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Cambridge IGCSE Physics 0625 · Core and Extended · Topic 3

Waves

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1IGPC-037-03-01Sound and ultrasound Paper 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.✓ 1Answer 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 ultrasound Paper 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.✓ 1Answer B

Answer: B

Exam tip: Learn the pairs: loudness goes with amplitude, and pitch goes with frequency.

3IGPC-037-03-03Sound and ultrasound Paper 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.✓ 1Answer 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 ultrasound Paper 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.

(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 .✓ 1
Part (b)
speed = distance ÷ time = 400 ÷ 1.18✓ 1
= 339 m/s✓ 1
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 .✓ 1

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 dispersion Paper 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.✓ 1Answer 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 dispersion Paper 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.✓ 1Answer 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 dispersion Paper 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.✓ 1Answer 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 dispersion Paper 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.✓ 1Answer 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 dispersion Paper 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°.

(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✓ 1
Part (b)
It bends towards the normal, so the angle of refraction is less than 40°✓ 1
Part (c)
The speed decreases✓ 1
Part (d)
The light must be travelling from the glass (denser medium) towards the air (less dense medium)✓ 1
The angle of incidence must be greater than the critical angle✓ 1

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 spectrum Paper 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.✓ 1Answer 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 spectrum Paper 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.✓ 1Answer 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 spectrum Paper 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.✓ 1Answer 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 spectrum Paper 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.✓ 1Answer 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 diffraction Paper 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.✓ 1Answer 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 diffraction Paper 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.✓ 1Answer 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 diffraction Paper 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.✓ 1Answer 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 diffraction Paper 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.✓ 1
Part (b)
f = 12 ÷ 60 = 0.20 Hz✓ 1
Part (c)
v = fλ = 0.20 × 15✓ 1
v = 3.0 m/s✓ 1
Part (d)
Energy, without transferring the water itself.✓ 1

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 spectrum Paper 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✓ 1
Part (b)
gamma rays / gamma radiation / γ✓ 1
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✓ 2
Part (d)
the speeds are the same (accept both 3.0 × 108 m/s)✓ 1

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 spectrum Paper 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✓ 1
Part (b)
internal heating of body cells / tissue (accept damage to or burning of internal tissue)✓ 1
Part (c)
λ = v / f = (3.0 × 108) / (2.45 × 109) ; λ = 0.12 m (accept 0.122 m)✓ 2

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 diffraction Paper CEasy5 marks

A dentist uses a small plane mirror to look at the back of a patient's teeth.

(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.✓ 1
Part (b)
50°✓ 1
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✓ 1
Second correct property✓ 1
Part (d)
(2.4 − 0.8) × 2 = 3.2 m✓ 1

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 dispersion Paper 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✓ 1
Part (b)
Least: red✓ 1
Most: violet✓ 1
Part (c)
White light is a mixture of different colours (of different frequencies / wavelengths)✓ 1
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✓ 1

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 ultrasound Paper 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 .✓ 2
Part (b)
The sound is louder . The amplitude of the wave is larger .✓ 2
Part (c)
Yes, because 196 Hz is within the human audible range of about 20 Hz to 20 000 Hz .✓ 1

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.

23IGPC-037-03-04Sound and ultrasound Paper 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?

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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.✓ 1Answer 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.

24IGPC-037-03-05Sound and ultrasound Paper 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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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.✓ 1Answer 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.

25IGPC-037-01-07Light, refraction, lenses and dispersion Paper 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?

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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.✓ 1Answer 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.

26IGPC-037-01-08Light, refraction, lenses and dispersion Paper 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.✓ 1Answer 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.

27IGPC-037-02-05The electromagnetic spectrum Paper 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.✓ 1Answer 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.

28IGPC-037-00-04Wave properties, reflection, refraction and diffraction Paper 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.✓ 1Answer A

Answer: A

Exam tip: When a wave enters a new medium, its frequency never changes. Speed and wavelength change together.

29IGPC-037-00-09Wave properties, reflection, refraction and diffraction Paper 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?

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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.✓ 1Answer C

Answer: C

Common error: A crest-to-trough distance is half a wavelength. One full wavelength runs from crest to crest.

30IGPC-037-03-14Sound and ultrasound Paper 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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Marking pointMarkNotes
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.

31IGPC-037-02-15The electromagnetic spectrum Paper 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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Marking pointMarkNotes
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.

32IGPC-037-00-12Wave properties, reflection, refraction and diffraction Paper CMedium6 marks

A student uses a ripple tank to study water waves. A vibrating bar produces straight wavefronts on the water surface.

(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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Marking pointMarkNotes
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.

33IGPC-037-03-10Sound and ultrasound Paper 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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Marking pointMarkNotes
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.✓ 1Answer 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.

34IGPC-037-02-10The electromagnetic spectrum Paper 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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Marking pointMarkNotes
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.✓ 1Answer 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.

35IGPC-037-03-13Sound and ultrasound Paper EEasy5 marks

Hospitals use ultrasound scanners to produce images of structures inside the body.

(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)
Show mark scheme
Marking pointMarkNotes
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.

36IGPC-037-03-06Sound and ultrasound Paper 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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Marking pointMarkNotes
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.✓ 1Answer 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.

37IGPC-037-03-07Sound and ultrasound Paper EMedium1 mark

A sound wave travels through air.
Which statement describes a rarefaction?

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Marking pointMarkNotes
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.✓ 1Answer 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.

38IGPC-037-03-08Sound and ultrasound Paper EMedium1 mark

Which statement about ultrasound is correct?

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Marking pointMarkNotes
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.✓ 1Answer 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.

39IGPC-037-01-04Light, refraction, lenses and dispersion Paper 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?

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Marking pointMarkNotes
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.✓ 1Answer 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.

40IGPC-037-01-05Light, refraction, lenses and dispersion Paper 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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Marking pointMarkNotes
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).✓ 1Answer 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.

41IGPC-037-01-09Light, refraction, lenses and dispersion Paper 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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Marking pointMarkNotes
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.✓ 1Answer 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.

42IGPC-037-02-06The electromagnetic spectrum Paper 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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Marking pointMarkNotes
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.
✓ 1Answer B

Answer: B

Common error: Mega means 10⁶. Write 95 MHz as 95 × 10⁶ Hz before substituting into λ = v/f.

43IGPC-037-02-07The electromagnetic spectrum Paper EMedium1 mark

Mobile phones communicate with nearby masts using microwaves. Why are microwaves suitable?

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Marking pointMarkNotes
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.✓ 1Answer 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.

44IGPC-037-02-08The electromagnetic spectrum Paper 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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Marking pointMarkNotes
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.✓ 1Answer 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.

45IGPC-037-00-06Wave properties, reflection, refraction and diffraction Paper 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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Marking pointMarkNotes
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.✓ 1Answer A

Answer: A

Common error: A wider gap does not give more diffraction. What matters is how the gap width compares with the wavelength.

46IGPC-037-00-16Wave properties, reflection, refraction and diffraction Paper EMedium6 marks

Sea waves of wavelength 12 m approach a harbour. The harbour wall has a single entrance gap 15 m wide.

(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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Marking pointMarkNotes
Part (a)
The spreading out of a wave as it passes through a gap or around the edge of an obstacle.✓ 1
Part (b)
The waves spread out a lot / become curved and reach most of the harbour.✓ 1
The wavelength stays the same (12 m).✓ 1
Part (c)
There is much less spreading; the waves continue mostly straight ahead and spread only at the edges.✓ 1
This is because the gap is now much larger than the wavelength.✓ 1
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.✓ 1

Exam tip: Say that the wavelength is unchanged by diffraction. Link the amount of spreading to how the gap size compares with the wavelength.

47IGPC-037-03-15Sound and ultrasound Paper 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.

(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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Marking pointMarkNotes
Part (a)
distance travelled = 5900 × 0.12 × 10−3 = 0.708 m✓ 1
The pulse travels the length of the bar twice, so divide by 2✓ 1
length = 0.354 m ≈ 0.35 m✓ 1
Part (b)
distance = 5900 × 0.040 × 10−3 ÷ 2✓ 1
= 0.118 m ≈ 0.12 m✓ 1
Part (c)
Any one valid advantage , e.g.:✓ 1
• the bar is not damaged, so it can still be used;—
• cracks inside the bar can be found without destroying it;—
• 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.

48IGPC-037-01-12Light, refraction, lenses and dispersion Paper 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°.

(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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Marking pointMarkNotes
Part (a)
n = sin i / sin r = sin 40° / sin 29°✓ 1
n = 0.643 / 0.485 = 1.33✓ 1
Part (b)
sin c = 1/n = 1/1.326 = 0.754 (accept 1/1.33 = 0.752)✓ 1
c = 49°✓ 1
Part (c)
The ray is travelling from the denser medium (water) and 55° is greater than the critical angle of 49°✓ 1
so the ray is totally internally reflected back into the water✓ 1

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.

49IGPC-037-00-14Wave properties, reflection, refraction and diffraction Paper 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.

(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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Marking pointMarkNotes
Part (a)
Light travels more slowly in the plastic than in air (accept: the plastic is optically denser, so the light slows down).✓ 1
Part (b)
n = sin i ÷ sin r = sin 55° ÷ sin 33°✓ 1
n = 1.50 (accept 1.5)✓ 1
Part (c)
speed in plastic = speed in air ÷ n = 3.0 × 108 ÷ 1.50✓ 1
= 2.0 × 108 m/s✓ 1
Part (d)
Parallel to the incident ray (but displaced sideways).✓ 1

Exam tip: Keep your calculator in degree mode. Carry the unrounded value of n into the next part.

50IGPC-037-03-09Sound and ultrasound Paper 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?

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Marking pointMarkNotes
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.✓ 1Answer 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.

51IGPC-037-03-16Sound and ultrasound Paper 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.

(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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Marking pointMarkNotes
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 .✓ 2
Part (b)
time in air = 1020 ÷ 340 = 3.00 s✓ 1
time in steel = 1020 ÷ 5100 = 0.200 s✓ 1
time between bangs = 3.00 − 0.200 = 2.80 s✓ 1
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 .✓ 2

Common error: Do not divide the distance by the difference in speeds. Work out each travel time separately, then subtract the two times.

52IGPC-037-01-10Light, refraction, lenses and dispersion Paper 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?

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Marking pointMarkNotes
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.✓ 1Answer 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.

53IGPC-037-01-14Light, refraction, lenses and dispersion Paper 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.

(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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Marking pointMarkNotes
Part (a)
sin c = 1/1.52 = 0.658✓ 1
c = 41°✓ 1
Part (b)
sin r = sin 30° / 1.52 = 0.329✓ 1
r = 19°✓ 1
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°✓ 1
= 71°✓ 1
71° is greater than the critical angle of 41°, so the ray is totally internally reflected and does not escape✓ 1
Part (d)
Any one from: telecommunications or broadband data transmission; an endoscope for viewing inside the body✓ 1

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.

54IGPC-037-02-09The electromagnetic spectrum Paper 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.
✓ 1Answer 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.

55IGPC-037-02-16The electromagnetic spectrum Paper 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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Marking pointMarkNotes
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✓ 2
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✓ 2
Part (d)
f = v/λ = 3.0 × 108 / 1.0 × 10−10 ; f = 3.0 × 1018 Hz✓ 2

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.

56IGPC-037-00-08Wave properties, reflection, refraction and diffraction Paper 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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Marking pointMarkNotes
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.
✓ 1Answer D

Answer: D

Exam tip: Check that your answer is greater than 1. A refractive index below 1 cannot be right.

57IGPC-037-00-10Wave properties, reflection, refraction and diffraction Paper 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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Marking pointMarkNotes
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.✓ 1Answer 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.

58IGPC-037-01-16Light, refraction, lenses and dispersion Paper 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.

(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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Marking pointMarkNotes
Part (a)
sin r = sin 60° / 1.50 = 0.577✓ 1
r = 35.3° (accept 35°)✓ 1
Part (b)
60°✓ 1
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✓ 1
Part (c)
distance = thickness / cos r = 4.0 / cos 35.3°✓ 1
= 4.9 cm✓ 1
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✓ 1

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.

59IGPC-037-00-15Wave properties, reflection, refraction and diffraction Paper 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.

(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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Marking pointMarkNotes
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✓ 1
c = 41° (41.1°)✓ 1
Part (c)
The ray is totally internally reflected, so it stays inside the fibre.✓ 1
This is because 48° is greater than the critical angle of 41°.✓ 1

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.