÷ f = 1.6 ÷ 0.80
✓ 1
A1 2.0 m
✓ 1
Part (b)
The waves spread out into the region beyond the gap, with circular wavefronts
✓ 1
because the gap is smaller than the wavelength
✓ 1
Part (c)
Little spreading; the waves mostly continue straight through, spreading only slightly at the edges
✓ 1
Part (d)
The wavelength is unchanged
✓ 1
The speed is unchanged
✓ 1
Exam tip: Diffraction changes the direction of the waves, not their speed, frequency or wavelength.
363IGPC3-3-09r36-07Light, refraction, lenses and dispersionPaper EHard1 mark
Light travelling inside a block of glass of refractive index 1.60 meets the boundary with air at an angle of incidence of 42°. What happens to the light?
Diagram not to scale
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Notes
The critical angle satisfies sin c = 1/1.60 = 0.625, so c = 38.7°. The angle of incidence, 42°, is greater than this, so total internal reflection occurs and the angle of reflection equals the angle of incidence, 42°. Refraction into the air, or along the boundary, only happens at or below the critical angle. The 48° option measures the angle from the boundary instead of from the normal.
✓ 1
Answer C
Answer: C
Common error: Angles of incidence and reflection are measured from the normal, not from the surface of the boundary.
364IGPC3-3-09r36-08Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Light of frequency 5.0×1014 Hz travels in air at 3.0×108 m/s and enters glass of refractive index 1.5. What is the wavelength of the light in the glass?
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Notes
The speed in the glass is 3.0×108 / 1.5 = 2.0×108 m/s and the frequency is unchanged, so the wavelength is 2.0×108 / 5.0×1014 = 4.0×10−7 m. 6.0×10−7 m is the wavelength in air, which ignores the change of speed. 9.0×10−7 m multiplies by the refractive index instead of dividing. 2.7×10−7 m assumes the frequency increases by the factor 1.5.
✓ 1
Answer A
Answer: A
Exam tip: Find the new speed from n = c/v first, then use the unchanged frequency in v = fλ.
365IGPC3-3-09r36-16Light, refraction, lenses and dispersionPaper EHard7 marks
A ray of light in air strikes one face of a rectangular glass block of refractive index 1.52 at an angle of incidence of 60°. The opposite face of the block is parallel to the first face.
Diagram not to scale
(a)
Calculate the angle of refraction at the first face. Give your answer to 3 significant figures.
(2)
(b)
State the angle at which the ray leaves the block through the opposite face, and explain your answer.
(2)
(c)
Calculate the critical angle of the glass. Give your answer to 3 significant figures.
(2)
(d)
Explain why the ray does not undergo total internal reflection at the second face.
(1)
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Notes
Part (a)
n = sin i / sin r, so sin r = sin 60° / 1.52
✓ 1
r = 34.7°
✓ 1
Part (b)
60°
✓ 1
the faces are parallel so the angle of incidence at the second face is 34.7° and the ray follows the reverse path, so the emerging ray is parallel to the incident ray
✓ 1
Part (c)
sin c = 1 / n
✓ 1
c = sin−1(1 / 1.52) = 41.1°
✓ 1
Part (d)
the angle of incidence (34.7°) is smaller than the critical angle (41.1°) (ECF from (a) and (c))
✓ 1
Exam tip: For total internal reflection compare the angle of incidence with the critical angle: it needs to be larger and the light must be travelling in the denser material.
366IGPC3-3-09r37-08Light, refraction, lenses and dispersionPaper EHard1 mark
A ray of light in air meets the flat surface of a pot of honey of refractive index 1.49. The angle of incidence is 50.0°. What is the angle of refraction in the honey?
Diagram not to scale
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Notes
Using n = sin i ÷ sin r, sin r = sin 50.0° ÷ 1.49 = 0.514, so r = 30.9°. 74.5° multiplies the angle by the refractive index, which is not allowed because the law connects the sines, and the ray also bends towards the normal so r must be smaller than i. 33.6° divides the angle itself, 50.0° ÷ 1.49, rather than the sine. 20.2° divides by the refractive index twice (n squared).
✓ 1
Answer B
Answer: B
Common error: Snell's law relates the sines of the angles, so do not divide or multiply the angles themselves by the refractive index.
367IGPC3-3-09r37-15Light, refraction, lenses and dispersionPaper EHard8 marks
An endoscope used in a hospital contains thin glass fibres. The glass has a refractive index of 1.52. For this question assume that each fibre is surrounded by air. The speed of light in air is 3.00 × 108 m/s.
(a)
Calculate the critical angle for light travelling from the glass into air. Give your answer to 3 significant figures.
(2)
(b)
A ray of light inside a fibre meets the wall at an angle of incidence of 52°. State what happens to the ray at the wall and explain why.
(2)
(c)
Another ray meets the wall at an angle of incidence of 30°. Describe what happens to this ray at the wall.
(2)
(d)
Calculate the speed of light in the glass. Give your answer to 3 significant figures.
(2)
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Part (a)
sin c = 1 ÷ n = 1 ÷ 1.52
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c = 41.1°
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Part (b)
the ray is totally internally reflected
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because the angle of incidence is greater than the critical angle (ECF from (a))
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Part (c)
most of the light refracts out of the glass into the air, bending away from the normal
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a small amount is partially reflected back into the glass
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Part (d)
speed in glass = speed in air ÷ refractive index = 3.00 × 108 ÷ 1.52
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1.97 × 108 m/s
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Exam tip: Total internal reflection needs two conditions: light moving towards a less dense medium, and an angle of incidence greater than the critical angle.
368IGPC3-3-08r40-08Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Water waves in a swimming pool travel from the deep end to the shallow end. In the deep end the waves have speed 0.30 m/s and wavelength 0.060 m. In the shallow end the wavelength is 0.040 m. What is the wave speed in the shallow end?
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The frequency is set by the source and does not change, so speed is proportional to wavelength: 0.30 × 0.040 ÷ 0.060 = 0.20 m/s. 0.30 m/s assumes the speed does not change. 0.45 m/s uses the ratio of wavelengths upside down. 0.10 m/s comes from halving the speed without using the wavelengths.
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Answer D
Answer: D
Common error: When a wave is refracted its frequency stays the same; it is the speed and wavelength that change together.
369IGPC3-3-08r40-15Light, refraction, lenses and dispersionPaper EHard7 marks
A narrow beam of white light is directed at 45° to the normal onto a glass prism. The refractive index of the glass is 1.51 for red light and 1.53 for violet light. The speed of light in air is 3.00 × 108 m/s.
(a)
State what is meant by dispersion.
(1)
(b)
Calculate the speed of red light and the speed of violet light in the glass. Give each answer to 3 significant figures.
(2)
(c)
Calculate the angle of refraction in the glass for red light and for violet light. Give each angle to 3 significant figures.
(2)
(d)
Use your answers to explain why violet light is deviated more than red light by the prism.
(2)
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Part (a)
the splitting of white light into its different colours (spectrum) because each colour is refracted by a different amount
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Part (b)
speed = c ÷ n
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red 1.99 × 108 m/s and violet 1.96 × 108 m/s
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Part (c)
sin r = sin 45° ÷ n
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red 27.9° and violet 27.5°
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Part (d)
violet light travels more slowly in the glass than red light / violet has the larger refractive index
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so violet bends more towards the normal (smaller angle of refraction) and is deviated more
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Exam tip: Link the idea of dispersion to speed: the colour that slows down more in the glass is the one that is bent more.
370IGPC3-3-09r42-07Wave properties, reflection, refraction and diffractionPaper EHard1 mark
A sound wave of frequency 1200 Hz travels from air, where its speed is 340 m/s, into water, where its speed is 1500 m/s. Which row gives its frequency and wavelength in the water?
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Frequency is set by the source and does not change at a boundary, so it stays at 1200 Hz. The wavelength is speed ÷ frequency = 1500 ÷ 1200 = 1.25 m. The rows with 0.28 m use the speed in air (340 ÷ 1200). The 5300 Hz row assumes frequency scales with speed while the wavelength stays fixed, and the 270 Hz row assumes the frequency falls when the wave slows.
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Answer C
Answer: C
Common error: When a wave crosses a boundary, the frequency stays the same; speed and wavelength change together.
371IGPC3-3-09r42-14Wave properties, reflection, refraction and diffractionPaper EHard9 marks
A ripple tank has a deep region and a shallow region. A vibrating dipper produces plane waves of frequency 5.0 Hz. In the deep region the wavelength is 3.2 cm. In the shallow region the wavelength is 2.0 cm.
Diagram not to scale
(a)
Calculate the speed of the waves in the deep region.
(2)
(b)
State the frequency of the waves in the shallow region.
(1)
(c)
Calculate the speed of the waves in the shallow region. Use your answer to (b).
(2)
(d)
The waves approach the boundary between the regions at an angle. State and explain the change in direction of the wavefronts as they enter the shallow region.
(2)
(e)
A barrier with a gap of width 1.0 cm is placed in the deep region. Describe what happens to the waves that pass through the gap and explain why.
(2)
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Part (a)
v = f × λ
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= 5.0 × 3.2 = 16 cm/s
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Part (b)
5.0 Hz
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Part (c)
v = f × λ
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= 5.0 × 2.0 = 10 cm/s (ECF from (b))
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Part (d)
wavefronts bend towards the normal
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because the waves slow down / speed decreases in shallow water
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Part (e)
waves spread out into semicircular wavefronts beyond the gap (diffraction)
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the gap is smaller than the wavelength (3.2 cm), so diffraction is significant
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Common error: Frequency is fixed by the source; only the speed and wavelength change when the depth changes.
372IGPC3-3-09r42-15Light, refraction, lenses and dispersionPaper EHard9 marks
In a gaming headset, a ray of light in air strikes a flat plastic surface with an angle of incidence of 55°. It refracts at an angle of 33°. The speed of light in a vacuum is 3.00 × 108 m/s.
Diagram not to scale
(a)
Calculate the refractive index of the plastic.
(2)
(b)
Calculate the speed of light in the plastic. Use n = 1.50.
(2)
(c)
Calculate the critical angle for the plastic–air boundary. Use n = 1.50.
(3)
(d)
A ray inside the plastic reaches the plastic–air boundary with an angle of incidence of 50°. State and explain what happens to the ray.
(2)
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Part (a)
n = sin i ÷ sin r
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= sin 55° ÷ sin 33° = 1.50
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Part (b)
v = c ÷ n
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= 3.00 × 108 ÷ 1.50 = 2.00 × 108 m/s
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Part (c)
sin c = 1 ÷ n
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sin c = 1 ÷ 1.50 = 0.667
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c = 41.8°
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Part (d)
total internal reflection occurs
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50° is greater than the critical angle of 41.8° (ECF from (c))
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Common error: Use the refractive index from the stem when told to; do not mix rounded and unrounded values.
373IGPC3-3-08r43-12Light, refraction, lenses and dispersionPaper EHard7 marks
Optical fibres carry data between library buildings. The glass core of the fibre has a refractive index of 1.50.
Diagram not to scale
(a)
State two conditions needed for total internal reflection at the boundary of the core.
(2)
(b)
Calculate the critical angle for the core.
(3)
(c)
A ray inside the core meets the boundary at an angle of incidence of 50°. State what happens to the ray and explain your answer.
(2)
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Notes
Part (a)
light travelling in the optically denser material (glass) towards a less dense material
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angle of incidence greater than the critical angle
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Part (b)
M1 sin c = 1 ÷ n
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M1 c = sin−1(1 ÷ 1.50)
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A1 41.8°
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Part (c)
the ray is totally internally reflected
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because 50° is greater than the critical angle (41.8°, ECF from (b))
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Exam tip: Total internal reflection needs both conditions: dense to less dense, and an angle above the critical angle.
374IGPC3-3-08r43-15Sound and ultrasoundPaper EHard8 marks
An offshore wind farm is built by driving steel piles into the seabed. The noise from a pile-driver travels 2040 m to a boat. Sound can reach the boat through the air, and also through the water. Use 340 m/s for the speed of sound in air and 1500 m/s for the speed of sound in water.
(a)
Calculate the time for the sound to travel 2040 m through the air.
(2)
(b)
Calculate the time for the sound to travel 2040 m through the water.
(2)
(c)
Calculate how much earlier the sound arrives through the water than through the air.
(2)
(d)
Explain why sound travels faster in water than in air.
(2)
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Part (a)
M1 time = distance ÷ speed = 2040 ÷ 340
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A1 6.0 s
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Part (b)
M1 time = 2040 ÷ 1500
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A1 1.36 s
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Part (c)
M1 6.0 − 1.36
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A1 4.64 s (ECF from (a) and (b))
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Part (d)
particles in water are much closer together
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so vibrations are passed from particle to particle more quickly
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Exam tip: For a time difference, work out each time separately and subtract; keep the units in seconds throughout.
375IGPC3-3-08r45-08Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Water waves in the deep end of a swimming pool have frequency 8.0 Hz and speed 2.0 m/s. They move into the shallow end, where their speed is 1.2 m/s. What is the wavelength in the shallow end?
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Notes
The frequency stays at 8.0 Hz, so the new wavelength is 1.2 ÷ 8.0 = 0.15 m. 0.25 m is the deep-end wavelength, which would apply only if the wavelength did not change. 0.42 comes from scaling the wavelength by 2.0/1.2 the wrong way round, and 0.09 from applying the speed ratio twice.
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Answer B
Answer: B
Common error: When waves enter a new medium, the frequency is fixed; find the new wavelength from λ = v / f.
376IGPC3-3-08r45-13Light, refraction, lenses and dispersionPaper EHard9 marks
A ray of light in a library display room passes from air into a glass block. The angle of incidence is 50° and the angle of refraction is 30°.
Diagram not to scale
(a)
Calculate the refractive index of the glass. Give your answer to 3 significant figures.
(3)
(b)
The speed of light in a vacuum is 3.0 × 108 m/s. Calculate the speed of light in the glass. Give your answer to 3 significant figures. (Accept use of either the rounded or the unrounded refractive index.)
(2)
(c)
Explain what is meant by total internal reflection.
(2)
(d)
A different glass has a refractive index of 1.60. Calculate its critical angle, to 3 significant figures.
(2)
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Notes
Part (a)
n = sin i ÷ sin r
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sin 50° ÷ sin 30°
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1.53
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Part (b)
v = c ÷ n
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1.96 × 108 m/s
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Part (c)
light travelling in a denser medium (glass) towards a less dense medium (air)
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angle of incidence greater than the critical angle, so all the light is reflected
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Part (d)
sin c = 1 ÷ n
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c = 38.7°
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Exam tip: Critical angle: sin c = 1/n. Remember it applies only for light going from the denser to the less dense medium.
377IGPC3-3-08r46-06Light, refraction, lenses and dispersionPaper EHard1 mark
A signal light is sent along a glass fibre beside a railway line. The refractive index of the glass is 1.50. What is the critical angle for light travelling from the glass to air?
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Notes
sin c = 1/n = 1/1.50 = 0.667, so c = 42°. The value 48° is 90° minus the critical angle and comes from using the wrong angle in the working. 34° results from using tan c = 1/n and 56° from using tan c = n, in both cases with the wrong function.
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Answer A
Answer: A
Exam tip: The critical angle satisfies sin c = 1/n; check that your answer is less than 90° and decreases as n increases.
378IGPC3-3-08r46-15Sound and ultrasoundPaper EHard9 marks
A sports scientist uses an ultrasound scanner to examine a muscle. The scanner emits short pulses and detects reflections. The speed of ultrasound in muscle is 1500 m/s.
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
An echo from a bone returns to the scanner 52 μs (5.2 × 10−5 s) after the pulse is emitted. Calculate the depth of the bone below the scanner. Give your answer in centimetres.
(3)
(c)
Explain why some of the pulse is reflected at the boundary between the muscle and the bone.
(2)
(d)
The scanner uses a frequency of 5.0 MHz. Calculate the wavelength of the ultrasound in the muscle.
(2)
(e)
Suggest why a scanner using a higher frequency can show smaller details.
(1)
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Notes
Part (a)
sound with a frequency higher than the upper limit of human hearing (above 20 000 Hz)
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Part (b)
distance travelled = speed × time = 1500 × 5.2 × 10−5 = 0.078 m
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depth = half the distance because the pulse travels there and back = 0.039 m
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3.9 cm
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Part (c)
the ultrasound meets a boundary between two media of different density
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part of the wave is reflected and part continues (transmitted)
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Part (d)
wavelength = speed ÷ frequency = 1500 ÷ 5.0 × 106
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3.0 × 10−4 m
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Part (e)
shorter wavelength, so smaller objects can be resolved (less diffraction)
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Exam tip: In echo problems the time recorded is for the journey to the object and back, so halve the distance.
379IGPC3-3-08r46-16Sound and ultrasoundPaper EHard9 marks
A railway engineer listens for an approaching train by pressing an ear to the steel rail. Sound from the train travels along the rail and through the air. The speed of sound is 5100 m/s in steel and 340 m/s in air.
(a)
State the type of wave that sound is, and describe how the particles of the medium move.
(1)
(b)
Explain why sound travels faster in steel than in air.
(2)
(c)
Calculate the time taken for sound to travel 255 m along the rail.
(2)
(d)
The sound arrives through the rail 0.75 s before it arrives through the air. Calculate the distance of the train from the engineer. Give your answer to 3 significant figures.
(4)
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Notes
Part (a)
longitudinal, with the particles vibrating parallel to the direction the wave travels
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Part (b)
the particles in steel are much closer together (stronger forces between them)
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so vibrations are passed on more quickly from particle to particle
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Part (c)
time = distance ÷ speed = 255 ÷ 5100
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0.050 s
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Part (d)
time in air = d ÷ 340 and time in rail = d ÷ 5100
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difference in times: d/340 − d/5100 = 0.75
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d (1/340 − 1/5100) = 0.75 so d = 0.75 ÷ 0.0027451
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d = 273 m
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Common error: Both times depend on the same unknown distance, so write them in terms of d and set up one equation using the difference.
380IGPC3-3-09r46-13Light, refraction, lenses and dispersionPaper EHard7 marks
Data from the turbines in an offshore wind farm are sent to a control centre along optical fibres. The glass core of a fibre has refractive index 1.52.
Diagram not to scale
(a)
Calculate the critical angle for the glass core in air. Give your answer to 3 significant figures.
(3)
(b)
State the two conditions needed for total internal reflection to occur at a boundary.
(2)
(c)
A ray inside the core meets the core boundary at an angle of incidence of 52°. Explain what happens to the ray.
(2)
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Notes
Part (a)
M1 n = 1 ÷ sin c
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M1 sin c = 1 ÷ 1.52 = 0.658
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A1 c = 41.1°
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Part (b)
B1 light travels in the more dense medium towards the less dense medium
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B1 angle of incidence is greater than the critical angle
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Part (c)
B1 52° is greater than the critical angle (41.1°)
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B1 so the ray is totally internally reflected, staying in the core, at an angle of reflection of 52°
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Exam tip: Quote the value of the critical angle and compare it with the angle of incidence in your explanation.
381IGPC3-3-09r48-07Wave properties, reflection, refraction and diffractionPaper EHard1 mark
Red light of wavelength 600 nm in air enters a glass of refractive index 1.5. What is the wavelength of the light in the glass?
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The frequency stays the same but the speed falls by a factor of 1.5, so the wavelength also falls by 1.5: 600 ÷ 1.5 = 400 nm. 900 nm multiplies instead of dividing. 600 nm assumes the wavelength is unchanged because the frequency is unchanged. 267 nm divides by 1.5 twice (by n2).
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Answer D
Answer: D
Common error: On entering a denser medium frequency stays constant while speed and wavelength both decrease.
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