International GCSE Physics · Pearson Edexcel 4PH1 · Topic 3
Waves
263 free, high-quality past-paper-style questions on Waves for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (c) The Electromagnetic Spectrum; (b) Properties of Waves; (d) Light and Sound. For when you have finished the past papers, or want to practise one topic at a time.Notes 3(b) →Notes 3(c) →Notes 3(d) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,345 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IGCSE-PHY-3-0013(c) The Electromagnetic SpectrumPapers 1 & 2Grade 4–51 mark
Which list shows four electromagnetic radiations in order of increasing frequency?
Show mark scheme
B (radio waves, infrared, visible light, ultraviolet);
A is incorrect because this is the order of decreasing frequency (increasing wavelength)
C is incorrect because radio waves have a lower frequency than microwaves, so they should come first
D is incorrect because infrared has a lower frequency than visible light, so it should come before visible light
3IGCSE-PHY-3-0033(c) The Electromagnetic SpectrumPapers 1 & 2Grade 4–56 marks
A hospital uses several types of electromagnetic radiation.
(a)
Name the type of electromagnetic radiation used for each of the following. (i) sterilising surgical instruments sealed in plastic bags (ii) night-vision cameras used by security staff (iii) the tubes of fluorescent lamps in the corridors
(3)
(b)
A nurse spends a lot of time outdoors in strong sunlight. State one harmful effect of excessive exposure to ultraviolet radiation and one simple way the nurse can protect against it.
(2)
(c)
State one property that all electromagnetic waves share when they travel through free space (a vacuum).
5IGCSE-PHY-3-0053(b) Properties of WavesPapers 1 & 2Grade 4–57 marks
A buoy floats on the sea. Water waves pass the buoy, making it move up and down. The time period of the waves is 4.0 s. The wavelength of the waves is 36 m.
(a)
State what is meant by the amplitude of a wave.
(1)
(b)
Calculate the frequency of the waves.
(2)
(c)
Calculate the speed of the waves.
(2)
(d)
Water waves are transverse. Explain the difference between transverse waves and longitudinal waves.
(2)
Show mark scheme
(a)
maximum displacement (of a point on the wave) from its rest / undisturbed / equilibrium position;
Allow
height of a crest from the middle (rest) position
Do not accept
distance from top of crest to bottom of trough
height of the wave
(b)
substitution;
evaluation;
e.g. f = 1 ÷ 4.0
(f =) 0.25 (Hz)
Allow
1/4 seen for substitution
Do not accept
4.0 Hz
(c)
substitution;
evaluation;
e.g. v = 0.25 × 36
(v =) 9.0 (m/s)
Allow
ecf from (b)
v = λ ÷ T = 36 ÷ 4.0 for substitution
Do not accept
144 (m/s) (using f = 4.0)
(d)
transverse: oscillations / vibrations perpendicular to the direction of energy transfer / wave travel;
longitudinal: oscillations / vibrations parallel to the direction of energy transfer / wave travel;
Allow
a correctly labelled diagram for either point
Do not accept
transverse waves go up and down and longitudinal waves go side to side (no reference to direction of travel)
6IGCSE-PHY-3-0213(b) Properties of WavesPapers 1 & 2Grade 4–51 mark
A student shakes one end of a long rope up and down. The graph shows the displacement of the rope against distance along the rope at one instant. Which option gives the amplitude and the wavelength of the wave?Show mark scheme
C amplitude 3 cm, wavelength 1.2 m;
A is incorrect: 6 cm is the crest-to-trough height, which is twice the amplitude
B is incorrect: 0.6 m is the distance from a crest to the next trough, which is half a wavelength
D is incorrect: 6 cm is crest to trough and 3.0 m is the whole length of rope shown (2.5 wavelengths)
7IGCSE-PHY-3-0223(c) The Electromagnetic SpectrumPapers 1 & 2Grade 4–51 mark
Sunlight travels through space to the Earth. Sunlight contains all the colours of the visible spectrum. Which statement about red light and violet light travelling through space is correct?
Show mark scheme
A Red light has a longer wavelength than violet light, and both travel at the same speed;
B is incorrect: red has the lower frequency of the two (frequency increases from red to violet)
C is incorrect: all electromagnetic waves travel at the same speed in free space; a longer wavelength does not mean a faster wave
D is incorrect: this reverses the order; violet has the shorter wavelength and higher frequency
8IGCSE-PHY-3-0233(c) The Electromagnetic SpectrumPapers 1 & 2Grade 4–58 marks
A food-processing factory uses several types of electromagnetic radiation.
(a)
Sealed packets of dried herbs are exposed to gamma rays. Explain why gamma rays can be used to sterilise the herbs after the packets have been sealed.
(2)
(b)
Jars of baby food pass through a scanner that checks for small pieces of glass or metal inside the jars. Name the type of electromagnetic radiation used in this scanner.
(1)
(c(i))
Staff heat their lunch in a microwave oven. State the harmful effect on the human body of excessive exposure to microwaves.
(1)
(c(ii))
The glass door of the microwave oven contains a metal mesh. Suggest why.
(1)
(d)
Infrared heaters keep cooked food warm on the serving counter. State one harmful effect of excessive exposure to infrared radiation.
(1)
(e)
Workers operate the gamma ray sterilising machine every day. State one harmful effect of excessive exposure to gamma rays and one way the workers can be protected.
(2)
Show mark scheme
(a)
gamma rays kill bacteria / microorganisms (on the herbs);
gamma rays pass through / penetrate the packaging (so herbs are sterilised inside the sealed packet);
Allow
so the food cannot be re-contaminated after sterilising (for MP2 if penetration implied)
Do not accept
gamma rays clean the herbs
gamma rays heat / cook the herbs
(b)
X-rays;
Allow
X-radiation
Do not accept
gamma rays
ultraviolet
(c(i))
internal heating of body tissue / cells;
Allow
burns to internal organs
heats water inside the body
Do not accept
skin cancer
radiation sickness
(c(ii))
to stop microwaves escaping (through the door) / to reflect or absorb microwaves so users are not exposed;
Do not accept
to stop heat escaping
to make the door stronger
(d)
(skin) burns;
Allow
damage to skin cells by heating
Do not accept
skin cancer
blindness
(e)
cancer / mutation (of cells / DNA);
any suitable protection, e.g. lead / concrete shielding around the source / operate the machine remotely or from behind a barrier / limit exposure time / keep a large distance from the source;
Allow
damage to DNA for MP1
wear a film badge / dosimeter to monitor dose (accept as part of protection only if linked to limiting exposure)
9IGCSE-PHY-3-0243(b) Properties of WavesPapers 1 & 2Grade 4–58 marks
At a football stadium, spectators make a 'Mexican wave'. As the wave reaches each spectator, they stand up and raise their arms, then sit down again. The spectators keep making waves, so a regular series of waves travels around the stadium. Each wave travels around the stadium at 12 m/s. The distance between the centres of two neighbouring waves (where spectators are fully standing) is 30 m.
(a)
Explain why the Mexican wave is an example of a transverse wave.
(2)
(b)
Explain how the Mexican wave shows that a wave can transfer energy without transferring matter.
(1)
(c)
State the wavelength of the Mexican wave.
(1)
(d)
Calculate the frequency of the Mexican wave.
(2)
(e)
Calculate the time between one spectator standing up and the same spectator standing up again.
(2)
Show mark scheme
(a)
spectators move up and down / vertically;
(which is) perpendicular / at right angles to the direction the wave travels (around the stadium);
Allow
oscillation perpendicular to direction of energy transfer for 2 marks
Do not accept
spectators move in the same direction as the wave
(b)
spectators stay in their seats / do not move around the stadium but the wave (energy / pattern) does;
10IGCSE-PHY-3-0253(d) Light and SoundPapers 1 & 2Grade 4–57 marks
A periscope is a tube containing two plane mirrors. Each mirror is fixed at 45° to the sides of the tube. The diagram shows a horizontal ray of light from an object entering the top of the periscope and striking the top mirror. The dashed line is the normal to the top mirror.
(a)
State the name of the process that happens to the light at the top mirror.
(1)
(b)
Determine the angle of incidence of the ray at the top mirror.
(1)
(c)
State the angle of reflection at the top mirror. Give a reason for your answer.
(2)
(d)
Describe the path of the ray from the top mirror until it reaches the observer's eye.
(3)
Show mark scheme
(a)
reflection;
Allow
(light is) reflected
Do not accept
refraction
total internal reflection
(b)
45(°);
(c)
45(°);
angle of incidence = angle of reflection / law of reflection;
Allow
ecf from (b)
(d)
(ray travels) vertically downwards (through the tube) to the bottom mirror;
(strikes bottom mirror with) angle of incidence 45° / reflected through 90° at the bottom mirror;
leaves horizontally / parallel to the original ray / through the bottom opening (to the eye);
Allow
correct description of turning through 90° at each mirror
11IGPE-052-00-013(d) Light and SoundPaper 1PEasy1 mark
The vibration motor in a games controller completes one full vibration every 0.025 s. What is the frequency of the vibration?
Show mark scheme
Marking point
Mark
Notes
Frequency = 1 ÷ time period = 1 ÷ 0.025 s = 40 Hz. Choosing 0.025 Hz treats the time period as the frequency. Choosing 25 Hz comes from misreading 0.025 s as 25 ms and then using 1 ÷ 0.04. Choosing 400 Hz is a power-of-ten slip when dividing.
✓ 1
Answer C
Answer: C
Exam tip: f = 1/T only works with T in seconds. Convert ms to s before dividing.
12IGPE2-056-01-013(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
Surgical instruments are sealed inside plastic packaging. They are then sterilised without opening the packaging. Which type of electromagnetic radiation is used for this?
Show mark scheme
Marking point
Mark
Notes
Gamma rays are highly penetrating, so they pass through the sealed plastic. They kill the bacteria on the instruments inside. Infrared is a distractor based on the idea that sterilising means heating. Microwaves are a distractor based on microwave ovens heating food. Ultraviolet does kill bacteria, but it cannot penetrate the packaging.
✓ 1
Answer A
Answer: A
Common error: UV does kill microorganisms, but only on exposed surfaces. When the item is sealed inside packaging, the radiation must be penetrating, so the answer is gamma.
13IGPE2-056-01-053(c) The Electromagnetic SpectrumPaper 1PEasy4 marks
Skiers at high altitude are exposed to large amounts of ultraviolet (UV) radiation from the Sun.
(a)
State two harmful effects of excessive exposure to ultraviolet radiation.
(2)
(b)
State one use of ultraviolet radiation.
(1)
(c)
Suggest why the risk from ultraviolet radiation is greater for a skier on a snowy mountain than for a person at sea level.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Any two different effects from: skin cancer; sunburn / damage to surface cells of the skin; eye damage (e.g. cataracts, snow blindness); premature ageing of the skin.
—
First correct effect
✓ 1
Second correct effect
✓ 1
Part (b)
Any one correct use, e.g. security marking / detecting forged banknotes (the marks fluoresce under UV); fluorescent lamps; sterilising water or medical equipment; sun-tanning lamps; insect traps
✓ 1
Part (c)
Any one correct reason, either alone or with the other: there is less atmosphere above the skier to absorb the UV, OR snow reflects UV onto the skier, so the exposure is greater
✓ 1
Common error: Do not say that UV causes internal heating. That is the hazard from microwaves. UV damages surface cells and the eyes.
15IGPE2-056-01-113(c) The Electromagnetic SpectrumPaper 1PEasy4 marks
A television remote control has an LED on the front. When a button is pressed, the LED sends signals that the human eye cannot see.
(a)
State which part of the electromagnetic spectrum the remote control uses.
(1)
(b)
A button on the remote control is held down. The LED is viewed on the screen of a mobile phone camera, and a flashing light is seen. Suggest why the light can be seen on the phone screen but not with the eye alone.
(1)
(c)
The LED is switched on and off 38 000 times every second. Calculate the time period of this switching.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
infrared / IR
✓ 1
Part (b)
The camera sensor detects infrared (which the eye cannot) and shows it on the screen as visible light
✓ 1
Part (c)
T = 1 ÷ f = 1 ÷ 38 000
✓ 1
T = 2.6 × 10−5 s (accept 2.63 × 10−5 s)
✓ 1
Exam tip: 'Times per second' is a frequency in hertz. The time period is found from T = 1 ÷ f.
16IGPE2-056-00-013(b) Properties of WavesPaper 1PEasy1 mark
A hospital ultrasound probe produces waves of frequency 3.5 MHz. The speed of these waves in soft tissue is 1540 m/s. What is the wavelength of the ultrasound in soft tissue?
Show mark scheme
Marking point
Mark
Notes
Rearrange wave speed = frequency × wavelength: λ = v / f = 1540 / (3.5 × 106) = 4.4 × 10−4 m. Option B treats MHz as kHz. Option C divides frequency by speed. Option D multiplies speed by frequency.
✓ 1
Answer A
Answer: A
Common error: M means mega, ×10⁶, not ×10³. Convert MHz to Hz before substituting.
17IGPE2-056-00-033(b) Properties of WavesPaper 1PEasy1 mark
At a football stadium, fans make a 'Mexican wave'. Each fan stands up and sits down again just after the person beside them. The wave travels round the stadium. Which statement correctly describes the Mexican wave?
Show mark scheme
Marking point
Mark
Notes
The fans move vertically while the wave travels horizontally. The oscillation is at right angles to the direction of travel, so the wave is transverse. Each fan stays in their own seat: the wave transfers energy (the disturbance) without transferring matter. Options A and D confuse the direction of travel with the direction of oscillation. Option B wrongly suggests that matter moves along with the wave.
✓ 1
Answer C
Answer: C
Exam tip: To classify a wave, compare the direction of the oscillations with the direction of energy transfer: perpendicular means transverse, parallel means longitudinal.
18IGPE2-056-00-053(b) Properties of WavesPaper 1PEasy7 marks
A wave-energy buoy floats out at sea. As ocean waves pass, the buoy rises and falls. It makes 12 complete up-and-down oscillations every minute. The vertical distance from its highest point to its lowest point is 3.0 m.
(a)
State the amplitude of the waves.
(1)
(b)
Calculate the frequency of the waves.
(2)
(c)
Calculate the period of the waves.
(1)
(d)
The distance between neighbouring wave crests is 40 m. Calculate the speed of the waves.
(2)
(e)
Explain why the buoy is not carried towards the shore by the waves.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1.5 m (half of 3.0 m)
✓ 1
Part (b)
12 oscillations in 60 s . f = 12 / 60 = 0.20 Hz .
✓ 2
Part (c)
T = 1 / f = 1 / 0.20 = 5.0 s
✓ 1
Part (d)
v = f × λ = 0.20 × 40 . v = 8.0 m/s .
✓ 2
Part (e)
Waves transfer energy, not matter, so the water and the buoy only oscillate about a fixed position .
✓ 1
Common error: Amplitude is measured from the rest position to a crest, not from crest to trough. Halve the total vertical distance.
19IGPE2-056-00-113(b) Properties of WavesPaper 1PEasy6 marks
In a PE lesson, a student ties one end of a long skipping rope to a post. She moves the other end up and down, sending waves along the rope. She makes 2.5 complete up-and-down movements each second.
(a)
State whether the waves on the rope are transverse or longitudinal.
(1)
(b)
State what is meant by the frequency of a wave.
(1)
(c)
State what is meant by the amplitude of a wave.
(1)
(d)
The student moves her hand up and down more quickly. The speed of the waves stays the same. State what happens to the wavelength.
(1)
(e)
Calculate the period of the waves when the frequency is 2.5 Hz.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Transverse
✓ 1
Part (b)
The number of waves passing a point (or produced) each second .
✓ 1
Part (c)
The maximum displacement of a point from its rest (undisturbed) position .
✓ 1
Part (d)
The wavelength decreases .
✓ 1
Part (e)
T = 1 / f = 1 / 2.5 . T = 0.40 s .
✓ 2
Exam tip: Define amplitude from the rest position, not from the lowest point of the wave.
20IGPE2-056-03-013(b) Properties of WavesPaper 1PEasy1 mark
At a football match, spectators create a 'stadium wave'. Each spectator stands up and then sits down again as the wave passes. The wave travels sideways around the stadium. Which row correctly describes this wave?
Show mark scheme
Marking point
Mark
Notes
The spectators move up and down. The wave travels sideways, so the oscillation is perpendicular to the direction of travel, which makes it transverse. No spectator changes seat. The wave transfers energy around the stadium, not the people. Option A wrongly assumes the medium travels with the wave. Option B classifies the wave by its direction of travel rather than by the direction of oscillation. Option D describes a longitudinal wave, but the motion here is up and down, not sideways.
✓ 1
Answer C
Answer: C
Common error: Classify a wave by comparing the direction of oscillation with the direction of energy transfer. The direction the wave travels on its own tells you nothing.
21IGPE2-056-03-043(b) Properties of WavesPaper 1PEasy7 marks
An earthquake produces P-waves, which are longitudinal, and S-waves, which are transverse. These waves travel through the Earth.
(a)
Describe the difference between a longitudinal wave and a transverse wave.
(2)
(b)
A P-wave travels at 6.0 km/s and has a frequency of 2.0 Hz. Calculate the wavelength of the P-wave.
(3)
(c)
A seismometer is 480 km from the earthquake. Calculate the time taken for the P-wave to reach the seismometer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
In a longitudinal wave the particles oscillate parallel to the direction of energy transfer . In a transverse wave the particles oscillate perpendicular to the direction of energy transfer .
✓ 2
Part (b)
λ = v ÷ f
✓ 1
= 6000 ÷ 2.0
✓ 1
= 3000 m (accept 3.0 km)
✓ 1
Part (c)
t = d ÷ v = 480 000 ÷ 6000 or 480 ÷ 6.0
✓ 1
= 80 s
✓ 1
Common error: Mixing km and m in the same calculation causes errors of a factor of 1000. Keep the units consistent, either both in km and km/s or both in m and m/s.
23IGPE2-056-03-063(b) Properties of WavesPaper 1PEasy6 marks
A wave machine in a leisure pool produces 40 waves every minute. The distance between neighbouring crests is 1.5 m.
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the speed of the waves.
(2)
(c)
A small float rests on the water as the waves pass. Describe the motion of the float and explain why it does not travel across the pool with the waves.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
f = 40 ÷ 60
✓ 1
= 0.67 Hz (accept 0.667 Hz or 2/3 Hz)
✓ 1
Part (b)
v = f × λ = 0.67 × 1.5
✓ 1
= 1.0 m/s (accept 1 m/s)
✓ 1
Part (c)
The float oscillates up and down about the same position (accept moves in small circles or loops about a fixed point) .
✓ 1
Waves transfer energy, not matter, so the water and the float are not carried along with the wave .
✓ 1
Common error: Using 40 Hz as the frequency is wrong. Frequency is the number of waves per second, so divide the 40 waves by 60 s.
24IGPE2-056-03-073(b) Properties of WavesPaper 1PEasy6 marks
A tsunami-warning buoy records ocean waves. The vertical distance from a crest to a trough is 0.80 m. The horizontal distance from one crest to the next is 12 m.
(a)
State the amplitude of the waves.
(1)
(b)
State the wavelength of the waves.
(1)
(c)
Waves transfer energy. State one other thing that waves can transfer.
(1)
(d)
The buoy records a time period of 8.0 s for the waves. Calculate the speed of the waves.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.40 m
✓ 1
Part (b)
12 m
✓ 1
Part (c)
Information (accept signals or data; do not accept matter, water or particles)
✓ 1
Part (d)
f = 1 ÷ T = 1 ÷ 8.0 = 0.125 Hz
✓ 1
v = f × λ = 0.125 × 12
✓ 1
= 1.5 m/s
✓ 1
Allow the direct method v = λ ÷ T = 12 ÷ 8.0 = 1.5 m/s for full marks.
—
Common error: Amplitude is measured from the rest position to a crest. It is half the crest-to-trough distance, not the whole distance.
25IGPE2-056-02-013(d) Light and SoundPaper 1PEasy1 mark
A bat produces sound waves with a frequency of 45 kHz to locate insects. A person standing nearby cannot hear these sounds. Which statement explains why the person cannot hear them?
Show mark scheme
Marking point
Mark
Notes
The range of human hearing is about 20 Hz to 20 000 Hz. A frequency of 45 kHz is 45 000 Hz, which is above 20 000 Hz, so it is ultrasound and cannot be heard. Amplitude affects loudness, not whether a frequency is audible. A high frequency gives a short wavelength, not a long one. Ultrasound does travel through air, which is how the bat uses its echoes.
✓ 1
Answer A
Answer: A
Exam tip: Convert kHz to Hz before comparing with the 20–20 000 Hz hearing range: 1 kHz = 1000 Hz.
26IGPE2-056-02-043(d) Light and SoundPaper 1PEasy6 marks
A car has a parking sensor in its rear bumper. The sensor sends out pulses of ultrasound and detects the echoes that reflect from objects behind the car. Speed of sound in air = 340 m/s
(a)
State what is meant by ultrasound.
(1)
(b)
An echo from a wall is detected 4.0 ms after the pulse is sent out. Calculate the distance between the sensor and the wall.
(3)
(c)
The ultrasound has a frequency of 40 kHz. Calculate its wavelength in air.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Sound with a frequency above 20 000 Hz / above the upper limit of human hearing
✓ 1
Part (b)
Total distance travelled = 340 × 4.0×10−3 = 1.36 m
✓ 1
The pulse travels to the wall and back, so divide by 2
✓ 1
Distance = 0.68 m
✓ 1
Part (c)
λ = v / f = 340 / 40 000
✓ 1
λ = 8.5×10−3 m
✓ 1
Common error: Forgetting to halve the echo distance. The time measured is for the journey there and back.
27IGPE2-056-02-053(d) Light and SoundPaper 1PEasy6 marks
A dentist uses a small plane mirror to see the back of a patient's teeth.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
A ray of light from a lamp strikes the mirror at an angle of 25° to the mirror surface. Calculate the angle of reflection.
(2)
(c)
Describe the image of a tooth formed by the plane mirror.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Angle of incidence = angle of reflection (both measured from the normal)
✓ 1
Part (b)
Angle of incidence = 90° − 25° = 65°
✓ 1
Angle of reflection = 65°
✓ 1
Part (c)
Any three of these properties: virtual / cannot be projected onto a screen; upright / the right way up; same size as the tooth; the same distance behind the mirror as the tooth is in front; laterally inverted.
—
First correct property
✓ 1
Second correct property
✓ 1
Third correct property
✓ 1
Common error: Giving 25° as the angle of reflection. Angles of incidence and reflection are always measured from the normal, not from the mirror surface.
28IGPE2-056-02-123(b) Properties of WavesPaper 1PEasy7 marks
Marine biologists use underwater microphones to record the low calls of fin whales. Speed of sound in seawater = 1500 m/s
(a)
State whether sound is a transverse or a longitudinal wave.
(1)
(b)
Describe how the particles of seawater move as the sound passes through.
(1)
(c)
A fin whale call has a frequency of 20 Hz. Calculate the wavelength of the call in seawater.
(3)
(d)
Some of the sound passes from the sea into the air, where it travels at 340 m/s. State what happens to the frequency and to the wavelength of the sound.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Longitudinal
✓ 1
Part (b)
They vibrate back and forth parallel to the direction in which the wave transfers energy
✓ 1
Part (c)
λ = v / f
✓ 1
λ = 1500 / 20
✓ 1
λ = 75 m
✓ 1
Part (d)
Frequency stays the same
✓ 1
Wavelength decreases
✓ 1
Exam tip: When a wave enters a new medium, its frequency is fixed by the source. Only the speed and wavelength change.
29IGPE-052-00-043(b) Properties of WavesPaper 1PEasy8 marks
Spectators in a sports stadium make a 'crowd wave'. Each spectator stands up and sits down again as the wave reaches them, and the wave travels sideways around the stand.
(a)
State whether the crowd wave is a transverse or a longitudinal wave. Explain your answer.
(2)
(b)
The wave travels 120 m along the stand in 12 s. Calculate the speed of the wave.
(2)
(c)
The distance from one group of standing spectators to the next group of standing spectators is 15 m. This distance is the wavelength of the crowd wave. Calculate the frequency of the crowd wave.
(3)
(d)
Explain how the crowd wave shows that waves transfer energy without transferring matter.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Transverse . The spectators move up and down, which is perpendicular to the direction in which the wave travels .
✓ 2
Part (b)
speed = distance ÷ time = 120 ÷ 12 ; = 10 m/s
✓ 2
Part (c)
f = v ÷ λ ; f = 10 ÷ 15 ; = 0.67 Hz
✓ 3
Part (d)
The wave (energy) moves around the stadium, but each spectator stays at their own seat, so no matter is transferred .
✓ 1
Exam tip: Carry your wave speed from the earlier part into v = fλ, and show your working. A correct method using your own earlier value can still earn credit even if that value was wrong.
30IGPE-052-00-053(c) The Electromagnetic SpectrumPaper 1PEasy7 marks
A cereal factory uses electromagnetic waves for several jobs, including checking sealed cardboard boxes for small stones or metal fragments.
(a)
The electromagnetic spectrum, in order of increasing frequency, is: radio waves, microwaves, ______, visible light, ultraviolet, ______, gamma rays. Name the two missing groups of waves.
(2)
(b)
State one property that all electromagnetic waves share.
(1)
(c)
Workers near the ultraviolet lamps used to sterilise conveyor belts must be protected. State one harmful effect of ultraviolet on the human body, and one harmful effect of gamma rays.
(2)
(d)
Suggest why X-rays, and not visible light, are used to check inside the sealed boxes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
infrared ; X-rays
✓ 2
Part (b)
Any one of: they all travel at the same speed in a vacuum / they are all transverse / they can all travel through a vacuum .
✓ 1
Part (c)
Ultraviolet: damage to surface cells / sunburn / skin cancer / eye damage . Gamma rays: cell mutation / cancer / damage to cells inside the body .
✓ 2
Part (d)
X-rays pass through the cardboard and the cereal, but visible light does not . Stones or metal absorb more X-rays than cereal, so they show up on the image .
✓ 2
Common error: Saying a wave is 'dangerous' earns nothing. Name a specific effect, such as skin cancer or cell mutation.
31IGPE-052-00-133(d) Light and SoundPaper 1PEasy6 marks
On a building site, a laser alignment tool sends a narrow beam of light towards a flat plane mirror fixed to a wall.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
State what is meant by the normal to the mirror.
(1)
(c)
The laser beam strikes the mirror at an angle of 35° to the mirror surface. Calculate the angle of reflection.
(2)
(d)
A worker looks at his reflection in the same mirror. Describe two properties of the image he sees.
(2)
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Part (a)
The angle of incidence equals the angle of reflection .
✓ 1
Part (b)
A line drawn at 90° (perpendicular) to the mirror surface at the point where the ray hits it .
✓ 1
Part (c)
angle of incidence = 90° − 35° = 55° ; angle of reflection = 55°
✓ 2
Part (d)
Any two of: virtual / upright / the same size as the object / the same distance behind the mirror as the object is in front / laterally inverted .
✓ 2
Common error: Angles in reflection and refraction are always measured from the normal. If you are given the angle to the surface, subtract it from 90° first.
32IGPE3-3-14r21-013(b) Properties of WavesPaper 1PEasy1 mark
A bat emits ultrasound of frequency 40 kHz. The speed of sound in air is 340 m/s. What is the time period of the sound?
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T = 1/f = 1/(40 × 103 Hz) = 2.5 × 10−5 s, so B is correct. A is the wavelength 340 ÷ 40 000 = 8.5 × 10−3 m with the wrong unit. C comes from using 40 Hz instead of 40 kHz. D is 40 × 1000, the frequency in hertz mistaken for the period.
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Answer B
Answer: B
Common error: Convert kHz to Hz (× 1000) before using T = 1/f.
33IGPE3-3-14r21-023(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
Red light and violet light from the same distant star travel through space to a telescope. Which statement is correct?
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All electromagnetic waves travel at 3.0 × 108 m/s in a vacuum, and since v = fλ red light (lower frequency) has the longer wavelength, so C is correct. A is wrong because speed does not depend on frequency in a vacuum. B wrongly links speed to wavelength. D is wrong because different frequencies at the same speed need different wavelengths.
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Answer C
Answer: C
Exam tip: In a vacuum every colour of light has the same speed; only frequency and wavelength differ.
35IGPE3-3-14r22-013(b) Properties of WavesPaper 1PEasy1 mark
Water waves in a ripple tank travel from a deep region into a shallow region. Which row describes what happens to the speed, wavelength and frequency of the waves?
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In shallow water the waves travel more slowly. The frequency is set by the source and does not change at the boundary, so v = fλ means the wavelength must decrease along with the speed. The row with all three decreasing is wrong because the frequency cannot change at a boundary. The row with a constant wavelength is wrong because a smaller speed at the same frequency gives a smaller wavelength. The row with constant speed is wrong because the speed does change when the depth changes.
✓ 1
Answer B
Answer: B
Common error: Thinking frequency changes on refraction; it stays the same, and only speed and wavelength change.
36IGPE3-3-14r22-043(b) Properties of WavesPaper 1PEasy1 mark
A farmer uses a device that emits sound of frequency 25 kHz to deter pests. The speed of sound in the air is 350 m/s. What is the wavelength of the sound?
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Wavelength = speed ÷ frequency = 350 ÷ 25 000 = 0.014 m. The value 14 m comes from using 25 instead of 25 000 Hz. The value 71 m comes from dividing frequency by speed. The value 8.8 × 106 m comes from multiplying speed by frequency.
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Answer C
Answer: C
Common error: Forgetting to convert kHz to Hz before substituting into v = fλ.
37IGPE3-3-14r22-093(d) Light and SoundPaper 1PEasy6 marks
A music student connects a microphone to an oscilloscope to display a note from a flute. The display shows exactly 4 complete waves in a time of 9.2 ms.
Oscilloscope trace. Diagram not to scale
(a)
Calculate the time for one complete wave. Give your answer in milliseconds.
(2)
(b)
Calculate the frequency of the note. Use a period of 2.3 ms (0.0023 s). Give your answer to 3 significant figures.
(2)
(c)
The flute player plays the same note more loudly. State how the trace on the oscilloscope changes.
(2)
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Part (a)
M1 9.2 ÷ 4
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A1 2.3 ms
✓ 1
Part (b)
M1 f = 1 ÷ T = 1 ÷ 0.0023
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A1 435 Hz (434.78...)
✓ 1
Part (c)
B1 the amplitude (height of the waves) increases
✓ 1
B1 the frequency (number of waves in 9.2 ms) stays the same
✓ 1
Common error: Using the time for all four waves as the period; divide by the number of waves first.
38IGPE3-3-15r22-013(b) Properties of WavesPaper 1PEasy1 mark
A scientific buoy bobs up and down as ocean waves pass. In 120 s the buoy completes 18 complete up-and-down oscillations. What is the frequency of the waves?
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Frequency = number of oscillations ÷ time = 18 ÷ 120 = 0.15 Hz, so 0.15 Hz is correct. 6.7 comes from 120 ÷ 18, which is the time period in seconds, not a frequency. 2160 comes from multiplying 18 by 120. 0.015 Hz is a power-of-ten slip in the division.
✓ 1
Answer C
Answer: C
Common error: Dividing time by the number of oscillations gives the time period, not the frequency. Check the unit: Hz means per second.
39IGPE3-3-15r22-023(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
A food factory uses microwaves to heat meals, infrared lamps to brown pastry, ultraviolet lamps to sterilise work surfaces and gamma rays to sterilise sealed packs. Which statement about these radiations is correct?
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All electromagnetic waves travel at 3.0×108 m/s in a vacuum, so the first statement is correct. Gamma rays do have the highest frequency, but that does not make them faster, so the second statement is wrong. Ultraviolet has a higher frequency than infrared, so the third statement is reversed. Microwaves have a much longer wavelength than gamma rays, so the fourth statement is reversed.
✓ 1
Answer A
Answer: A
Common error: Higher frequency means higher photon energy, not higher speed. In a vacuum every electromagnetic wave has the same speed.
40IGPE3-3-15r22-053(b) Properties of WavesPaper 1PEasy7 marks
A scientific buoy floats on the sea and moves up and down as waves pass. A sensor records that 18 wave crests pass the buoy in 90 s. The vertical distance between the highest and lowest positions of the buoy is 1.8 m. The distance between two neighbouring crests is 35 m.
(a)
Calculate the amplitude of the waves.
(1)
(b)
Calculate the time period of the waves.
(2)
(c)
Calculate the frequency of the waves.
(1)
(d)
Calculate the speed of the waves, using the frequency from (c).
(3)
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Part (a)
0.90 m
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Part (b)
time period = 90 ÷ 18
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= 5.0 s
✓ 1
Part (c)
f = 1 ÷ 5.0 = 0.20 Hz (ECF from (b))
✓ 1
Part (d)
v = f × λ
✓ 1
v = 0.20 × 35 (ECF from (c))
✓ 1
= 7.0 m/s
✓ 1
Exam tip: Amplitude is measured from the rest position to a crest, so it is half of the total vertical movement.
41IGPE3-3-15r22-063(c) The Electromagnetic SpectrumPaper 1PEasy6 marks
In a food factory, packed vegetables pass through a beam of gamma rays that kills bacteria on the food.
(a)
State one property that gamma rays share with all other electromagnetic waves.
(1)
(b)
Explain why gamma rays can kill bacteria but workers must be shielded from the beam.
(2)
(c)
Gamma rays used in the factory have a frequency of 6.0×1019 Hz. The speed of electromagnetic waves in a vacuum is 3.0×108 m/s. Calculate the wavelength of these gamma rays.
(3)
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Part (a)
they travel at the same speed in a vacuum (3.0×108 m/s) / they are transverse waves / they transfer energy without transferring matter
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Part (b)
gamma rays are ionising
✓ 1
they can damage or kill living cells (or cause mutations or cancer), including those of workers
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Part (c)
wave speed = frequency × wavelength, or λ = v ÷ f
✓ 1
λ = (3.0×108) ÷ (6.0×1019)
✓ 1
= 5.0×10−12 m
✓ 1
Exam tip: Write powers of ten into your calculator with the EXP key and check that a gamma-ray wavelength comes out tiny.
42IGPE3-3-14r26-013(b) Properties of WavesPaper 1PEasy1 mark
A float bobs up and down on the surface of a swimming pool as waves from a wave machine pass it. The waves travel along the length of the pool. Which statement is correct?
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The float only moves up and down about a fixed position, so the water is not carried along the pool; the waves are transverse and transfer energy in the direction they travel. The first option is wrong because water is not carried along with the wave, the second because no water is transferred along the pool, and the last because the oscillation is at right angles to the direction of travel.
✓ 1
Answer C
Answer: C
Common error: A wave transfers energy, not matter; the particles of the medium only oscillate about fixed positions.
43IGPE3-3-14r26-023(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
A library uses radio waves in its book-tag gates, red light in its barcode scanners, microwaves in its wireless network and infrared in its remote-controlled blinds. Which row lists these four radiations in order of increasing frequency?
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In the electromagnetic spectrum frequency increases from radio waves to microwaves to infrared to visible light, so this order is correct. The second option lists them in order of decreasing frequency, the third swaps radio waves and microwaves, and the last swaps microwaves and infrared.
✓ 1
Answer A
Answer: A
Exam tip: Learn the spectrum in order (radio, microwave, infrared, visible, ultraviolet, X-ray, gamma) and remember that wavelength decreases as frequency increases.
45IGPE3-3-14r26-143(d) Light and SoundPaper 1PEasy6 marks
A microphone in a library is connected to an oscilloscope. The time-base is set to 2.0 ms per cm. The trace of a steady note shows exactly 4 complete waves across 8.0 cm of the screen.
Oscilloscope screen, 1 cm grid. Diagram not to scale
(a)
Calculate the period of the note.
(2)
(b)
Calculate the frequency of the note.
(2)
(c)
The same note is played more loudly. Describe how the trace changes.
(1)
(d)
A note of the same loudness but higher pitch is played. Describe how the trace changes.
(1)
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Part (a)
M1 time for 4 waves = 8.0 × 2.0 = 16 ms
✓ 1
A1 16 ÷ 4 = 4.0 ms
✓ 1
Part (b)
M1 f = 1 ÷ T with T = 0.0040 s (ECF from (a))
✓ 1
A1 250 Hz
✓ 1
Part (c)
The waves are taller (greater amplitude) with the same number of waves across the screen
✓ 1
Part (d)
More waves across the screen (waves closer together) with the same height
✓ 1
Common error: The period is the time for one wave, not for all the waves on the screen; divide the total time by the number of waves.
46IGPE3-3-16r28-013(b) Properties of WavesPaper 1PEasy1 mark
A wave machine in a leisure-centre pool makes waves that travel at 1.5 m/s. The distance between neighbouring crests is 3.0 m. How long is there between two successive crests passing a fixed point at the edge of the pool?
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The time between crests is the period, T = wavelength ÷ speed = 3.0 ÷ 1.5 = 2.0 s. The value 0.50 is speed ÷ wavelength, which is the frequency in Hz, not a time. Multiplying speed by wavelength (4.5) has no meaning here. The value 3.0 gives the wavelength as a number of seconds.
✓ 1
Answer C
Answer: C
Common error: Time between crests is the period; do not confuse it with frequency, which is its reciprocal.
47IGPE3-3-16r28-023(b) Properties of WavesPaper 1PEasy1 mark
In a stadium, spectators stand up and sit down in turn so that a Mexican wave travels around the stand. Which statement is correct?
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Each spectator moves vertically while the pattern travels horizontally, so the oscillation is perpendicular to the direction of travel and the wave is transverse. The idea that people travel round the stand is wrong because they stay in place. The idea of pushing is wrong because no pushing is needed and the motion is not parallel to the travel. Equating wave speed with one person's standing speed confuses the speed of one person's movement with the speed of the pattern.
✓ 1
Answer A
Answer: A
Common error: In any wave, the medium (here, the people) oscillates about a fixed position; only the pattern and energy move along.
48IGPE3-3-16r28-053(c) The Electromagnetic SpectrumPaper 1PEasy6 marks
A coffee-roasting plant uses an infrared sensor to measure the temperature of beans from the radiation they emit. The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
State whether infrared radiation has a longer or a shorter wavelength than visible light.
(1)
(b)
Microwaves, X-rays, visible light and radio waves are electromagnetic waves. Write them in order of increasing frequency.
(2)
(c)
Visible light from the roaster window has a wavelength of 6.0 × 10−7 m. Calculate its frequency.
(3)
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Part (a)
Longer
✓ 1
Part (b)
radio waves lowest and X-rays highest
✓ 1
microwaves before visible light in the middle, i.e. radio waves, microwaves, visible light, X-rays
✓ 1
Part (c)
M1 f = v ÷ λ
✓ 1
M1 3.0 × 108 ÷ 6.0 × 10−7
✓ 1
A1 5.0 × 1014 Hz
✓ 1
Exam tip: Frequency increases as wavelength decreases along the spectrum, because all electromagnetic waves travel at the same speed in a vacuum.
50IGPE3-3-16r32-013(b) Properties of WavesPaper 1PEasy1 mark
A leak-detection unit on a water main produces ultrasound of frequency 25 kHz. What is the time period of the wave?
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T = 1 ÷ f = 1 ÷ 25 000 = 4.0 × 10⁻⁵ s. 2.5 × 10⁴ s is the frequency value used as the period. 4.0 × 10⁻² s forgets to convert kHz to Hz (1 ÷ 25). 4.0 × 10⁻⁴ s is a slip of one power of ten.
✓ 1
Answer C
Answer: C
Common error: Convert kHz to Hz (× 1000) before using T = 1/f.
51IGPE3-3-16r32-043(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
A pharmacy steriliser uses gamma rays. Which list shows electromagnetic waves in order of increasing wavelength?
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Wavelength increases from gamma rays through ultraviolet and infrared to radio waves. The first list is the reverse order (decreasing wavelength). The second swaps ultraviolet and infrared. The third puts ultraviolet before gamma rays, but gamma rays have the shortest wavelength.
✓ 1
Answer D
Answer: D
Exam tip: Remember the order of the spectrum: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma, with wavelength decreasing along the list.
53IGPE3-3-16r32-063(b) Properties of WavesPaper 1PEasy6 marks
A leisure-centre pool has a wave machine at one end. It produces waves of frequency 0.25 Hz. The distance between neighbouring crests is 6.0 m. The pool is 24 m long.
(a)
Calculate the speed of the waves.
(2)
(b)
Calculate the time taken for a wave crest to travel the length of the pool.
(2)
(c)
A float on the surface moves up and down as the waves pass. State what this shows about the type of wave.
(2)
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Part (a)
M1 v = f × λ = 0.25 × 6.0
✓ 1
A1 1.5 m/s
✓ 1
Part (b)
M1 t = distance ÷ speed = 24 ÷ 1.5 (ECF from (a))
✓ 1
A1 16 s
✓ 1
Part (c)
the float moves up and down, at right angles to the direction the wave travels
✓ 1
so the wave is transverse
✓ 1
Exam tip: The wavelength is the distance between neighbouring crests, not the distance between a crest and a trough.
54IGPE3-3-16r34-013(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
Red light and violet light from a distant star travel through space to a telescope in orbit. Which statement compares the two types of light in a vacuum?
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All electromagnetic waves travel at the same speed in a vacuum. Violet light has a shorter wavelength than red light, so from v = fλ with the same v it must have a higher frequency. A higher frequency does not mean a higher speed, and a longer wavelength does not mean a higher speed. The two colours cannot have the same frequency if their wavelengths differ while the speed is the same.
✓ 1
Answer C
Answer: C
Common error: Do not link wave speed to frequency or wavelength alone in a vacuum: speed is the same, so a shorter wavelength means a higher frequency.
55IGPE3-3-16r34-023(b) Properties of WavesPaper 1PEasy1 mark
Which statement correctly describes a sound wave travelling through air?
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Sound is a longitudinal wave: particles vibrate back and forth along the direction of energy transfer, which produces regions of higher pressure (compressions) and lower pressure (rarefactions). Vibration at right angles describes a transverse wave such as light. The particles only oscillate about fixed positions and do not travel to the ear. Sound needs a medium, so it cannot cross a vacuum however loud the source.
✓ 1
Answer A
Answer: A
Common error: Energy is transferred by a wave, not matter. Sound is longitudinal, so the particles move to and fro along the direction of the wave.
57IGPE3-3-16r35-013(b) Properties of WavesPaper 1PEasy1 mark
An ultrasonic sensor on a library book-return robot emits sound of frequency 40 kHz. What is the time period of the sound wave?
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The frequency is 40 000 Hz, so T = 1 ÷ f = 1 ÷ 40 000 = 2.5 × 10−5 s. 2.5 × 10−2 s results from using 40 Hz instead of 40 kHz. 4.0 × 104 s is the frequency in hertz, which is the reciprocal of the period. 2.5 × 10−6 s is a power-of-ten slip of one place.
✓ 1
Answer C
Answer: C
Exam tip: Convert kHz to Hz before using T = 1/f, and check that a high frequency gives a very small period.
58IGPE3-3-16r35-033(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
A food factory needs to kill bacteria on food that is already sealed in plastic packets. Which electromagnetic radiation is best suited to this?
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Gamma rays are ionising and penetrate the packaging, so they kill bacteria inside sealed packets. Visible light does not kill bacteria and is simply reflected or transmitted without harm. Radio waves carry too little energy per photon to damage cells. Infrared mainly heats the surface and would cook the food rather than sterilise it inside the packet.
✓ 1
Answer D
Answer: D
Exam tip: Link each radiation to a property: ionising radiation such as gamma rays kills cells, while infrared transfers thermal energy.
59IGPE3-3-16r35-053(d) Light and SoundPaper 1PEasy6 marks
A wind farm operator uses a detector to record bat calls near the turbines. A bat produces a call of frequency 50 kHz. The speed of sound in air is 340 m/s.
(a)
Explain why a person cannot hear this call.
(2)
(b)
Calculate the time period of the sound wave.
(2)
(c)
Calculate the wavelength of the sound in air.
(2)
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Part (a)
B1 the range of human hearing is about 20 Hz to 20 000 Hz
✓ 1
B1 50 kHz is above the upper limit, so it is ultrasound
✓ 1
Part (b)
M1 T = 1 ÷ f
✓ 1
A1 1 ÷ 50 000 = 2.0 × 10−5 s
✓ 1
Part (c)
M1 λ = v ÷ f
✓ 1
A1 340 ÷ 50 000 = 6.8 × 10−3 m
✓ 1
Common error: Frequencies given in kHz must be converted to Hz before substituting into v = f λ.
61IGPE3-3-15r38-013(b) Properties of WavesPaper 1PEasy1 mark
A loudspeaker in a library produces a note. The speed of sound in air is 340 m/s. The distance from the centre of a compression to the centre of the next rarefaction is 0.40 m. What is the frequency of the note?
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A compression and the next rarefaction are half a wavelength apart, so the wavelength is 0.80 m. Then f = v ÷ λ = 340 ÷ 0.80 = 425 Hz, which is the correct value. 850 Hz comes from treating 0.40 m as a whole wavelength. 136 Hz comes from multiplying the speed by 0.40 m, and 272 Hz from multiplying the speed by 0.80 m; both multiply instead of dividing.
✓ 1
Answer C
Answer: C
Common error: A compression to the next rarefaction is only half a wavelength; a full wavelength runs from one compression to the next compression.
62IGPE3-3-15r38-023(b) Properties of WavesPaper 1PEasy1 mark
In a stadium, spectators stand up and sit down one after another so that a Mexican wave travels horizontally round the stands. Which statement explains why this is a transverse wave?
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In a transverse wave the vibrations are perpendicular to the direction of energy transfer; standing up and sitting down is vertical motion while the wave travels horizontally, so the first statement is correct. Moving round the stadium with the wave means the spectators themselves would be travelling, which a wave does not require. Back-and-forth motion parallel to the travel direction describes a longitudinal wave. Pushing on neighbours is not how the wave is passed on; each spectator simply reacts to the one before.
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Answer A
Answer: A
Exam tip: To decide between transverse and longitudinal, compare the direction in which the particles (or people) vibrate with the direction in which the wave travels.
65IGPE3-3-15r39-013(d) Light and SoundPaper 1PEasy1 mark
A ray of light strikes a plane mirror. The angle of incidence is 35°. What is the angle between the incident ray and the reflected ray?
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The angle of reflection equals the angle of incidence, so the reflected ray is also 35° from the normal and the angle between the two rays is 35° + 35° = 70°. 35° is only one of the two angles. 55° is the angle between the ray and the mirror surface. 110° is twice that surface angle.
✓ 1
Answer B
Answer: B
Common error: Angles of incidence and reflection are always measured from the normal, not from the mirror surface.
66IGPE3-3-15r39-023(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
A television remote control sends signals using radiation of wavelength 9.4×10−7 m. Which part of the electromagnetic spectrum is this?
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A wavelength of 9.4×10−7 m is longer than that of red light (about 7×10−7 m) but much shorter than microwaves, so it is infrared. Ultraviolet has wavelengths shorter than violet light. Visible light spans roughly 4×10−7 m to 7×10−7 m. Microwaves have wavelengths of the order of millimetres to centimetres.
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Answer C
Answer: C
Exam tip: Learn the order of the spectrum by wavelength: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma, and place any given wavelength between its neighbours.
67IGPE3-3-15r39-053(c) The Electromagnetic SpectrumPaper 1PEasy7 marks
A smart home contains four devices that use electromagnetic waves. The wavelength of the radiation used by each device is shown. Doorbell chime: 0.69 m Presence radar sensor: 0.0125 m Thermal motion sensor: 1.0×10−5 m Air-purifier lamp: 2.5×10−7 m speed of electromagnetic waves in a vacuum = 3.0×108 m/s.
(a)
Identify the type of radiation used by each of the four devices.
(4)
(b)
State which device uses radiation of the highest frequency.
(1)
(c)
Calculate the frequency of the radiation used by the presence radar sensor.
(2)
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Part (a)
doorbell chime: radio waves
✓ 1
presence radar sensor: microwaves
✓ 1
thermal motion sensor: infrared
✓ 1
air-purifier lamp: ultraviolet
✓ 1
Part (b)
air-purifier lamp / the shortest wavelength
✓ 1
Part (c)
M1 f = v ÷ λ = 3.0×108 ÷ 0.0125
✓ 1
A1 2.4×1010 Hz
✓ 1
Exam tip: Radiation with a shorter wavelength has a higher frequency because the wave speed is the same for all electromagnetic waves in a vacuum.
68IGPE3-3-15r39-063(d) Light and SoundPaper 1PEasy7 marks
A coffee roaster uses a plane mirror to look at the beans inside a roasting drum. A ray of light from the beans hits the mirror at an angle of 25° to the surface of the mirror.
Diagram not to scale
(a)
State the law of reflection.
(1)
(b)
Calculate the angle of incidence.
(2)
(c)
Calculate the angle between the incident ray and the reflected ray.
(2)
(d)
State two features of the image formed by a plane mirror.
(2)
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Part (a)
angle of incidence = angle of reflection
✓ 1
Part (b)
M1 angle of incidence measured from the normal: 90° − 25°
✓ 1
A1 65°
✓ 1
Part (c)
M1 2 × 65°
✓ 1
A1 130° (ECF from (b))
✓ 1
Part (d)
any two from: virtual / same size as object / same distance behind mirror as object is in front / laterally inverted
✓ 1
second correct feature
✓ 1
Common error: The 25° is measured from the mirror surface, so subtract it from 90° before using the law of reflection.
70IGPE3-3-16r39-013(c) The Electromagnetic SpectrumPaper 1PEasy1 mark
A robot vacuum cleaner has a sensor that emits radiation of wavelength 8.5×10−7 m. Which type of radiation is this?
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8.5×10−7 m is longer than the wavelength of red light (about 7×10−7 m), so it is infrared. Visible light and ultraviolet have shorter wavelengths. Radio waves have wavelengths longer than about a tenth of a metre.
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Answer C
Answer: C
Common error: Wavelengths just beyond red light look close to visible light, but anything longer than about 7×10⁻⁷ m is infrared.
71IGPE3-3-16r39-033(d) Light and SoundPaper 1PEasy1 mark
A student measures the speed of sound by clapping in time with echoes from a wall. Which change would improve the accuracy of the measurement?
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Timing many intervals and dividing makes the error from human reaction time a much smaller fraction of the total time. Standing closer makes each interval shorter, so the same timing error matters more. Timing a single interval gives the largest percentage error. A shorter tape measure does not remove any error and cannot measure the longer distance.
✓ 1
Answer D
Answer: D
Exam tip: Timing many repeats and dividing is the standard way to reduce the effect of reaction time.
72IGPE3-3-16r39-053(c) The Electromagnetic SpectrumPaper 1PEasy8 marks
A smart home has four devices that use electromagnetic waves. The wavelength of the radiation each uses is shown. Wi-Fi router: 0.12 m Smart curtain light sensor: 6.0×10−7 m Hallway motion detector: 1.0×10−5 m Wardrobe sterilising lamp: 2.6×10−7 m speed of electromagnetic waves in a vacuum = 3.0×108 m/s.
(a)
Identify the type of radiation used by each device.
(4)
(b)
List the four devices in order of increasing frequency of radiation.
(2)
(c)
Calculate the frequency of the radiation detected by the curtain light sensor.
(2)
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Part (a)
Wi-Fi router: microwaves
✓ 1
curtain light sensor: visible light
✓ 1
motion detector: infrared
✓ 1
sterilising lamp: ultraviolet
✓ 1
Part (b)
M1 recognises shortest wavelength has highest frequency / reverse order of wavelength
73IGPE3-3-16r39-103(c) The Electromagnetic SpectrumPaper 1PEasy4 marks
A fashion company makes clothing from a fabric that blocks ultraviolet radiation. The ultraviolet radiation from the Sun has a wavelength of 3.0×10−7 m. speed of electromagnetic waves in a vacuum = 3.0×108 m/s.
(a)
State one harmful effect of too much exposure to ultraviolet radiation on human skin.
(1)
(b)
Calculate the frequency of the ultraviolet radiation.
(2)
(c)
State how the speed of ultraviolet radiation in a vacuum compares with the speed of visible light in a vacuum.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
any one from: sunburn; skin cancer; premature ageing of skin
✓ 1
Part (b)
M1 f = v ÷ λ = 3.0×108 ÷ 3.0×10−7
✓ 1
A1 1.0×1015 Hz
✓ 1
Part (c)
they are the same
✓ 1
Exam tip: All electromagnetic waves travel at the same speed in a vacuum, so only wavelength and frequency differ.
74IGPE3-3-15r43-013(d) Light and SoundPaper 1PEasy1 mark
A doctor uses an endoscope containing optical fibres to look inside a patient. Which statement explains why light does not escape through the sides of a fibre?
Show mark scheme
Marking point
Mark
Notes
Total internal reflection needs the angle of incidence to be greater than the critical angle, so B is correct. C reverses the condition: below the critical angle some light always refracts out. A describes absorption and re-emission, which is not how a fibre carries light. D is wrong because refraction along the wall happens only at exactly the critical angle, not for every ray.
✓ 1
Answer B
Answer: B
Common error: Total internal reflection needs an angle of incidence bigger than the critical angle, not smaller.
75IGPE3-3-15r43-023(b) Properties of WavesPaper 1PEasy1 mark
A bird-scaring device on a farm emits sound of frequency 40 kHz. The speed of sound in air is 340 m/s. What is the wavelength of the sound?
Show mark scheme
Marking point
Mark
Notes
Using wavelength = speed ÷ frequency = 340 ÷ 40 000 = 8.5 × 10−3 m, so C is correct. A comes from dividing by 40 instead of 40 000 (kHz not converted). B is frequency ÷ speed, the wrong way round. D multiplies speed by frequency.
✓ 1
Answer C
Answer: C
Common error: Convert kHz to Hz (× 1000) before using v = fλ.
78IGPE3-3-16r43-013(b) Properties of WavesPaper 1PEasy1 mark
A cooling fan in a data centre produces sound that travels through the air to a technician. Which statement about the sound waves in the air is correct?
Show mark scheme
Marking point
Mark
Notes
Sound in air is a longitudinal wave, so the particles vibrate parallel to the direction of travel, making A correct. B describes a transverse wave such as light on a rope or water ripples. C is wrong because only the energy is transferred; the particles oscillate about fixed positions. D has no basis in the wave pattern.
✓ 1
Answer A
Answer: A
Common error: Sound is longitudinal: the particles do not travel with the wave, they only vibrate.
79IGPE3-3-16r43-023(d) Light and SoundPaper 1PEasy1 mark
A cycle reflector contains plastic prisms. Light enters a prism and meets a sloping face with an angle of incidence of 45°. The critical angle of the plastic in air is 42°. What happens to the light at this face?
Show mark scheme
Marking point
Mark
Notes
The angle of incidence (45°) is greater than the critical angle (42°), so all of the light is totally internally reflected: D. A and C contradict this, since above the critical angle no light is refracted out. B has the right outcome but states the condition the wrong way round.
✓ 1
Answer D
Answer: D
Common error: Total internal reflection needs the angle of incidence to be greater than the critical angle.
80IGPE3-3-16r43-053(d) Light and SoundPaper 1PEasy4 marks
A cycle reflector contains a plastic prism with a flat rear face and two sloping faces at right angles to each other. Light from a car headlight enters the rear face along the normal. Each sloping face meets the light at an angle of incidence of 45°. The critical angle of the plastic is 42°.
Diagram not to scale
(a)
Describe the path of the light from entering the rear face until it leaves the prism.
(3)
(b)
State why this is useful for a cyclist at night.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
it passes straight through the rear face without changing direction
✓ 1
it is totally internally reflected at the first sloping face, because 45° is greater than 42°
✓ 1
it is totally internally reflected at the second sloping face and leaves through the rear face, travelling back parallel to the incoming ray
✓ 1
Part (b)
light is sent back towards the driver, so the cyclist can be seen
✓ 1
Exam tip: Link the angle of incidence (45°) to the critical angle (42°) when you describe total internal reflection.
82IGCSE-PHY-3-0063(d) Light and SoundPapers 1 & 2Grade 6–71 mark
The critical angle for light travelling from water into air is 49°. A ray of light travelling upwards through water in a pond hits the water surface. The ray makes an angle of 38° with the water surface. Which statement describes what happens to the ray at the surface?
Show mark scheme
A (totally internally reflected, angle of reflection 52°);
B is incorrect because the angle of reflection is measured from the normal, so it equals the angle of incidence, 52°, not the 38° to the surface
C is incorrect because it treats 38° as the angle of incidence; the angle of incidence is 52°, which is greater than the critical angle
D is incorrect because it uses 38° as the angle of incidence and also bends the ray the wrong way: light entering air from water bends away from the normal
83IGCSE-PHY-3-0073(c) The Electromagnetic SpectrumPapers 1 & 2Grade 6–77 marks
A radio station broadcasts using radio waves of frequency 97.6 MHz. A television company sends signals to a satellite in orbit 36 000 km above the Earth's surface. The satellite immediately sends the signals back down to a receiver directly below it.
(a)
State the speed of electromagnetic waves in free space.
(1)
(b)
Calculate the wavelength of the radio waves.
(3)
(c)
Calculate the time taken for a signal to travel from the ground up to the satellite and back down to the ground. Assume the signals travel at the speed of electromagnetic waves in free space.
(2)
(d)
Microwaves, not radio waves, are used for satellite transmissions. Suggest one reason why.
(1)
Show mark scheme
(a)
3.0 × 10⁸ (m/s);
Allow
300 000 000 (m/s)
3 × 10⁵ km/s
(b)
conversion of 97.6 MHz to 9.76 × 10⁷ Hz;
rearrangement OR substitution;
evaluation;
e.g. λ = v ÷ f = 3.0 × 10⁸ ÷ 9.76 × 10⁷
(λ =) 3.07 (m)
Allow
3.1 (m)
ecf from (a)
3.07 × 10⁶ m or 3.07 × 10⁻⁶ m scores 2 (POT error)
Do not accept
λ = f ÷ v or f × v
(c)
substitution into time = distance ÷ speed with distance 7.2 × 10⁷ m;
evaluation;
e.g. t = 7.2 × 10⁷ ÷ 3.0 × 10⁸
(t =) 0.24 (s)
Allow
0.12 s (one-way distance only) scores 1
ecf from (a)
(d)
any one from: (microwaves) pass through the atmosphere / ionosphere (radio waves may be reflected by it); can be sent as a narrow (focused) beam towards the satellite;
84IGCSE-PHY-3-0083(d) Light and SoundPapers 1 & 2Grade 6–79 marks
A student investigates the refraction of light using a rectangular glass block, a ray box, a protractor and a sheet of paper. The student's results are shown. Angle of incidence i / °: 15, 30, 45, 60, 75 Angle of refraction r / °: 10, 19, 28, 36, 40
(a)
Describe how the student could obtain one pair of values for i and r.
(4)
(b)
State the independent variable in this investigation.
(1)
(c)
Calculate the refractive index of the glass using the result for an angle of incidence of 45°.
(2)
(d)
The student plots a graph of sin i on the y-axis against sin r on the x-axis. State how the student can find the refractive index from this graph.
(1)
(e)
Suggest one way the student could improve the accuracy of the measured angles.
(1)
Show mark scheme
(a)
any four from:
MP1. place block on paper and draw around it;
MP2. draw a normal at the point where the ray enters the block;
MP3. shine a (narrow) ray from the ray box at the block at a measured angle to the normal (using protractor);
MP4. mark the path of the incident and emerging rays (e.g. with crosses or dots);
MP5. remove block and join the entry and exit points to show the refracted ray;
MP6. measure angle between refracted ray and normal with a protractor;
Allow
marks can be scored from a labelled diagram
Do not accept
measuring angles from the surface of the block
(b)
angle of incidence;
Allow
i
Do not accept
angle of refraction
(c)
substitution;
evaluation;
e.g. n = sin 45 ÷ sin 28
(n =) 1.51 (no unit)
Allow
1.5
answers in range 1.50 to 1.51
Do not accept
0.66 (sin 28 ÷ sin 45)
1.61 (45 ÷ 28)
(d)
refractive index = gradient (of the line);
Allow
calculate gradient = change in sin i ÷ change in sin r
Do not accept
area under the graph
(e)
any one from: use a narrower / finer ray; mark rays with points far apart; use a sharp pencil / thin lines; repeat each angle and take a mean; use a darkened room;
Do not accept
use a bigger protractor without explanation
repeat the experiment (with no mention of averaging)
85IGCSE-PHY-3-0093(d) Light and SoundPapers 1 & 2Grade 6–79 marks
Road-side reflectors use right-angled prisms made of acrylic. The refractive index of the acrylic is 1.49. The diagram shows a ray of light entering the long face of the prism at 90° to the surface.
(a)
Explain why the ray does not change direction as it enters the long face.
(1)
(b)
Calculate the critical angle for the acrylic.
(3)
(c)
Explain why the ray is totally internally reflected when it reaches the first short face.
(2)
(d)
The ray is totally internally reflected again at the second short face. Describe the direction of the ray when it leaves the prism through the long face, compared with the incident ray.
(1)
(e)
The manufacturer wants to make the prism from a cheaper plastic. The prism only works if the critical angle of the plastic is less than 45°. Calculate the minimum refractive index the plastic must have.
(2)
Show mark scheme
(a)
ray travels along the normal / angle of incidence is 0°;
Allow
ray hits the surface at 90°
Do not accept
the speed does not change
(b)
substitution;
rearrangement;
evaluation;
e.g. sin c = 1 ÷ 1.49
c = sin⁻¹(1 ÷ 1.49)
(c =) 42.2 (°)
Allow
42 (°)
Do not accept
0.671 (°) (no inverse sine)
(c)
angle of incidence at short face is 45°;
45° is greater than the critical angle (of 42.2°);
Allow
ecf from (b)
(d)
parallel to the incident ray but travelling in the opposite direction;
86IGCSE-PHY-3-0103(b) Properties of WavesPapers 1 & 2Grade 6–77 marks
A loudspeaker produces a sound of frequency 680 Hz. The diagram shows the positions of air particles at one instant. The distance marked is between the centres of two compressions.
(a)
Describe how the air particles move as the sound wave passes.
(2)
(b)
Determine the wavelength of the sound wave.
(1)
(c)
Calculate the speed of the sound wave.
(2)
(d)
State one difference, other than the direction of oscillation, between sound waves and light waves.
(1)
(e)
State one everyday observation which shows that sound waves can be reflected.
(1)
Show mark scheme
(a)
vibrate / oscillate (backwards and forwards) about a fixed position;
parallel to the direction of wave travel / energy transfer;
Do not accept
particles move from the loudspeaker to the end of the diagram
up and down
(b)
0.50 (m);
Allow
0.5 m
Do not accept
1.00 m
0.25 m
(c)
substitution;
evaluation;
e.g. v = 680 × 0.50
(v =) 340 (m/s)
Allow
ecf from (b)
680 (m/s) (using λ = 1.00 m) scores 1
(d)
any one from: sound needs a medium / cannot travel through a vacuum; sound travels much more slowly; light is electromagnetic;
Do not accept
sound is longitudinal (this is the direction of oscillation)
87IGCSE-PHY-3-0113(b) Properties of WavesPapers 1 & 2Grade 6–79 marks
A student uses a ripple tank to study water waves. A vibrating bar produces plane waves of frequency 8.0 Hz. In the deep water the wavelength is 2.5 cm. In the shallow water the waves travel at 0.12 m/s. The diagram shows the wavefronts as they cross a straight boundary between deep and shallow water.
(a)
State what is meant by a wavefront.
(1)
(b)
Calculate the speed of the waves in the deep water, in m/s.
(2)
(c)
State what happens to the frequency of the waves when they enter the shallow water.
(1)
(d)
Calculate the wavelength of the waves in the shallow water.
(3)
(e)
Explain why the waves change direction at the boundary.
(2)
Show mark scheme
(a)
line joining points (on the wave) that are in phase / in step / e.g. a line along a crest;
Do not accept
the front of the wave
(b)
substitution with λ in m;
evaluation;
e.g. v = 8.0 × 0.025
(v =) 0.20 (m/s)
Allow
20 cm/s seen scores MP1 only if not converted to m/s
(c)
stays the same;
Allow
8.0 Hz
Do not accept
decreases
(d)
rearrangement;
substitution;
evaluation;
e.g. λ = v ÷ f = 0.12 ÷ 8.0
(λ =) 0.015 (m)
Allow
1.5 cm
λ = 2.5 × (0.12 ÷ 0.20) for full marks
Do not accept
0.96 (m) (v × f)
(e)
(waves slow down in shallow water and) one part of the wavefront enters the shallow water / slows down before the rest;
so the wavefront turns / changes direction (refraction);
Allow
refraction for MP2 if linked to change in speed
Do not accept
waves bend because the water gets shallower (no reference to speed)
88IGCSE-PHY-3-0263(d) Light and SoundPapers 1 & 2Grade 6–71 mark
A laser beam in air is directed at the flat surface of the water in an aquarium. The refracted beam in the water makes an angle of 30° with the normal. The refractive index of water is 1.33. What is the angle of incidence of the laser beam?
Show mark scheme
C 41.7°;
A is incorrect: 22.1° comes from sin i = sin 30° ÷ 1.33 (equation rearranged the wrong way); the angle in air must be larger than the angle in water
B is incorrect: 39.9° = 1.33 × 30° multiplies the angle instead of its sine
D is incorrect: 48.3° = 90° − 41.7° is the angle between the beam and the water surface, not the normal
89IGCSE-PHY-3-0273(b) Properties of WavesPapers 1 & 2Grade 6–79 marks
A drone flies at constant speed in a straight line. It passes just above a microphone on the ground. The drone's motors produce a steady humming sound. A computer connected to the microphone records the frequency of the sound received. The graph shows how the recorded frequency changes with time. The speed of sound in air is 340 m/s.
(a)
Use the graph to determine the time at which the drone passes the microphone.
(1)
(b)
Explain why the frequency recorded while the drone is approaching is higher than the frequency recorded while it is moving away.
(3)
(c)
Calculate the wavelength of the sound reaching the microphone while the drone is approaching.
(3)
(d)
State the name of this change in the observed frequency and wavelength.
(1)
(e)
The drone flies past again at a greater speed. State how the frequency recorded while it approaches will compare with the value shown on the graph.
(1)
Show mark scheme
(a)
4.0 (s);
Allow
3.8 to 4.2 (s)
(b)
(when approaching) the drone moves towards the microphone between emitting successive wavefronts / wavefronts are bunched together in front of the drone;
so the wavelength (reaching the microphone) is shorter when approaching / longer when moving away;
speed of sound is the same so more waves arrive per second / f = v ÷ λ is higher;
Allow
reverse argument for moving away
Do not accept
the drone makes a higher-pitched sound when approaching
sound travels faster when the drone approaches
(c)
(frequency read from graph =) 510 (Hz);
(λ =) v ÷ f / 340 ÷ 510;
0.67 (m);
Allow
505–510 Hz for MP1
0.666–0.673 (m)
ecf from frequency read
0.75 (m) using the 'moving away' value (453 Hz) scores 2
90IGCSE-PHY-3-0283(d) Light and SoundPapers 1 & 2Grade 6–79 marks
A student investigates refraction using a triangular glass prism. Each angle of the prism is 60°. The refractive index of the glass is 1.50. She places the prism on a sheet of paper and directs a narrow ray of light from a ray box at one face. The angle of incidence is 40°. The diagram shows the ray entering the prism and travelling through the glass to the second face. The normals at both faces are shown as dashed lines.
(a)
Describe how the student can record the path of the ray on the paper so that she can measure the angles after removing the prism.
(2)
(b)
Calculate the angle of refraction at the first face.
(3)
(c)
The ray inside the prism meets the second face with an angle of incidence of 34.6°. Calculate the angle between the ray that leaves the prism and the normal at the second face.
(3)
(d)
Explain why the ray bends away from the normal as it leaves the prism.
(1)
Show mark scheme
(a)
draw around the prism (outline);
mark points / crosses (far apart) along the incident and emerging rays and join with a ruler;
Allow
use pins to mark the ray for MP2
a diagram showing these steps
Do not accept
trace the ray by eye
measure the angles while the ray is shining (without marking)
(b)
sin r = sin 40 ÷ 1.50;
(sin r =) 0.429;
(r =) 25.4(°);
Allow
25(°) for MP3
rearrangement and substitution in one step scores MP1
Do not accept
26.7(°) (40 ÷ 1.50)
(c)
1.50 = sin e ÷ sin 34.6 / sin e = 1.50 × sin 34.6;
(sin e =) 0.852;
(e =) 58.4(°);
Allow
58(°) for MP3
Do not accept
22.2 / 22.3(°) (sin 34.6 ÷ 1.50, equation used the wrong way round)
51.9(°) (1.50 × 34.6)
(d)
light speeds up / travels faster in air than in glass;
Allow
air has a lower refractive index / is less dense optically
91IGCSE-PHY-3-0293(d) Light and SoundPapers 1 & 2Grade 6–76 marks
A museum protects a display cabinet with a laser alarm. A laser beam reflects off a small plane mirror fixed to the cabinet door and travels to a detector. If the beam stops reaching the detector, an alarm sounds. The diagram shows the beam striking the mirror with an angle of incidence of 25°.
(a)
State the angle of reflection.
(1)
(b)
A thief opens the door slightly. This turns the mirror so that the angle of incidence of the laser beam increases to 35°. Determine the angle through which the reflected beam turns.
(3)
(c)
The detector is several metres from the mirror. Use your answer to (b) to explain why the alarm is triggered even when the door moves only a very small amount.
(2)
Show mark scheme
(a)
25(°);
Do not accept
65(°)
(b)
(angle between beams before =) 50(°);
(angle between beams after =) 70(°);
(beam turns through) 20(°);
Allow
mirror (normal) turns through 10° so reflected beam turns through 2 × 10 = 20° scores 3
angle of reflection changes from 25° to 35° scores MP1 and MP2 if linked to both sides of the normal
Do not accept
10(°)
(c)
reflected beam turns through twice the angle the mirror turns;
a small angle at a large distance (several metres) gives a large movement of the beam at the detector / beam misses the detector;
92IGCSE-PHY-3-0303(c) The Electromagnetic SpectrumPapers 1 & 2Grade 6–710 marks
A smart home uses four electromagnetic radiations, P, Q, R and S. P is used by a digital radio. Q carries satellite television signals. R is emitted by a lamp on a night-vision security camera. S is emitted by an insect-killer lamp. The table shows their wavelengths. Radiation / name: P, Q, R, S Wavelength / m: 1.5, 0.025, 8.5 × 10⁻⁷, 3.6 × 10⁻⁷ All electromagnetic waves travel at 3.0 × 10⁸ m/s in air.
(a)
Name the part of the electromagnetic spectrum to which each radiation belongs.
(2)
(b)
State which of the four radiations has the highest frequency.
(1)
(c)
Visible light has wavelengths from 4.0 × 10⁻⁷ m to 7.0 × 10⁻⁷ m. Use the data to explain why a burglar cannot see the light from the camera's lamp R.
(2)
(d)
Calculate how many times greater the frequency of S is than the frequency of P.
(3)
(e)
Explain how the night-vision camera can produce an image of a person in complete darkness.
(2)
Show mark scheme
(a)
P radio (waves), Q microwaves, R infrared, S ultraviolet
two or three correct;
all four correct;
Allow
IR / UV
all four correct scores 2; two or three correct scores 1; zero or one correct scores 0
Do not accept
Q radio waves
S X-rays
(b)
S;
Allow
ultraviolet
(c)
8.5 × 10⁻⁷ m is longer than 7.0 × 10⁻⁷ m / longer than the longest visible (red) wavelength;
so it is (infrared and) outside the visible range / the eye cannot detect it;
Do not accept
R is too weak to see
R travels too fast to be seen
(d)
f = v ÷ λ used for P or S, e.g. 3.0 × 10⁸ ÷ 1.5 = 2.0 × 10⁸ (Hz);
(f_S =) 3.0 × 10⁸ ÷ 3.6 × 10⁻⁷ = 8.3 × 10¹⁴ (Hz);
(ratio =) 4.2 × 10⁶;
Allow
ratio of wavelengths 1.5 ÷ 3.6 × 10⁻⁷ = 4.2 × 10⁶ scores 3 (same speed so f ∝ 1/λ)
4.17 × 10⁶
Do not accept
2.4 × 10⁻⁷ (ratio inverted)
(e)
camera detects infrared (radiation);
emitted by the (warm) person / reflected from the person when lit by R;
Allow
the camera converts infrared into a visible image for MP1 if infrared is mentioned
93IGCSE-PHY-3-0313(d) Light and SoundPapers 1 & 2Grade 6–76 marks
Doctors use an endoscope to see inside a patient's stomach without surgery. An endoscope is a long, flexible tube containing two bundles of very thin glass optical fibres. A lamp outside the patient is connected to one bundle. A camera is connected to the other bundle.
(a)
Explain how the endoscope allows the doctor to see an image of the inside of the stomach. Your answer should refer to total internal reflection and the critical angle.
(6)
Show mark scheme
(a)
Indicative content:
MP1. light from the lamp travels along the first bundle (of fibres) into the stomach;
MP2. light hits the inside surface / boundary of the fibre at an angle of incidence greater than the critical angle;
MP3. so it is totally internally reflected (many times) along the fibre;
MP4. critical angle is the angle of incidence (in the glass) at which the angle of refraction is 90°;
MP5. light reflected (scattered) from the stomach lining enters the second bundle;
MP6. light carried back to the camera / eyepiece by total internal reflection in the same way;
MP7. each fibre carries light from one small part of the stomach / fibres are in the same arrangement at both ends so an image forms;
MP8. fibres can bend (gently) and light still travels along them;
MP9. if a fibre is bent too sharply the angle of incidence becomes less than the critical angle and light escapes (refracts out);
MP10. very little light is lost so the image is bright;
Allow
award 1 mark per MP to a maximum of 6, but use the levels for the overall mark
a labelled diagram of a ray zig-zagging along a fibre with angle > c may credit MP2 and MP3
Do not accept
light is refracted along the fibre
light reflects off the outside of the fibre
critical angle described as the angle at which light is reflected (without reference to refraction at 90°)
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated points, e.g. light travels down the fibre or light is reflected inside the fibre; little or no reference to the critical angle.
Level 2 (3–4)
Explains total internal reflection along a fibre with angle of incidence greater than the critical angle, and describes light going to and returning from the stomach; some gaps in the chain.
Level 3 (5–6)
Full, logically linked explanation: illumination bundle and imaging bundle, TIR at angles greater than c (with c correctly defined), and how an image is formed / limits of bending.
94IGPE-052-00-023(d) Light and SoundPaper 1PMedium1 mark
A ray of red light passes from air into a glass block. Which row describes what happens to the light as it enters the glass?
Show mark scheme
Marking point
Mark
Notes
The frequency is set by the source, so it does not change when the light crosses a boundary. Light travels more slowly in glass. Since v = fλ and f is constant, the wavelength must also decrease. The options that change the frequency assume the wave 'slows down' by vibrating less often. The option with wavelength increasing gets the link between v and λ the wrong way round.
✓ 1
Answer B
Answer: B
Common error: When a wave enters a new medium, frequency never changes. Only speed and wavelength change, and they change in the same direction.
95IGPE-052-00-033(d) Light and SoundPaper 1PMedium1 mark
A type of flint glass used in a camera prism has a refractive index of 1.65. What is the critical angle for light inside this glass meeting a glass–air boundary?
Show mark scheme
Marking point
Mark
Notes
sin c = 1/n = 1/1.65 = 0.606, so c = sin−1(0.606) = 37.3°. Writing 0.61° gives the sine value as if it were the angle. Choosing 31.2° uses tan−1 instead of sin−1. Choosing 52.7° gives the angle measured from the surface (90° − c) rather than from the normal.
✓ 1
Answer C
Answer: C
Exam tip: After finding sin c, press the inverse sine key. Check your calculator is in degree mode.
96IGPE2-056-01-023(b) Properties of WavesPaper 1PMedium1 mark
A Wi-Fi router transmits electromagnetic waves with a frequency of 5.0 GHz. The speed of electromagnetic waves is 3.0 × 108 m/s. What is the wavelength of these waves?
Show mark scheme
Marking point
Mark
Notes
Rearrange v = f × λ to get λ = v ÷ f. λ = 3.0 × 108 ÷ 5.0 × 109 = 0.060 m. 17 m comes from calculating f ÷ v. 60 m comes from treating GHz as MHz. 1.5 × 1018 m comes from multiplying v by f.
✓ 1
Answer A
Answer: A
Exam tip: Convert the prefixes before substituting: k = 10³, M = 10⁶ and G = 10⁹.
97IGPE2-056-01-033(c) The Electromagnetic SpectrumPaper 1PMedium1 mark
A distant star explodes. It emits a burst of gamma rays and a flash of visible light at the same instant. A space telescope above the atmosphere detects both. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
All electromagnetic waves travel at the same speed in a vacuum, 3.0 × 108 m/s. Both waves cover the same distance through space, so they arrive together. Wave speed does not depend on frequency or wavelength, so the first two options are wrong. Gamma rays and visible light have very different wavelengths, so the last option is wrong.
✓ 1
Answer C
Answer: C
Common error: Higher frequency does not mean faster. In a vacuum, every part of the electromagnetic spectrum travels at the same speed.
99IGPE2-056-01-073(b) Properties of WavesPaper 1PMedium6 marks
A speed gun at a tennis tournament measures the speed of a serve. It emits electromagnetic waves of wavelength 1.25 cm towards the ball, and the waves reflect off the ball back to the gun. The speed of electromagnetic waves is 3.0 × 108 m/s.
(a)
State which part of the electromagnetic spectrum the speed gun uses.
(1)
(b)
The ball moves towards the speed gun. Explain why the waves reflected from the ball have a higher frequency than the waves emitted by the gun.
(2)
(c)
Calculate the frequency of the waves emitted by the gun.
(3)
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Part (a)
microwaves
✓ 1
Part (b)
The ball acts as a moving source of the reflected waves: each reflected wavefront leaves from a point closer to the gun, so the wavefronts are closer together / the wavelength is shorter
✓ 1
The wave speed is unchanged, so more waves reach the gun each second and the frequency is higher (Doppler effect)
✓ 1
Part (c)
f = v ÷ λ
✓ 1
f = 3.0 × 108 ÷ 0.0125
✓ 1
f = 2.4 × 1010 Hz
✓ 1
Common error: Convert 1.25 cm to 0.0125 m before substituting. If you use 1.25 directly, your answer will be 100 times too small.
100IGPE2-056-01-123(c) The Electromagnetic SpectrumPaper 1PMedium4 marks
Scientists measure the distance to the Moon by firing pulses of green laser light at reflectors left on the Moon's surface. Each pulse returns to the observatory 2.56 s after it is sent. The speed of light is 3.0 × 108 m/s.
(a)
State which part of the electromagnetic spectrum the laser light belongs to.
(1)
(b)
Calculate the distance from the observatory to the Moon.
(3)
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Part (a)
visible light
✓ 1
Part (b)
distance = speed × time
✓ 1
The 2.56 s covers the journey there and back, so use 1.28 s for one way, or halve the total distance
✓ 1
d = 3.0 × 108 × 1.28 = 3.8 × 108 m (accept 3.84 × 108 m)
✓ 1
Common error: The pulse travels to the Moon and back. If you forget to halve the time, your distance will be twice the true value.
101IGPE2-056-00-023(b) Properties of WavesPaper 1PMedium1 mark
A racing car drives straight towards a spectator standing beside the track. The engine gives out a sound of constant frequency. Compared with the sound heard when the car is stationary, what does the spectator hear as the car approaches?
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The car moves towards the spectator between emitting one wavefront and the next, so the wavefronts arrive bunched together. The wavelength is shorter. The speed of sound in air does not change, so by v = fλ the observed frequency is higher. This is the Doppler effect. Option A wrongly pairs a higher frequency with a longer wavelength. Option C describes a source moving away, and gets even that only half right. Option D wrongly assumes the speed of sound depends on the speed of the source.
✓ 1
Answer B
Answer: B
Exam tip: In the Doppler effect the wave speed stays the same. Frequency and wavelength always change in opposite directions.
102IGPE2-056-00-043(b) Properties of WavesPaper 1PMedium6 marks
A seismometer records primary (P) waves from an earthquake. P-waves are longitudinal waves. A P-wave travels 240 km through rock in 40 s.
(a)
Describe how the particles of rock move as a longitudinal wave passes through them.
(2)
(b)
Calculate the speed of the P-wave in m/s.
(2)
(c)
The P-wave has a frequency of 2.0 Hz. Calculate its wavelength.
(2)
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Part (a)
The particles vibrate / oscillate parallel to the direction of energy transfer / wave travel . Any one of: they oscillate about fixed positions with no net movement along the wave / the motion produces compressions and rarefactions .
✓ 2
Part (b)
Use speed = distance / time: v = 240 000 / 40 . v = 6000 m/s .
✓ 2
Part (c)
Use λ = v / f = 6000 / 2.0 . λ = 3000 m . Allow error carried forward from the speed found in (b).
✓ 2
Common error: Convert km to m before calculating. Leaving 240 km unconverted gives a speed that is 1000 times too small.
103IGPE2-056-00-073(b) Properties of WavesPaper 1PMedium5 marks
In a school athletics 100 m race, a timekeeper stands at the finish line. The starter fires a pistol at the start line, 100 m away. The timekeeper starts a stopwatch when the bang is heard. The speed of sound in air is 340 m/s.
Diagram not to scale
(a)
Calculate the time taken for the sound to reach the timekeeper.
(2)
(b)
Explain the effect this delay has on the runners' recorded times.
(2)
(c)
Suggest how the timekeeper could reduce this error without new equipment.
(1)
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Part (a)
t = distance / speed = 100 / 340 . t = 0.29 s .
✓ 2
Part (b)
The stopwatch starts about 0.29 s after the runners have started . So every recorded time is about 0.29 s shorter than the true time .
✓ 2
Part (c)
Start the stopwatch on a visual signal at the start instead of the sound, because light reaches the timekeeper almost instantly. Accept the flash or smoke from the pistol, or the starter's arm dropping or a signal given at the moment of firing .
✓ 1
Exam tip: In 'explain the effect' questions, state the direction of the error (too short or too long) as well as its size.
104IGPE2-056-03-023(b) Properties of WavesPaper 1PMedium1 mark
A buoy marking the edge of an offshore wind farm rises and falls 12 complete times in 30 s. The distance between neighbouring wave crests is 6.0 m. What is the speed of the water waves?
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Frequency = 12 ÷ 30 = 0.40 Hz. Wave speed = frequency × wavelength = 0.40 × 6.0 = 2.4 m/s. Option B (15 m/s) multiplies the period (2.5 s) by the wavelength. Option C (72 m/s) multiplies the number of oscillations by the wavelength and ignores the time. Option D (0.42 m/s) divides the period by the wavelength.
✓ 1
Answer A
Answer: A
Exam tip: Turn 'number of waves in a time' into a frequency in Hz first. Only then use v = f × λ.
105IGPE2-056-03-033(b) Properties of WavesPaper 1PMedium1 mark
A radar speed gun at a tennis tournament sends microwaves towards a ball that is moving directly towards the gun. The microwaves reflect off the ball and return to the gun. Compared with the emitted microwaves, the reflected microwaves received by the gun have
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Notes
The ball moves towards the gun, so the reflected wavefronts are bunched together. This is the Doppler effect. The wavelength becomes shorter. The wave speed (the speed of light) does not change, so v = fλ means the frequency must rise. Option B ignores v = fλ, because frequency and wavelength cannot both increase at a fixed speed. Option C is the effect for an object moving away. Option D forgets that a shorter wavelength at constant speed requires a higher frequency.
✓ 1
Answer A
Answer: A
Exam tip: For the Doppler effect, an approaching source or reflector gives a shorter wavelength and a higher frequency. One moving away gives a longer wavelength and a lower frequency. Wave speed stays the same.
106IGPE2-056-03-113(c) The Electromagnetic SpectrumPaper 1PMedium7 marks
Engineers on Earth control a rover on Mars using radio waves. On one day, Mars is 2.25 × 1011 m from Earth. Radio waves travel at 3.0 × 108 m/s.
(a)
Calculate the time taken for a radio signal to travel from Earth to the rover.
(2)
(b)
The radio signal has a frequency of 8.4 GHz. Calculate its wavelength.
(3)
(c)
Suggest why the rover cannot be steered in real time with a joystick on Earth.
(2)
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Part (a)
t = d ÷ v = 2.25 × 1011 ÷ 3.0 × 108
✓ 1
= 750 s
✓ 1
Part (b)
λ = v ÷ f
✓ 1
= 3.0 × 108 ÷ 8.4 × 109
✓ 1
= 0.036 m
✓ 1
Part (c)
Commands arrive about 12.5 minutes after they are sent, and the images take as long to return . The rover could hit an obstacle before a correction arrives, so its moves must be planned in advance or it must steer itself .
✓ 2
Exam tip: Write powers of ten carefully on your calculator, for example 8.4 × 10⁹ entered as 8.4E9. A slip here changes the answer by whole powers of ten.
107IGPE2-056-02-023(d) Light and SoundPaper 1PMedium1 mark
A jeweller tests a cut gemstone in air and finds that its critical angle is 35°. What is the refractive index of the gemstone?
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For a material surrounded by air, sin c = 1/n, so n = 1/sin c = 1/sin 35° = 1/0.574 = 1.74. Option A is sin 35° itself, because the equation was not rearranged. Option B is 1/cos 35°, which comes from measuring the angle from the surface instead of from the normal. Option D is 1/35, where the sine was left out. A refractive index is always greater than 1, which rules out A and D straight away.
✓ 1
Answer C
Answer: C
Common error: Forgetting to take the sine of the angle. Check that your calculator is in degree mode, and remember that a refractive index can never be less than 1.
108IGPE2-056-02-083(d) Light and SoundPaper 1PMedium6 marks
An engineer is testing a new acrylic plastic for car rear-light covers. A student measures the refraction of light in a rectangular block of the plastic and records these results: angle of incidence 20°, angle of refraction 13.3° angle of incidence 40°, angle of refraction 25.6° angle of incidence 60°, angle of refraction 35.5°
(a)
Describe how the student could use a ray box, the block, paper and a protractor to find an angle of refraction.
(3)
(b)
Suggest why taking results at three different angles is better than taking only one.
(1)
(c)
Use the results for an angle of incidence of 40° to calculate the refractive index of the acrylic.
(2)
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Part (a)
Draw around the block on paper and shine a ray into it at a measured angle to a drawn normal
✓ 1
Mark the incoming and outgoing rays with crosses, remove the block and join the entry and exit points to show the ray inside the block
✓ 1
Measure the angle between this refracted ray and the normal using the protractor
✓ 1
Part (b)
Allows a mean refractive index to be found / anomalous results can be spotted, so the effect of random errors is reduced
✓ 1
Part (c)
n = sin i / sin r = sin 40° / sin 25.6°
✓ 1
n = 0.643 / 0.432 = 1.49
✓ 1
Common error: Dividing the angles themselves (40 ÷ 25.6) instead of their sines. The refractive index uses sin i ÷ sin r.
110IGPE-052-00-063(c) The Electromagnetic SpectrumPaper 1PMedium7 marks
The pilot of a radio-controlled model glider uses a transmitter that sends signals at a frequency of 2.4 GHz. Speed of electromagnetic waves = 3.0 × 108 m/s.
(a)
State the equation linking wave speed, frequency and wavelength.
(1)
(b)
Calculate the wavelength of the signal.
(3)
(c)
The glider is 600 m from the pilot. Calculate the time for a signal to travel from the transmitter to the glider.
(2)
(d)
Suggest why the pilot does not notice any delay between moving the controls and the glider responding.
(1)
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Part (a)
wave speed = frequency × wavelength, or v = f × λ
✓ 1
Part (b)
f = 2.4 × 109 Hz ; λ = (3.0 × 108) ÷ (2.4 × 109) ; λ = 0.125 m, accept 0.13 m
✓ 3
Part (c)
t = distance ÷ speed = 600 ÷ (3.0 × 108) ; t = 2.0 × 10−6 s
✓ 2
Part (d)
The travel time (a few millionths of a second) is far too short for a human to detect / it is much shorter than human reaction time
✓ 1
Common error: 1 GHz is 10⁹ Hz, not 10⁶ Hz. Mixing up giga and mega makes the wavelength 1000 times too large.
111IGPE3-3-14r21-033(b) Properties of WavesPaper 1PMedium1 mark
Water waves of frequency 4 Hz and wavelength 6.0 cm travel across a ripple tank. The frequency of the waves is increased to 8 Hz and the depth of water does not change. What is the new wavelength?
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The speed is 4 × 6.0 = 24 cm/s and stays the same because the depth is unchanged, so λ = 24 ÷ 8 = 3.0 cm, so C is correct. A multiplies wavelength by the frequency ratio. B assumes wavelength is unchanged. D assumes the speed also halves (12 ÷ 8).
✓ 1
Answer C
Answer: C
Common error: If the medium does not change, wave speed stays constant, so doubling the frequency halves the wavelength.
112IGPE3-3-14r21-043(d) Light and SoundPaper 1PMedium1 mark
Archaeologists survey a sunken harbour wall using a sonar pulse. The pulse is reflected from the wall and detected 0.080 s after it was sent. The speed of sound in the water is 1500 m/s. How far is the wall from the sonar?
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The pulse travels to the wall and back, so distance = 1500 × 0.080 ÷ 2 = 60 m, so B is correct. A forgets to halve. C halves twice. D divides speed by time.
✓ 1
Answer B
Answer: B
Exam tip: For an echo, the total time covers the journey there and back, so use half of it.
114IGPE3-3-14r21-083(b) Properties of WavesPaper 1PMedium7 marks
Fans in a data centre produce sound of frequency 1.2 kHz. The speed of sound in air is 340 m/s.
(a)
Describe how the sound travels through the air from a fan to a technician's ear.
(3)
(b)
Calculate the wavelength of the sound. Give your answer to 3 significant figures.
(2)
(c)
The technician wears ear defenders that make the sound quieter. State what happens to the amplitude and to the frequency of the sound reaching the ear.
(2)
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Part (a)
B1 air particles vibrate parallel to the direction in which the sound travels
✓ 1
B1 this forms compressions and rarefactions
✓ 1
B1 energy is passed on without the air as a whole moving from the fan to the ear
✓ 1
Part (b)
M1 λ = 340 ÷ 1200
✓ 1
A1 0.283 m
✓ 1
Part (c)
B1 amplitude decreases
✓ 1
B1 frequency stays the same
✓ 1
Exam tip: Longer answers about sound should use the terms compression, rarefaction and parallel to the direction of travel.
116IGPE3-3-14r21-123(d) Light and SoundPaper 1PMedium6 marks
A cyclist stands 425 m from a long cliff face and claps once. An echo is heard 2.50 s after the clap.
Diagram not to scale
(a)
Calculate the speed of sound in air.
(3)
(b)
Suggest one way to make the timing in this experiment more accurate.
(1)
(c)
The cyclist then stands 5.0 m from a wall. Explain why the echo cannot be heard as a separate sound. The ear cannot separate sounds less than about 0.1 s apart.
(2)
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Part (a)
M1 distance travelled = 2 × 425 = 850 m
✓ 1
M1 speed = distance ÷ time = 850 ÷ 2.50
✓ 1
A1 340 m/s
✓ 1
Part (b)
B1 repeat and average, or use a longer distance, or use an electronic timer
✓ 1
Part (c)
M1 time for echo = 2 × 5.0 ÷ 340 = 0.029 s
✓ 1
A1 this is less than about 0.1 s so the echo merges with the original sound
✓ 1
Exam tip: For echoes, the sound covers twice the distance to the reflector.
117IGPE3-3-14r22-023(b) Properties of WavesPaper 1PMedium1 mark
A radio station broadcasts at a frequency of 100 MHz. Radio waves travel at 3.0 × 108 m/s in air. What is the wavelength of the radio waves?
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Wavelength = speed ÷ frequency = 3.0 × 108 ÷ 1.0 × 108 = 3.0 m. The value 0.33 m comes from dividing frequency by speed. The value 30 m comes from converting 100 MHz to 107 Hz instead of 108 Hz. The value 3.0 × 1016 m comes from multiplying speed by frequency.
✓ 1
Answer D
Answer: D
Exam tip: Convert MHz to Hz (× 10⁶) before using v = fλ, and rearrange to λ = v ÷ f.
118IGPE3-3-14r22-053(b) Properties of WavesPaper 1PMedium7 marks
A ripple tank is used to study water waves. A vibrating bar produces straight waves of frequency 6.0 Hz. In the deep region the wavelength is 3.0 cm. The waves then move into a shallow region where the wavelength is 1.8 cm.
(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.
(2)
(d)
The waves reach the boundary between the regions at an angle. Describe what happens to their direction and explain why.
(2)
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Part (a)
M1 v = f × λ = 6.0 × 3.0
✓ 1
A1 18 cm/s
✓ 1
Part (b)
B1 6.0 Hz
✓ 1
Part (c)
M1 6.0 × 1.8
✓ 1
A1 10.8 cm/s
✓ 1
Part (d)
B1 the waves bend towards the normal (change direction)
✓ 1
B1 because they slow down, so one part of each wavefront slows before another
✓ 1
Exam tip: Write the equation v = fλ first, then substitute; keep the units the same as the wavelength units given (cm gives cm/s).
119IGPE3-3-14r22-063(b) Properties of WavesPaper 1PMedium7 marks
A passenger aircraft receives internet data by microwaves sent through a satellite in a geostationary orbit. The satellite is 3.6 × 107 m above the aircraft. Microwaves travel at 3.0 × 108 m/s in air and in space.
(a)
State two properties that are the same for all electromagnetic waves.
(2)
(b)
The microwaves have a frequency of 12 GHz. Calculate their wavelength. (1 GHz = 1.0 × 109 Hz)
(3)
(c)
A message is sent from the satellite down to the aircraft and the aircraft replies to the satellite straight away. Calculate the time taken for the signal to travel down to the aircraft and the reply to travel back up to the satellite.
(2)
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Part (a)
B1 any one from: transverse waves; travel at the same speed in a vacuum; transfer energy without transferring matter
121IGPE3-3-14r22-123(d) Light and SoundPaper 1PMedium8 marks
A greenhouse roof is made from a clear plastic sheet. A ray of light in air meets the sheet at an angle of incidence of 40°, and the angle of refraction is 25°. The refractive index n of a material is given by n = sin i ÷ sin r.
Diagram not to scale
(a)
Calculate the refractive index of the plastic. Give your answer to 3 significant figures.
(3)
(b)
Another ray meets the sheet at an angle of incidence of 60°. Use n = 1.52 to 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 plastic.
(2)
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Part (a)
M1 n = sin 40° ÷ sin 25°
✓ 1
M1 0.6428 ÷ 0.4226
✓ 1
A1 1.52 (1.521)
✓ 1
Part (b)
M1 sin r = sin 60° ÷ 1.52
✓ 1
M1 r = sin−1(0.5697)
✓ 1
A1 34.7°
✓ 1
Part (c)
B1 light travels more slowly in the plastic than in air
✓ 1
B1 so the wavefront changes direction as it slows
✓ 1
Exam tip: Check that your angle of refraction is smaller than the angle of incidence when light enters a denser material.
122IGPE3-3-15r22-033(d) Light and SoundPaper 1PMedium1 mark
A hospital scanner sends a short pulse of ultrasound into the body. An echo from a boundary between two tissues returns to the probe 80 μs after the pulse was sent. The speed of ultrasound in the tissue is 1500 m/s. How far below the probe is the boundary?
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The pulse travels to the boundary and back, so the total distance is 1500 × 80×10−6 = 0.12 m. The boundary is half this distance, 0.060 m = 6.0 cm. 12 cm forgets that the pulse makes a round trip. 3.0 cm halves the distance twice. 60 cm comes from a factor-of-ten error when converting microseconds.
✓ 1
Answer D
Answer: D
Exam tip: In echo questions, the time recorded is for the journey there and back, so halve the distance travelled.
123IGPE3-3-15r22-073(b) Properties of WavesPaper 1PMedium6 marks
Waves travel towards a beach. Far out at sea, a buoy shows that the waves have a time period of 8.0 s and travel at 6.0 m/s. Near the shore the sea is shallower and the waves slow down to 3.0 m/s. The frequency of the waves does not change.
(a)
Calculate the wavelength of the waves far out at sea. Use the unrounded frequency in your working.
(3)
(b)
Explain why the frequency of the waves stays the same as they move into shallower water.
(1)
(c)
Calculate the wavelength of the waves near the shore.
(2)
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Part (a)
f = 1 ÷ 8.0 = 0.125 Hz
✓ 1
λ = v ÷ f = 6.0 ÷ 0.125
✓ 1
= 48 m
✓ 1
Part (b)
the frequency is set by the source of the waves / the same number of waves arrive each second as leave the deep water
✓ 1
Part (c)
λ = 3.0 ÷ 0.125
✓ 1
= 24 m
✓ 1
Common error: When a wave slows down, its frequency stays fixed and its wavelength becomes shorter. Do not swap the two.
124IGPE3-3-15r22-083(c) The Electromagnetic SpectrumPaper 1PMedium8 marks
A food factory uses microwaves of frequency 2.45 GHz to heat ready meals on a conveyor belt. The speed of electromagnetic waves in a vacuum is 3.00×108 m/s. 1 GHz = 1×109 Hz. Ultraviolet lamps are used above the belt to sterilise the packaging.
(a)
State one hazard to workers from exposure to ultraviolet radiation.
(1)
(b)
Explain why the microwaves make the food hot.
(2)
(c)
Infrared lamps are used to dry the coating on biscuits. Explain why infrared radiation can raise the temperature of the biscuits.
(2)
(d)
Calculate the wavelength of the microwaves. Give your answer to 3 significant figures.
(3)
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Part (a)
skin cancer / skin burns / eye damage
✓ 1
Part (b)
water molecules in the food absorb the microwaves
✓ 1
energy is transferred to the thermal energy store of the food / the particles gain kinetic energy, so the temperature rises
✓ 1
Part (c)
the infrared radiation is absorbed by the surface of the biscuits
✓ 1
energy is transferred to the thermal store, so the particles vibrate more
✓ 1
Part (d)
v = f × λ, or λ = v ÷ f
✓ 1
λ = (3.00×108) ÷ (2.45×109)
✓ 1
= 0.122 m
✓ 1
Exam tip: Convert the prefix before you substitute: 2.45 GHz is 2.45×10⁹ Hz.
125IGPE3-3-15r22-093(d) Light and SoundPaper 1PMedium7 marks
A student investigates the refractive index of a rectangular glass block using a ray box and a protractor. A ray enters the top face of the block and leaves through the bottom face. The student draws a normal at the point where the ray enters and measures the angle of incidence i and the angle of refraction r. Her results are: Reading 1: i = 25°, r = 16° Reading 2: i = 40°, r = 25° Reading 3: i = 55°, r = 32°
(a)
State the line from which both angles must be measured.
(1)
(b)
Suggest one way the student can improve the accuracy of her measurement of the path of the ray.
(1)
(c)
Explain why the student calculates the refractive index from three readings and finds the mean.
(1)
(d)
Use the readings to calculate the mean refractive index of the glass. Give your answer to 3 significant figures.
(4)
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Part (a)
the normal
✓ 1
Part (b)
use a narrower ray / use a sharp pencil to mark the path / put the marks on the path further apart
✓ 1
Part (c)
to reduce the effect of random errors / to spot an anomalous result
✓ 1
Part (d)
n = sin i ÷ sin r
✓ 1
at least two values of n correct to 2 d.p.: 1.53, 1.52, 1.55
✓ 1
mean = sum of the three values ÷ 3
✓ 1
= 1.53
✓ 1
Exam tip: Use sine of each angle, not the angle itself. A refractive index of glass should be a little above 1.5.
127IGPE3-3-15r22-143(d) Light and SoundPaper 1PMedium6 marks
During a Formula 1 test, a microphone is placed in the pit lane. The engine sound travels across the track, reflects from a grandstand and returns to the microphone 1.6 s after a short burst of engine noise. The speed of sound in air is 340 m/s.
(a)
State the range of frequencies that a healthy young person can hear.
(1)
(b)
As the driver accelerates, the frequency of the engine noise rises. State what happens to the pitch and to the wavelength of the sound. The speed of sound does not change.
(2)
(c)
Calculate the distance from the microphone to the grandstand. Give your answer to 3 significant figures.
(3)
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Part (a)
20 Hz to 20 000 Hz
✓ 1
Part (b)
the pitch gets higher
✓ 1
the wavelength gets shorter
✓ 1
Part (c)
time for one-way journey = 1.6 ÷ 2 = 0.80 s
✓ 1
distance = 340 × 0.80
✓ 1
= 272 m
✓ 1
Exam tip: State the number of significant figures in your final line: 272 m has 3 significant figures, so do not round it to 270 m.
128IGPE3-3-14r26-033(b) Properties of WavesPaper 1PMedium1 mark
Light from the air enters the water of a swimming pool. Which row shows what happens to the speed, wavelength and frequency of the light?
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Light slows down in water. The frequency is set by the source and does not change at a boundary, so with v = fλ the wavelength must decrease. The first and second options wrongly change the frequency, and the third has light speeding up in water.
✓ 1
Answer D
Answer: D
Common error: Frequency never changes when a wave crosses a boundary; speed and wavelength change together.
129IGPE3-3-14r26-043(b) Properties of WavesPaper 1PMedium1 mark
Water waves of wavelength 24 m travel at 6.0 m/s past a wind-farm turbine foundation. What is the time between two successive crests passing the foundation?
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The period is T = λ ÷ v = 24 ÷ 6.0 = 4.0 s. The value 0.25 is v ÷ λ, which is the frequency in hertz, not the period; 18 comes from subtracting the speed from the wavelength; 144 comes from multiplying them.
✓ 1
Answer B
Answer: B
Exam tip: Check the unit of your answer: a time in seconds comes from dividing a distance by a speed.
130IGPE3-3-14r26-063(b) Properties of WavesPaper 1PMedium8 marks
A loudspeaker in a library plays a steady note of frequency 480 Hz. The speed of sound in air is 340 m/s.
(a)
Explain how the vibrating cone of the loudspeaker produces compressions and rarefactions in the air.
(3)
(b)
State what is meant by a rarefaction.
(1)
(c)
Calculate the distance between the centres of two adjacent compressions. Give your answer to 3 significant figures.
(2)
(d)
The volume of the note is increased without changing its pitch. State what happens to the amplitude and to the wavelength of the sound.
(2)
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Part (a)
The cone moves forward and pushes air particles together, forming a compression
✓ 1
The cone moves backward, so particles spread out and form a rarefaction
✓ 1
This repeats as the cone vibrates, sending a series of compressions and rarefactions along the air
✓ 1
Part (b)
A region where the air particles are further apart than normal (lower pressure)
✓ 1
Part (c)
M1 λ = v ÷ f
✓ 1
A1 0.708 m
✓ 1
Part (d)
Amplitude increases
✓ 1
Wavelength stays the same
✓ 1
Exam tip: Compressions and rarefactions are regions along the direction of travel, so sound is a longitudinal wave; do not draw them as peaks and troughs.
131IGPE3-3-14r26-083(c) The Electromagnetic SpectrumPaper 1PMedium10 marks
A smart home contains devices that use electromagnetic radiation. The table gives the wavelength in a vacuum of the radiation each device uses. The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s. Wi-Fi router: 0.12 m TV remote control: 9.4 × 10−7 m Ultraviolet lamp for disinfecting: 2.6 × 10−7 m FM radio speaker: 3.0 m
(a)
Name the type of electromagnetic radiation used by each of the four devices.
(4)
(b)
Calculate the frequency of the radiation used by the Wi-Fi router. Give your answer in hertz in standard form.
(3)
(c)
State which of the four devices uses radiation of the highest frequency.
(1)
(d)
Explain why the ultraviolet lamp must not be used while people are in the room.
(2)
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Notes
Part (a)
Wi-Fi router: microwaves
✓ 1
TV remote control: infrared
✓ 1
Ultraviolet lamp: ultraviolet
✓ 1
FM radio speaker: radio waves
✓ 1
Part (b)
M1 v = fλ rearranged to f = v ÷ λ
✓ 1
M1 3.0 × 108 ÷ 0.12
✓ 1
A1 2.5 × 109 Hz
✓ 1
Part (c)
The ultraviolet lamp (it has the shortest wavelength)
✓ 1
Part (d)
Ultraviolet radiation is absorbed by skin and eyes and can damage cells
✓ 1
This can cause skin burns or cancer, or damage to the eyes
✓ 1
Exam tip: All electromagnetic waves travel at the same speed in a vacuum, so the shorter the wavelength, the higher the frequency.
133IGPE3-3-14r26-133(c) The Electromagnetic SpectrumPaper 1PMedium6 marks
A wind-farm operator sends control signals to a turbine using long-wave radio of frequency 200 kHz. A hill about 300 m wide lies between the transmitter and a turbine in a valley behind the hill. The speed of radio waves is 3.0 × 108 m/s.
(a)
State what is meant by diffraction.
(1)
(b)
Calculate the wavelength of the radio waves.
(2)
(c)
Explain why the long-wave signal can reach the turbine behind the hill, but a Wi-Fi signal of wavelength 0.12 m would not.
(3)
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Notes
Part (a)
The spreading of waves as they pass an edge or through a gap
✓ 1
Part (b)
M1 λ = v ÷ f with f = 200 × 103 Hz
✓ 1
A1 1500 m
✓ 1
Part (c)
Diffraction is greatest when the wavelength is similar to or larger than the size of the obstacle
✓ 1
The radio wavelength (1500 m) is larger than the hill, so the waves bend around it
✓ 1
The Wi-Fi wavelength (0.12 m) is much smaller than the hill, so there is very little diffraction and a shadow forms
✓ 1
Exam tip: Compare the wavelength with the size of the gap or obstacle: diffraction is significant when they are of similar size or the wavelength is larger.
134IGPE3-3-16r28-033(c) The Electromagnetic SpectrumPaper 1PMedium1 mark
Which row correctly matches a type of electromagnetic radiation with a use and a danger?
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Notes
Microwaves carry signals to and from satellites, and microwaves absorbed by water in body cells heat the tissue, so this row is correct. Night-vision cameras detect infrared, not ultraviolet. Food is cooked with microwaves or infrared, not X-rays. Radio waves are not ionising and do not sterilise equipment (gamma rays are used).
✓ 1
Answer D
Answer: D
Common error: Learn each radiation as a use plus a danger; ionising radiation (UV, X-rays, gamma) damages cells, while microwaves and infrared mainly heat.
135IGPE3-3-16r28-073(b) Properties of WavesPaper 1PMedium8 marks
At a football stadium, spectators in a long straight stand keep a Mexican wave going. The wave travels along the stand at 9.0 m/s, and each spectator stands up once every 2.0 s.
(a)
Explain why the Mexican wave is a transverse wave.
(2)
(b)
Calculate the wavelength of the wave.
(3)
(c)
The stand is 144 m long. Calculate the time taken for the wave to travel its whole length.
(2)
(d)
State what is transferred along the stand by the wave.
(1)
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Notes
Part (a)
Each spectator moves up and down
✓ 1
at right angles to the direction in which the wave travels along the stand
✓ 1
Part (b)
M1 frequency = 1 ÷ 2.0 = 0.50 Hz
✓ 1
M1 λ = v ÷ f = 9.0 ÷ 0.50
✓ 1
A1 18 m
✓ 1
Part (c)
M1 time = 144 ÷ 9.0
✓ 1
A1 16 s
✓ 1
Part (d)
Energy
✓ 1
Exam tip: A wave transfers energy, not matter: the spectators stay in their seats.
136IGPE3-3-16r28-093(c) The Electromagnetic SpectrumPaper 1PMedium6 marks
A space probe is 6.0 × 1010 m from Earth. It sends data to Earth using radio waves of frequency 8.4 GHz. The speed of radio waves in a vacuum is 3.0 × 108 m/s.
(a)
Calculate the time taken for a signal to reach Earth.
(2)
(b)
State one other property that radio waves share with all electromagnetic waves.
(1)
(c)
Calculate the wavelength of the radio waves. Give your answer to 2 significant figures.
138IGPE3-3-16r32-023(b) Properties of WavesPaper 1PMedium1 mark
Straight water waves in a ripple tank move from a deep region into a shallow region. Which row describes what happens to the frequency, wavelength and speed of the waves as they enter the shallow region? Frequency / wavelength / speed
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Notes
The source fixes the frequency, so it is unchanged. The waves travel more slowly in shallow water, and because v = fλ with f constant the wavelength decreases too. Frequency decreasing with constant speed is not correct because the source has not changed. Unchanged wavelength with a lower speed would break v = fλ. A faster wave in shallow water is the opposite of what happens.
✓ 1
Answer A
Answer: A
Common error: Frequency never changes when a wave crosses a boundary; speed and wavelength change together.
139IGPE3-3-16r32-073(b) Properties of WavesPaper 1PMedium7 marks
In a ripple tank, a vibrating bar produces straight waves of frequency 8.0 Hz. In the deep region the wavelength is 2.5 cm. The waves then pass into a shallow region above a glass plate, where the wavelength is 1.5 cm.
(a)
Calculate the speed of the waves in the deep region.
(2)
(b)
Calculate the speed of the waves in the shallow region.
(2)
(c)
The wavefronts meet the edge of the glass plate at an angle. Explain why the waves change direction.
(3)
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Notes
Part (a)
M1 v = f × λ = 8.0 × 2.5
✓ 1
A1 20 cm/s
✓ 1
Part (b)
M1 the frequency is still 8.0 Hz, so v = 8.0 × 1.5
✓ 1
A1 12 cm/s
✓ 1
Part (c)
one part of each wavefront reaches the shallow region first
✓ 1
and slows down while the rest is still moving faster
✓ 1
so the wavefront turns, bending towards the normal
✓ 1
Exam tip: Keep the same units as the data given: here cm and cm/s are fine, so there is no need to convert to metres.
140IGPE3-3-16r32-103(d) Light and SoundPaper 1PMedium6 marks
An ultrasound pulse is sent along a water-filled pipe and reflects from a closed valve 30 m from the transmitter. The echo returns to the transmitter after 0.0400 s.
(a)
Calculate the speed of sound in the water.
(2)
(b)
The pipe is drained and filled with air, where the speed of sound is 340 m/s. Calculate the time for an echo from the same valve. Give your answer to 3 significant figures.
(2)
(c)
Give two reasons why ultrasound is used for this job instead of audible sound.
(2)
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Notes
Part (a)
M1 total distance travelled = 2 × 30 = 60 m
✓ 1
A1 speed = 60 ÷ 0.0400 = 1500 m/s
✓ 1
Part (b)
M1 time = 60 ÷ 340
✓ 1
A1 0.176 s
✓ 1
Part (c)
it cannot be heard, so it does not disturb people nearby
✓ 1
its short wavelength gives a narrow beam and shows small features / it can be told apart from background noise (any valid second reason)
✓ 1
Common error: An echo travels to the reflector and back, so use twice the distance to the reflector.
141IGPE3-3-16r32-113(c) The Electromagnetic SpectrumPaper 1PMedium7 marks
A pharmacy cabinet kills bacteria on equipment using ultraviolet radiation of wavelength 2.5 × 10−7 m. Electromagnetic waves travel at 3.0 × 108 m/s in a vacuum.
(a)
Calculate the frequency of this radiation.
(2)
(b)
State two properties that all electromagnetic waves have in common.
(2)
(c)
State one hazard to people of exposure to ultraviolet radiation.
(1)
(d)
Visible light has a wavelength of 6.0 × 10−7 m. State, with a reason, whether visible light or this ultraviolet radiation has the higher frequency.
(2)
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Notes
Part (a)
M1 f = v ÷ λ = 3.0 × 108 ÷ 2.5 × 10−7
✓ 1
A1 1.2 × 1015 Hz
✓ 1
Part (b)
they are transverse waves
✓ 1
they travel at the same speed in a vacuum / they transfer energy (any valid second property)
✓ 1
Part (c)
skin cancer / eye damage / skin damage (any one)
✓ 1
Part (d)
the ultraviolet radiation
✓ 1
its wavelength is shorter and the speed is the same, so the frequency is higher
✓ 1
Exam tip: Write powers of ten carefully: dividing by 10⁻⁷ raises the power of ten.
142IGPE3-3-16r34-033(d) Light and SoundPaper 1PMedium1 mark
A diver's torch shines a ray of light from air into water. The angle of refraction in the water is 22°. The refractive index of water is 1.33. What is the angle of incidence in air?
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Notes
n = sin i ÷ sin r, so sin i = 1.33 × sin 22° = 0.498, giving i = 30°. 17° comes from dividing 22° by 1.33, but the angles themselves are not in proportion, only their sines. 22° assumes the ray does not bend. 60° is the angle between the incident ray and the water surface instead of the normal.
✓ 1
Answer D
Answer: D
Exam tip: Use n = sin i ÷ sin r with angles measured from the normal. Never multiply or divide the angles themselves by n.
143IGPE3-3-16r34-083(d) Light and SoundPaper 1PMedium7 marks
A marine biologist shines a narrow beam of light from a torch in air onto the calm surface of a tank of water. The refractive index of water is 1.33. The beam is refracted at an angle of 28° to the normal.
Diagram not to scale
(a)
Calculate the angle of incidence needed for this angle of refraction. Give your answer to 3 significant figures.
(4)
(b)
State what happens to the speed of the light as it enters the water.
(1)
(c)
Explain why the beam bends towards the normal as it enters the water.
(2)
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Notes
Part (a)
n = sin i ÷ sin r
✓ 1
sin i = 1.33 × sin 28° = 1.33 × 0.469
✓ 1
sin i = 0.624
✓ 1
i = 38.6°
✓ 1
Part (b)
the speed decreases
✓ 1
Part (c)
the light travels more slowly in water than in air
✓ 1
the part of each wavefront that enters first slows down first, so the wavefronts change direction (towards the normal)
✓ 1
Common error: Angles of incidence and refraction are always measured between the ray and the normal, not between the ray and the surface.
144IGPE3-3-16r34-103(d) Light and SoundPaper 1PMedium7 marks
A logistics company uses an ultrasonic sensor to measure the distance to the end wall of a warehouse. The sensor sends out a short pulse and detects the echo 0.10 s later. The speed of sound in the warehouse is 340 m/s.
(a)
Calculate the distance from the sensor to the wall.
(3)
(b)
The sensor uses sound of frequency 40 kHz. State why this sound cannot be heard by a person.
(1)
(c)
Calculate the wavelength of the 40 kHz sound in the warehouse.
(2)
(d)
On a hot day the speed of sound is slightly greater. State the effect on the time taken for the echo to return.
(1)
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Notes
Part (a)
distance travelled by the pulse = speed × time = 340 × 0.10 = 34 m
✓ 1
this is a journey to the wall and back, so halve it
✓ 1
17 m
✓ 1
Part (b)
the frequency is above the upper limit of human hearing (20 000 Hz)
✓ 1
Part (c)
λ = v ÷ f = 340 ÷ 40 000
✓ 1
0.0085 m (8.5 mm)
✓ 1
Part (d)
the time decreases (shorter)
✓ 1
Common error: An echo travels to the wall and back. Halve the distance travelled by the pulse to find the distance to the wall.
145IGPE3-3-16r34-133(c) The Electromagnetic SpectrumPaper 1PMedium6 marks
Microwaves, infrared radiation, ultraviolet radiation and X-rays are all part of the electromagnetic spectrum. A cargo ship uses a satellite link that works with microwaves of frequency 12 GHz (1 GHz = 1.0 × 109 Hz). The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s.
(a)
State which of the four types of radiation listed has the longest wavelength.
(1)
(b)
Calculate the wavelength of the 12 GHz microwaves.
(3)
(c)
State one danger to people of exposure to ultraviolet radiation.
(1)
(d)
State one use of infrared radiation.
(1)
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Notes
Part (a)
microwaves
✓ 1
Part (b)
f = 12 × 109 Hz and λ = v ÷ f
✓ 1
λ = 3.0 × 108 ÷ 12 × 109
✓ 1
0.025 m
✓ 1
Part (c)
skin cancer, sunburn or eye damage (any one)
✓ 1
Part (d)
any one of: remote controls, thermal imaging, cooking, heating, night-vision cameras
✓ 1
Common error: Convert prefixes before substituting: 12 GHz is 12 × 10⁹ Hz, not 12 Hz.
146IGPE3-3-16r35-023(b) Properties of WavesPaper 1PMedium1 mark
A wave machine in a swimming pool makes waves of frequency 0.25 Hz. The crests are 3.0 m apart. What is the speed of the waves?
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Notes
The distance between crests is the wavelength, so v = f λ = 0.25 × 3.0 = 0.75 m/s. 12 m/s comes from dividing the wavelength by the frequency. 0.083 m/s comes from dividing the frequency by the wavelength. 1.5 m/s treats the crest spacing as half a wavelength.
✓ 1
Answer A
Answer: A
Common error: The distance between two neighbouring crests is one whole wavelength, not half of it.
148IGPE3-3-16r35-073(c) The Electromagnetic SpectrumPaper 1PMedium7 marks
A food factory uses several types of electromagnetic radiation in its production line.
(a)
Name the radiation used to detect metal fragments in sealed packets and give a reason for your choice.
(2)
(b)
Gamma rays are also used to sterilise packaged food. State one hazard to workers from gamma rays and one precaution.
(2)
(c)
Microwaves of frequency 2.5 GHz are used to heat food. The speed of electromagnetic waves in a vacuum is 3.0 × 108 m/s. Calculate the wavelength of these microwaves.
(3)
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Notes
Part (a)
B1 X-rays
✓ 1
B1 they pass through the food and packaging but are absorbed by metal, so fragments show up
✓ 1
Part (b)
B1 hazard: ionising radiation can damage cells or cause cancer
✓ 1
B1 precaution: thick lead or concrete shielding, operating the equipment remotely, or limiting exposure time
✓ 1
Part (c)
M1 convert 2.5 GHz to 2.5 × 109 Hz
✓ 1
M1 λ = v ÷ f
✓ 1
A1 3.0 × 108 ÷ 2.5 × 109 = 0.12 m
✓ 1
Common error: 1 GHz is 10⁹ Hz, not 10⁶ Hz. Convert before substituting.
151IGPE3-3-16r35-143(d) Light and SoundPaper 1PMedium6 marks
A microphone in a quiet library study room is connected to an oscilloscope. The time base is set to 2.0 ms per division. One complete wave on the screen occupies 5.0 divisions horizontally.
Oscilloscope screen. Diagram not to scale
(a)
Calculate the time period of the sound wave.
(2)
(b)
Calculate the frequency of the sound.
(2)
(c)
The same note is played more loudly. Describe how the trace changes.
(2)
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Notes
Part (a)
M1 period = divisions × time base
✓ 1
A1 5.0 × 2.0 = 10 ms (0.010 s)
✓ 1
Part (b)
M1 f = 1 ÷ T
✓ 1
A1 1 ÷ 0.010 = 100 Hz (ECF from (a))
✓ 1
Part (c)
B1 the amplitude (height) of the wave increases
✓ 1
B1 the period and frequency stay the same
✓ 1
Common error: Convert the time period to seconds before using f = 1/T. The time base is in milliseconds.
152IGPE3-3-15r38-043(b) Properties of WavesPaper 1PMedium1 mark
A buoy rises to the top of a wave 6 times in 30 s as regular ocean waves pass it. The distance between neighbouring wave crests is 15 m. What is the speed of the waves?
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Notes
The period is 30 s ÷ 6 = 5.0 s, so the frequency is 0.20 Hz and v = fλ = 0.20 × 15 = 3.0 m/s, the correct value. 75 m/s comes from multiplying the wavelength by the period. 90 m/s comes from multiplying the wavelength by the number of waves. 0.33 m/s comes from dividing the period by the wavelength, the wrong way round.
✓ 1
Answer B
Answer: B
Exam tip: Find the period or frequency first from the number of waves and the time, then use v = fλ.
153IGPE3-3-15r38-073(b) Properties of WavesPaper 1PMedium6 marks
Spectators in a stadium perform a Mexican wave. Each person stands up and sits down in turn, and the wave moves round the stadium at 9.0 m/s. At one instant the distance between the centres of two neighbouring groups of standing spectators is 30 m. The stands form a loop 360 m long.
(a)
Calculate the frequency of the wave.
(2)
(b)
Calculate the time taken for the wave to travel once round the stadium.
(2)
(c)
The spectators now perform the wave at double the frequency from (a), but the speed stays at 9.0 m/s. Calculate the new wavelength.
(2)
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Notes
Part (a)
M1 f = v ÷ λ = 9.0 ÷ 30
✓ 1
A1 0.30 Hz
✓ 1
Part (b)
M1 t = d ÷ v = 360 ÷ 9.0
✓ 1
A1 40 s
✓ 1
Part (c)
M1 λ = v ÷ f = 9.0 ÷ (2 × 0.30)
✓ 1
A1 15 m (ECF from (a))
✓ 1
Common error: When the speed is fixed, doubling the frequency halves the wavelength; it does not double it.
154IGPE3-3-15r38-083(b) Properties of WavesPaper 1PMedium7 marks
A buoy beside an offshore wind farm rises to the top of a wave crest 18 times in 90 s as regular waves pass. The distance between neighbouring crests is 24 m.
(a)
Calculate the period of the waves.
(2)
(b)
Calculate the frequency of the waves.
(1)
(c)
Calculate the speed of the waves.
(2)
(d)
Describe the movement of the buoy as the waves pass, and state what the waves transfer along the surface.
(2)
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Notes
Part (a)
M1 T = time ÷ number of waves = 90 ÷ 18
✓ 1
A1 5.0 s
✓ 1
Part (b)
A1 f = 1 ÷ T = 1 ÷ 5.0 = 0.20 Hz (ECF from (a))
✓ 1
Part (c)
M1 v = f × λ
✓ 1
A1 0.20 × 24 = 4.8 m/s (ECF from (b))
✓ 1
Part (d)
the buoy moves up and down about a fixed position and does not travel along with the waves
✓ 1
the waves transfer energy
✓ 1
Exam tip: Period is the time for one wave, so divide the total time by the number of waves.
155IGPE3-3-15r38-123(c) The Electromagnetic SpectrumPaper 1PMedium5 marks
Wind turbines at an offshore wind farm send data to the shore using microwaves of frequency 1.5 × 109 Hz. The speed of electromagnetic waves in air is 3.0 × 108 m/s.
(a)
Calculate the wavelength of these microwaves.
(2)
(b)
State two types of electromagnetic wave that have a higher frequency than microwaves.
(2)
(c)
State one hazard of exposure to microwaves.
(1)
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Notes
Part (a)
M1 λ = v ÷ f = 3.0 × 108 ÷ 1.5 × 109
✓ 1
A1 0.20 m
✓ 1
Part (b)
any two from: infrared, visible light, ultraviolet, X-rays, gamma rays
✓ 2
Part (c)
internal heating of body cells / tissue damage by heating
✓ 1
Exam tip: Convert frequencies given in standard form carefully; a wrong power of ten changes the wavelength by a factor of ten.
157IGPE3-3-15r39-033(b) Properties of WavesPaper 1PMedium1 mark
A satellite television signal has a frequency of 12 GHz (1 GHz = 1×109 Hz). The speed of the wave is 3.0×108 m/s. What is its wavelength?
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Notes
wavelength = speed ÷ frequency = 3.0×108 ÷ 12×109 = 0.025 m. Dividing the numbers 12 ÷ 3 gives 4.0 and ignores the powers of ten. 2.5 m and 0.25 m come from slipping by one or two powers of ten when converting GHz to Hz.
✓ 1
Answer D
Answer: D
Common error: Convert GHz to Hz before substituting: 12 GHz is 12×10⁹ Hz, not 12×10⁶ Hz.
158IGPE3-3-15r39-073(c) The Electromagnetic SpectrumPaper 1PMedium7 marks
A space probe is near Neptune, at a distance of 4.5×109 km from Earth. Signals from the probe travel to Earth as electromagnetic waves. speed of electromagnetic waves in a vacuum = 3.0×108 m/s.
(a)
Suggest why sound waves cannot be used to send signals between the probe and Earth.
(1)
(b)
Calculate the time taken for a signal to travel from the probe to Earth.
(3)
(c)
The probe transmits at a frequency of 8.4 GHz. State which part of the electromagnetic spectrum this is.
(1)
(d)
Calculate the wavelength of this signal. Give your answer to 3 significant figures. (1 GHz = 1×109 Hz)
(2)
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Notes
Part (a)
sound needs a medium / cannot travel through the vacuum of space
✓ 1
Part (b)
M1 distance = 4.5×1012 m
✓ 1
M1 t = d ÷ v = 4.5×1012 ÷ 3.0×108
✓ 1
A1 1.5×104 s
✓ 1
Part (c)
microwaves
✓ 1
Part (d)
M1 λ = v ÷ f = 3.0×108 ÷ 8.4×109
✓ 1
A1 0.0357 m
✓ 1
Exam tip: Write the distance in metres before dividing by a speed in m/s.
159IGPE3-3-15r39-093(c) The Electromagnetic SpectrumPaper 1PMedium5 marks
A coffee roaster measures the temperature of the beans using a sensor that detects infrared radiation of wavelength 3.0×10−6 m. speed of electromagnetic waves in a vacuum = 3.0×108 m/s.
(a)
Explain why a hotter batch of beans gives a stronger signal at the sensor.
(2)
(b)
Calculate the frequency of the infrared radiation.
(2)
(c)
Dark roasted beans are good absorbers of infrared radiation. State one other property of a dark, matt surface in relation to infrared radiation.
(1)
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Notes
Part (a)
hotter objects emit more infrared radiation
✓ 1
so more radiation (energy per second) reaches the sensor
✓ 1
Part (b)
M1 f = v ÷ λ = 3.0×108 ÷ 3.0×10−6
✓ 1
A1 1.0×1014 Hz
✓ 1
Part (c)
it is a good emitter of infrared radiation
✓ 1
Exam tip: A surface that is a good absorber of radiation is also a good emitter of radiation.
160IGPE3-3-15r39-103(b) Properties of WavesPaper 1PMedium5 marks
A long-wave radio station broadcasts at a frequency of 252 kHz. A family living in a valley behind a hill can still receive the broadcast clearly. The speed of radio waves is 3.0×108 m/s.
(a)
Explain why the signal can be received in the valley.
(2)
(b)
State one advantage of transmitting the programme as a digital signal rather than an analogue signal.
(1)
(c)
Calculate the wavelength of the broadcast. Give your answer to 3 significant figures. (1 kHz = 1×103 Hz)
(2)
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Notes
Part (a)
the waves are diffracted at / around the hill
✓ 1
the wavelength is large (comparable with the size of the hill) so diffraction is significant
✓ 1
Part (b)
any one from: less affected by noise / interference; the signal can be regenerated without loss of quality; more information can be sent
✓ 1
Part (c)
M1 λ = v ÷ f = 3.0×108 ÷ 252×103
✓ 1
A1 1.19×103 m
✓ 1
Exam tip: Diffraction is greatest when the wavelength is similar to or larger than the gap or obstacle.
161IGPE3-3-15r39-123(d) Light and SoundPaper 1PMedium6 marks
A technician uses a small plane mirror on a pole to inspect the underside of a bridge beam. A ray of light from the beam is reflected so that the angle between the incident ray and the reflected ray is 76°.
Diagram not to scale
(a)
Calculate the angle of incidence.
(2)
(b)
Calculate the angle between the reflected ray and the surface of the mirror.
(2)
(c)
The mirror is tilted so that the angle of incidence increases by 10° from its first value. Calculate the new angle between the incident ray and the reflected ray.
(2)
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Notes
Part (a)
M1 76° ÷ 2
✓ 1
A1 38°
✓ 1
Part (b)
M1 90° − 38°
✓ 1
A1 52° (ECF from (a))
✓ 1
Part (c)
M1 new angle of incidence = 38° + 10° = 48°
✓ 1
A1 2 × 48° = 96°
✓ 1
Common error: The angle between the incident and reflected rays is twice the angle of incidence, not equal to it.
162IGPE3-3-16r39-043(d) Light and SoundPaper 1PMedium1 mark
In an experiment with a rectangular glass block, light enters the glass with an angle of incidence of 50° and an angle of refraction of 31°. What is the refractive index of the glass?
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Notes
n = sin i ÷ sin r = sin 50° ÷ sin 31° = 1.487, which is 1.49. 1.61 comes from dividing the angles (50 ÷ 31). 0.67 is the inverted ratio sin r ÷ sin i. 1.25 comes from using the angle to the surface for the incident ray (cos 50° ÷ sin 31°).
✓ 1
Answer A
Answer: A
Common error: Use the sines of the angles measured from the normal, with the angle in air on top.
163IGPE3-3-16r39-063(b) Properties of WavesPaper 1PMedium8 marks
A bat hunting a moth sends out a pulse of ultrasound. The echo from the moth returns to the bat 0.012 s later. The speed of sound in air is 340 m/s.
(a)
Explain why a human cannot hear the bat's pulse.
(1)
(b)
Calculate the distance from the bat to the moth. Give your answer to 3 significant figures.
(3)
(c)
The bat sends 25 pulses each second. Calculate the time between one pulse and the next.
(2)
(d)
The bat does not send a new pulse until the echo from the last one has returned. Calculate the greatest distance from the bat at which an object can be detected with a pulse rate of 25 per second.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
its frequency is above 20 000 Hz, the upper limit of human hearing
✓ 1
Part (b)
M1 total distance travelled = 340 × 0.012 = 4.08 m
✓ 1
M1 halve for the journey to the moth only: 4.08 ÷ 2
✓ 1
A1 2.04 m
✓ 1
Part (c)
M1 T = 1 ÷ f = 1 ÷ 25
✓ 1
A1 0.040 s
✓ 1
Part (d)
M1 340 × 0.040 ÷ 2
✓ 1
A1 6.8 m
✓ 1
Common error: Sound in an echo travels to the object and back, so divide the total distance by 2.
164IGPE3-3-16r39-073(d) Light and SoundPaper 1PMedium7 marks
A robotics club measures the speed of sound by clapping in time with the echoes from a flat wall. A student stands 62.0 m from the wall and claps so that each clap coincides with the echo of the previous clap. A timer shows 11.0 s for 30 intervals between claps.
Diagram not to scale
(a)
Explain why the student times 30 intervals rather than one interval.
(1)
(b)
Calculate the time for one interval between claps. Give your answer to 3 significant figures.
(2)
(c)
Calculate the speed of sound. Use the unrounded value of the time in your calculation and give your answer to 3 significant figures.
(3)
(d)
Suggest one way to check that the result is reliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
reduces the effect of reaction-time error / percentage uncertainty in the time
✓ 1
Part (b)
M1 11.0 ÷ 30
✓ 1
A1 0.367 s
✓ 1
Part (c)
M1 distance travelled by sound in one interval = 2 × 62.0 = 124 m
✓ 1
M1 v = d ÷ t = 124 ÷ (11.0 ÷ 30)
✓ 1
A1 338 m/s
✓ 1
Part (d)
repeat the measurement and calculate a mean
✓ 1
Exam tip: The sound travels to the wall and back between claps, so the distance used in v = d ÷ t is twice the distance to the wall.
165IGPE3-3-16r39-093(b) Properties of WavesPaper 1PMedium7 marks
A fishing boat uses sonar to find shoals of fish. The sonar sends a pulse of ultrasound downwards and the echo from a shoal returns after 0.16 s. The speed of sound in sea water is 1500 m/s.
(a)
State what is meant by ultrasound.
(1)
(b)
Calculate the depth of the shoal below the boat.
(3)
(c)
The sonar pulse has a frequency of 50 kHz. Calculate the wavelength of the ultrasound in sea water. (1 kHz = 1×103 Hz)
(2)
(d)
Suggest why radio waves are not suitable for finding fish under the sea.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
sound with a frequency above 20 000 Hz
✓ 1
Part (b)
M1 total distance = 1500 × 0.16 = 240 m
✓ 1
M1 halve for one way: 240 ÷ 2
✓ 1
A1 120 m
✓ 1
Part (c)
M1 λ = v ÷ f = 1500 ÷ 50 000
✓ 1
A1 0.030 m
✓ 1
Part (d)
radio waves are strongly absorbed by sea water / do not travel far in water
✓ 1
Exam tip: State the formula, then substitute with all quantities in SI units before calculating.
166IGPE3-3-16r39-113(d) Light and SoundPaper 1PMedium6 marks
An ultrasonic distance sensor on a warehouse robot sends out a pulse every 0.50 s. The speed of sound in air is 340 m/s. The echo from a box returns 0.0090 s after one pulse is sent. The robot moves towards the box at a constant speed, and the echo from the next pulse returns after 0.0070 s.
(a)
Calculate the distance from the robot to the box at the first pulse.
(2)
(b)
Calculate the distance from the robot to the box at the second pulse.
(1)
(c)
Calculate the speed of the robot.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 340 × 0.0090 ÷ 2
✓ 1
A1 1.53 m
✓ 1
Part (b)
A1 340 × 0.0070 ÷ 2 = 1.19 m
✓ 1
Part (c)
M1 distance moved = 1.53 − 1.19 = 0.34 m (ECF from (a), (b))
✓ 1
M1 speed = distance ÷ time = 0.34 ÷ 0.50
✓ 1
A1 0.68 m/s
✓ 1
Exam tip: Each distance comes from half of speed × echo time, and the robot's speed uses the time between pulses.
167IGPE3-3-15r43-033(d) Light and SoundPaper 1PMedium1 mark
A microphone detects a note from a ukulele and is connected to an oscilloscope. The time base is set to 2.0 ms per division. The screen is 10 divisions wide and shows exactly 2.5 complete waves across its full width. What is the frequency of the note?
Show mark scheme
Marking point
Mark
Notes
The screen shows 10 × 2.0 = 20 ms, so one wave takes 20 ÷ 2.5 = 8.0 ms and f = 1 ÷ 0.0080 = 125 Hz, so A is correct. B treats a single division (2.0 ms) as the period. C gives the period in ms as if it were the frequency. D uses the whole screen time and ignores that it holds 2.5 waves.
✓ 1
Answer A
Answer: A
Exam tip: Find the period of one wave first, then use f = 1/T.
168IGPE3-3-15r43-073(d) Light and SoundPaper 1PMedium6 marks
A student tunes a ukulele string using a microphone connected to an oscilloscope. The time base is set to 0.50 ms per division. One complete wave on the screen is 4.5 divisions wide.
Oscilloscope screen. Diagram not to scale
(a)
Calculate the period of the wave. Give your answer in seconds.
(2)
(b)
Calculate the frequency of the note. Use the unrounded period from (a) and give your answer to 3 significant figures.
(2)
(c)
The student plucks the string harder. Describe two changes, or lack of change, in the trace.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 period = 4.5 × 0.50 ms
✓ 1
A1 2.25 × 10−3 s (accept 0.00225 s)
✓ 1
Part (b)
M1 f = 1/T = 1/(2.25 × 10−3)
✓ 1
A1 444 Hz
✓ 1
Part (c)
the height (amplitude) of the trace increases
✓ 1
the width of one wave (period) stays the same, so the frequency and pitch are unchanged
✓ 1
Common error: A harder pluck changes loudness (amplitude), not pitch (frequency).
169IGPE3-3-15r43-103(d) Light and SoundPaper 1PMedium5 marks
An engineer inspects the inside of an aircraft engine with an endoscope. The fibres in it are 1.2 m long and light travels along them at 2.0 × 108 m/s.
(a)
Explain why light travelling along a fibre does not escape through the wall.
(3)
(b)
Calculate the time taken for a pulse of light to travel the full length of a fibre in a straight line.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
light travels from a more dense material (glass) to a less dense material at the wall
✓ 1
the angle of incidence at the wall is greater than the critical angle
✓ 1
so all the light is totally internally reflected back into the fibre
✓ 1
Part (b)
M1 time = distance ÷ speed = 1.2 ÷ (2.0 × 108)
✓ 1
A1 6.0 × 10−9 s
✓ 1
Exam tip: For an explain question about fibres, mention the critical angle and say that all the light is reflected.
170IGPE3-3-15r43-113(d) Light and SoundPaper 1PMedium5 marks
A student displays the signal from a microphone on an oscilloscope with the time base set to 1.0 ms per division. The screen is 10 divisions wide. String A produces exactly 3 complete waves across the full width of the screen.
Oscilloscope screen. Diagram not to scale
(a)
Calculate the frequency of the note from string A.
(2)
(b)
String B produces a note of frequency 450 Hz. Calculate the number of complete waves that would fit across the full width of the screen with the same settings. The answer need not be a whole number.
(2)
(c)
State which string produces the note with the higher pitch.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total time on screen = 10 × 1.0 ms = 0.010 s and f = 3 ÷ 0.010
✓ 1
A1 300 Hz
✓ 1
Part (b)
M1 number of waves = frequency × time on screen = 450 × 0.010
✓ 1
A1 4.5
✓ 1
Part (c)
string B
✓ 1
Common error: Pitch depends on frequency: more waves on the screen means a higher pitch.
171IGPE3-3-15r43-123(d) Light and SoundPaper 1PMedium6 marks
A poultry farm uses a rodent deterrent that sweeps its sound frequency from 17 kHz to 30 kHz. The makers claim that the sound is completely inaudible to people. The speed of sound in air is 340 m/s.
(a)
State what is meant by ultrasound.
(1)
(b)
Evaluate the claim that the sound is completely inaudible to people.
(3)
(c)
Calculate the wavelength of the sound at 17 kHz.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
sound of frequency above 20 000 Hz (above the range of human hearing)
✓ 1
Part (b)
human hearing extends up to about 20 kHz
✓ 1
frequencies from 17 kHz to 20 kHz are within the audible range, so part of the sweep can be heard
✓ 1
only frequencies above 20 kHz are inaudible, so the claim is only partly true
✓ 1
Part (c)
M1 wavelength = 340 ÷ 17 000
✓ 1
A1 0.020 m
✓ 1
Exam tip: Evaluate means weigh the evidence, so use the numbers (17 kHz and 20 kHz) in your answer.
173IGPE3-3-16r43-073(d) Light and SoundPaper 1PMedium3 marks
An archaeologist is testing the water in a flooded excavation. In a laboratory test, a ray of light travels from the water into air. The angle of incidence in the water is 30° and the angle of refraction in the air is 41.7°.
Diagram not to scale
(a)
Calculate the refractive index of the water. Give your answer to 3 significant figures.
(2)
(b)
State whether the ray bends towards or away from the normal as it passes from water into air.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 n = sin 41.7° ÷ sin 30°
✓ 1
A1 1.33
✓ 1
Part (b)
away from the normal
✓ 1
Common error: For light going from water into air, the larger angle (in the air) goes on top of the fraction.
174IGPE3-3-16r43-093(d) Light and SoundPaper 1PMedium5 marks
A cycle reflector has sloping faces at which the light meets the plastic with an angle of incidence of 45°. Plastic A has a critical angle of 42° in air. A cheaper plastic B has a critical angle of 47° in air.
(a)
Explain why a reflector made from plastic A returns the light.
(2)
(b)
Explain what would happen at a sloping face if plastic B were used.
(2)
(c)
State which plastic has the higher refractive index.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
45° is greater than the critical angle of 42°
✓ 1
so the light is totally internally reflected at the sloping faces and none passes out
✓ 1
Part (b)
45° is less than the critical angle of 47°, so there is no total internal reflection and some light is refracted out of the face
✓ 1
less light is returned, so the reflector would appear dimmer
✓ 1
Part (c)
plastic A, because it has the smaller critical angle
✓ 1
Common error: A smaller critical angle means a higher refractive index, since sin c = 1/n.
178IGCSE-PHY-3-0123(d) Light and SoundPapers 1 & 2Grade 8–96 marks
A student is given a semicircular glass block, a ray box, a protractor and a sheet of plain paper.
(a)
Describe how the student could use this apparatus to determine the critical angle of the glass and hence the refractive index of the glass. You may draw a diagram to help your answer.
(6)
Show mark scheme
(a)
Indicative content:
MP1. place block on paper, draw around it and mark the centre of the flat face / draw normal at the centre;
MP2. direct a narrow ray through the curved face towards the centre of the flat face;
MP3. ray enters curved surface along a radius / along the normal so it does not refract there;
MP4. increase the angle of incidence (at the flat face) gradually;
MP5. until the refracted ray just travels along the flat face / angle of refraction is 90° / just before total internal reflection;
MP6. mark the ray (e.g. with crosses) and measure the angle between the ray and the normal with a protractor (this is c);
MP7. repeat (from the other side of the normal / several times) and calculate a mean;
MP8. calculate refractive index using n = 1 ÷ sin c;
MP9. use a darkened room so the faint refracted ray can be seen;
Allow
award 1 mark per MP to a maximum of 6, but use the levels for the overall mark
a labelled diagram may score MP1–MP3 and MP6
Do not accept
measuring the angle from the flat surface
using n = sin i ÷ sin r without identifying i = c and r = 90°
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated relevant steps, e.g. shining a ray into the block or measuring an angle; no clear method to identify the critical angle.
Level 2 (3–4)
A workable method that identifies the critical angle as the angle at which the refracted ray grazes the surface, with some detail of how the angle is measured or how n is calculated.
Level 3 (5–6)
A detailed, logically sequenced method: ray along a radius to the centre, angle increased until refracted ray grazes the flat face, angle measured from the normal, repeats/mean, and n = 1/sin c.
Model sketch
Ray aimed at the centre of the flat face. The critical angle c is the angle of incidence when the refracted ray just travels along the flat face. Diagram not to scale
179IGCSE-PHY-3-0133(b) Properties of WavesPapers 1 & 2Grade 8–99 marks
An earthquake produces two types of wave that travel through rock. P-waves are longitudinal and travel at 6.0 km/s. S-waves are transverse and travel at 3.5 km/s. A seismometer records the S-waves arriving 50 s after the P-waves. Both types of wave leave the earthquake at the same time and travel the same distance to the seismometer.
(a)
Describe the difference between the way the rock vibrates as a P-wave passes and as an S-wave passes.
(2)
(b(i))
Show that the S-waves take about 0.12 s longer than the P-waves to travel each 1 km through the rock.
(2)
(b(ii))
Use your answer to (b)(i) to calculate the distance from the earthquake to the seismometer, in km.
(2)
(c)
The S-waves have a frequency of 0.70 Hz. Calculate the wavelength of the S-waves, in m.
(2)
(d)
The rock at the seismometer is many kilometres from the earthquake, but the seismometer still records the waves. Explain how this shows that waves transfer energy without transferring matter.
(1)
Show mark scheme
(a)
P-wave: (rock) vibrates parallel to the direction of wave travel / energy transfer;
S-wave: (rock) vibrates perpendicular to the direction of wave travel / energy transfer;
Do not accept
P-waves go side to side, S-waves go up and down (no reference to direction of travel)
(b(i))
time for 1 km for both waves e.g. 1 ÷ 3.5 = 0.286 (s) and 1 ÷ 6.0 = 0.167 (s);
difference = 0.119 (s);
Allow
0.1190 (s)
use of 1000 m with speeds in m/s
Do not accept
0.12 (s) with no working (target value)
1 ÷ 2.5 = 0.40 s (speed difference)
(b(ii))
distance = 50 ÷ their (b)(i) e.g. 50 ÷ 0.119;
evaluation: 420 (km);
Allow
417 to 420 (km) (using 0.12)
ecf from (b)(i)
4.2 × 10⁵ m
algebraic route d ÷ 3.5 − d ÷ 6.0 = 50 for full marks
substitution with speed in m/s e.g. λ = 3500 ÷ 0.70;
evaluation;
(λ =) 5000 (m)
Allow
5.0 km if unit stated
5 (m) scores 1 (km/s not converted)
(d)
rock particles (at the seismometer) only vibrate about fixed positions / did not travel from the earthquake, but energy arrived (to move the seismometer);
180IGCSE-PHY-3-0143(d) Light and SoundPapers 1 & 2Grade 8–910 marks
An engineer is testing a plastic optical fibre with no outer coating. The refractive index of the plastic is 1.34. The diagram shows a ray of light in air striking the flat end of the fibre at an angle of incidence of 80°. The engineer claims that this ray will be trapped inside the fibre by total internal reflection.
(a)
Calculate the angle of refraction of the ray as it enters the end of the fibre.
(3)
(b)
Calculate the critical angle for the plastic.
(2)
(c)
The refracted ray then hits the side wall of the fibre. The side wall is at 90° to the end face. Use your answers to (a) and (b) to evaluate the engineer's claim.
(3)
(d)
Describe how total internal reflection allows optical fibres to transmit information.
(2)
Show mark scheme
(a)
substitution;
rearrangement;
evaluation;
e.g. 1.34 = sin 80 ÷ sin r
sin r = sin 80 ÷ 1.34 = 0.735
(r =) 47.3 (°)
Allow
47.2 to 47.4
47 (°)
Do not accept
0.735 given as the angle
(b)
substitution and rearrangement e.g. c = sin⁻¹(1 ÷ 1.34);
evaluation;
(c =) 48.3 (°)
Allow
48 (°)
(c)
angle of incidence at the side wall = 90 − (a) = 42.7(°);
comparison: 42.7° is less than the critical angle (48.3°);
conclusion: ray is not totally internally reflected / ray refracts out (leaves the fibre) so the claim is wrong;
Allow
ecf from (a) and (b) for all three marks
conclusion consistent with candidate's own comparison
Do not accept
claim wrong with no comparison of angles
(d)
(light / infrared) pulses carry the information / signal;
repeated total internal reflection keeps the light inside the fibre (even round bends) along its length;
181IGCSE-PHY-3-0153(b) Properties of WavesPapers 1 & 2Grade 8–910 marks
A toy boat on a pond has a small vibrating paddle that produces water waves of frequency 4.0 Hz. The waves travel at 0.30 m/s. The boat moves forward at a steady 0.10 m/s. The diagram shows the wavefronts produced by the moving boat, viewed from above.
(a)
Calculate the wavelength of the waves when the boat is not moving.
(2)
(b)
Explain why the wavefronts are closer together ahead of the boat than behind it.
(3)
(c)
In one time period of the waves, the boat moves forward a small distance. Use this idea to calculate the wavelength of the waves directly ahead of the moving boat.
(2)
(d)
A leaf floats on the water directly ahead of the boat. Calculate the frequency at which wavefronts reach the leaf.
(2)
(e)
State the name given to this change in observed frequency and wavelength.
(1)
Show mark scheme
(a)
rearrangement and substitution e.g. λ = 0.30 ÷ 4.0;
evaluation;
(λ =) 0.075 (m)
Allow
7.5 cm
Do not accept
1.2 m (v × f)
(b)
the boat (source) moves forward in the time between producing successive wavefronts / in one period;
so each wavefront is produced from a point further forward (centres of circles shift forward);
so the distance between wavefronts ahead is reduced / wavelength ahead is shorter (and longer behind);
Allow
the source catches up with the waves it has produced
Do not accept
the waves ahead travel faster
the frequency of the paddle increases
(c)
distance moved by boat in one period = 0.10 × 0.25 = 0.025 m;
182IGCSE-PHY-3-0323(d) Light and SoundPapers 1 & 2Grade 8–99 marks
A gemmologist tests a colourless stone to find out whether it is a diamond. She directs a narrow beam of light from air onto a flat polished face of the stone. Angle of incidence = 50.0° Angle of refraction = 18.4° The table shows the refractive indices of some colourless materials. Material / name: glass, sapphire, cubic zirconia, diamond Refractive index / –: 1.52, 1.77, 2.16, 2.42Diagram not to scale
(a)
Explain what is meant by the critical angle of a material.
(2)
(b)
Use the data to identify the stone.
(3)
(c)
A cut diamond sparkles because much of the light entering through its top face is totally internally reflected from its lower faces before leaving through the top. An imitation made of glass is cut to exactly the same shape but sparkles much less. Explain why. Use calculations in your answer.
(4)
Show mark scheme
(a)
(light travelling) from the material to air / from higher to lower refractive index;
angle of incidence at which the angle of refraction is 90° / refracted ray travels along the boundary;
Allow
the smallest angle of incidence at which total internal reflection occurs for MP2
Do not accept
the angle at which light is refracted
angle measured from the surface
(b)
(n =) sin 50.0 ÷ sin 18.4;
(n =) 2.43;
(stone is) diamond;
Allow
2.42–2.43
MP3 is dependent on a value of n between 2.3 and 2.5
Do not accept
2.72 (50 ÷ 18.4)
0.412 (sin r ÷ sin i)
(c)
c for diamond = 24.4(°);
c for glass = 41.1(°);
(diamond has a smaller critical angle so) more rays (inside) hit the lower faces at angles greater than c;
so more light is totally internally reflected (back out through the top) / in glass more light refracts out through the lower faces;
Allow
24(°) and 41(°)
ecf from n in (b) for c of the stone
MP3 and MP4 may be awarded without calculations
Do not accept
diamond reflects more because it is denser (no reference to critical angle)
183IGCSE-PHY-3-0333(d) Light and SoundPapers 1 & 2Grade 8–99 marks
People who live 2 km from a motorway notice that the traffic noise is louder on calm, clear nights than during the day. On these nights, the air close to the ground is colder than the air higher up. Sound travels faster in warmer air. Speed of sound at ground level = 336 m/s Speed of sound 60 m above the ground = 340 m/s
(a)
A wavefront of sound from the motorway travels away from the motorway. Calculate how much further the part of the wavefront 60 m above the ground travels in 2.0 s than the part at ground level.
(2)
(b)
Use your answer to (a) to explain why sound from the motorway that starts travelling slightly upwards is bent back down towards the ground on these nights, so the noise is louder at the houses.
(4)
(c)
During the day, the air close to the ground is warmer than the air higher up. Suggest why the traffic noise at the houses is quieter during the day.
(2)
(d)
Sound waves can also be reflected. Suggest how a wall built alongside the motorway reduces the noise reaching the houses.
184IGCSE-PHY-3-0343(b) Properties of WavesPapers 1 & 2Grade 8–910 marks
A leisure centre has a wave pool 36 m long. A wave machine at one end of the pool produces water waves that travel along the pool. A lifeguard measures that a wave crest takes 10 s to travel the 36 m length of the pool. The wave machine produces 15 waves every minute.
(a)
A swimmer floats in the middle of the pool. Explain why the swimmer moves up and down but is not carried along to the far end of the pool.
(2)
(b)
Calculate the wavelength of the waves.
(4)
(c)
The manager wants exactly 3 complete wavelengths to fit along the 36 m length of the pool at any instant. Calculate how many waves per minute the machine must produce. Assume the wave speed does not change.
(3)
(d)
Suggest how the lifeguard could measure the speed of the waves more accurately.
(1)
Show mark scheme
(a)
waves transfer energy but not matter / water;
water (particles) / swimmer oscillate (up and down) about a fixed position;
Do not accept
the swimmer is too heavy to be moved
(b)
(v =) 36 ÷ 10 = 3.6 (m/s);
(f =) 15 ÷ 60 = 0.25 (Hz);
(λ =) v ÷ f / 3.6 ÷ 0.25;
14.4 (m);
Allow
14 (m)
T = 60 ÷ 15 = 4.0 s and λ = v × T = 3.6 × 4.0 for MP2 and MP3
ecf from v and f
Do not accept
0.24 (m) (15 not converted to per second)
0.9 (m) (v × f)
(c)
(λ =) 36 ÷ 3 = 12 (m);
(f =) 3.6 ÷ 12 = 0.30 (Hz);
18 (waves per minute);
Allow
ecf from speed in (b)
number of crests ∝ frequency: 15 × (3 ÷ 2.5) = 18 scores 3
0.30 (Hz) with no conversion to waves per minute scores 2
185IGCSE-PHY-3-0353(d) Light and SoundPapers 1 & 2Grade 8–99 marks
A student investigates the refraction of light in a sugar solution. She fills a transparent semicircular dish with the solution and places it on a sheet of paper. She directs a ray of light from a ray box at the centre of the flat face of the dish. The ray refracts into the solution and leaves through the curved side without changing direction. She measures the angle of refraction r in the solution for different angles of incidence i. Angle of incidence i / °: 10, 20, 30, 40, 50, 60, 70 Angle of refraction r / °: 7, 14, 20, 27, 38, 37, 41 She calculates sin i and sin r and plots the graph shown.
(a)
Identify the anomalous result and suggest one possible cause.
(2)
(b)
Use the graph to determine the refractive index of the sugar solution.
(3)
(c)
The student writes: 'My results show that the angle of refraction is directly proportional to the angle of incidence.' Evaluate this statement. Use the data in your answer.
(3)
(d)
Explain why using the gradient of the graph gives a more reliable value for the refractive index than using one pair of readings.
(1)
Show mark scheme
(a)
i = 50° (r = 38°);
suitable cause, e.g. ray not aimed at the centre of the flat face / normal drawn in the wrong place / ray path marked inaccurately / protractor misread;
Do not accept
parallax (without explanation)
the block was the wrong shape
(b)
large triangle used / two points far apart read from the line, e.g. (0.60, 0.86);
(n = gradient =) Δ sin i ÷ Δ sin r;
1.43;
Allow
1.38–1.48 for MP3
values read from the line, not data points, for MP1
Do not accept
0.70 (sin r ÷ sin i)
(c)
statement is incorrect;
use of data, e.g. i doubles from 30° to 60° but r increases from 20° to 37° (not 40°) / i ÷ r = 1.50 at 30° but 1.71 at 70°;
the straight line through the origin shows sin i is proportional to sin r (not i to r);
Allow
any correct pair of values showing i ÷ r is not constant for MP2
Do not accept
statement is correct because the graph is a straight line
(d)
line of best fit averages many readings / reduces effect of random errors / anomalous point can be identified and ignored;
186IGPE2-056-01-103(c) The Electromagnetic SpectrumPaper 1PHard5 marks
A student uses a microwave oven of frequency 2.45 GHz to estimate the speed of electromagnetic waves. She removes the turntable and heats a flat bar of chocolate for a short time. Small melted spots appear on the chocolate. The distance between neighbouring melted spots is 6.2 cm, which is half of one wavelength of the microwaves.
(a)
Explain why microwaves can be harmful to the human body.
(2)
(b)
Determine the wavelength of the microwaves.
(1)
(c)
Calculate the speed of the microwaves.
(2)
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Part (a)
Microwaves are absorbed by water in body tissue
✓ 1
This causes internal heating of body cells, which can damage them
✓ 1
Part (b)
λ = 2 × 6.2 = 12.4 cm, or 0.124 m
✓ 1
Part (c)
v = f × λ = 2.45 × 109 × 0.124
✓ 1
v = 3.0 × 108 m/s (accept 3.04 × 108 m/s)
✓ 1
Common error: The spacing between the spots is only half a wavelength. Double it before you use v = f × λ, and convert both GHz and cm to SI units.
187IGPE2-056-01-133(c) The Electromagnetic SpectrumPaper 1PHard6 marks
An astronaut on a spacewalk outside a space station is exposed to infrared, ultraviolet, X-rays and gamma rays from the Sun. On Earth's surface, the atmosphere absorbs much of this radiation.
(a)
Explain the harm that each of these types of radiation can do to the astronaut's body.
(3)
(b)
Explain why the astronaut faces a greater risk from this radiation than a person on Earth's surface.
(1)
(c)
Explain how the design of the spacesuit and its helmet visor, and the planning of the spacewalk, can reduce these risks.
(2)
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Part (a)
• infrared is absorbed by the skin, heating it and causing burns
✓ 1
• ultraviolet damages surface cells of the skin, causing sunburn / skin cancer, OR damages the eyes
✓ 1
• X-rays and gamma rays are ionising and penetrate the body, damaging cells / DNA and causing mutations / cancer
✓ 1
Part (b)
• there is no atmosphere in space to absorb the UV, X-rays and gamma rays, so more radiation reaches the astronaut
✓ 1
Part (c)
• a reflective / shiny outer suit layer reflects infrared, OR a tinted / coated visor reflects infrared and absorbs UV, protecting the skin / eyes
✓ 1
• gamma rays (and some X-rays) are too penetrating to be stopped fully by the suit, so spacewalks are kept short / the dose is monitored / spacewalks avoid periods of high solar activity such as solar flares
✓ 1
Exam tip: Link each type of radiation to its specific harmful effect. For each protection, say which radiation it deals with and how it works: by reflecting it, absorbing it or limiting exposure time.
188IGPE2-056-00-083(b) Properties of WavesPaper 1PHard6 marks
A drone carries a loudspeaker that gives out a steady note of frequency 800 Hz. The drone flies at a constant speed and a constant height in a large horizontal circle. The air is still. Student A stands still on the ground directly below the centre of the circle. Student B stands still on the ground some distance outside the circle. The drone flies round the circle several times.
(a)
Explain what each student hears. Use ideas about wavefronts, wavelength and frequency in your answer.
(6)
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Part (a)
Student A: the distance from the drone to Student A stays constant, so the drone never moves towards or away from Student A .
✓ 1
So the wavefronts reach Student A evenly spaced, the wavelength is unchanged and Student A hears a steady 800 Hz with no change in pitch .
✓ 1
Student B: when the drone moves towards Student B, the wavefronts in front of it are bunched closer together .
✓ 1
So the wavelength is shorter and, because the speed of sound is unchanged, Student B hears a higher frequency / pitch, above 800 Hz .
✓ 1
When the drone moves away from Student B, the wavefronts are spread out, so the wavelength is longer and the frequency / pitch is lower, below 800 Hz .
✓ 1
Student B hears the pitch rise and fall repeatedly, once for each lap of the drone (accept: varies smoothly between a highest and a lowest pitch / this is the Doppler effect) .
✓ 1
Exam tip: The Doppler effect depends on motion towards or away from the observer. A source moving round an observer at a fixed distance produces no change in frequency.
189IGPE2-056-00-133(b) Properties of WavesPaper 1PHard9 marks
A railway engineer strikes a steel rail with a hammer. A colleague stands 510 m away along the track with one ear close to the rail. The colleague hears two separate bangs, 1.4 s apart. The speed of sound in air is 340 m/s.
Diagram not to scale
(a)
Explain why the colleague hears two bangs.
(2)
(b)
Calculate the time taken for the sound to travel through the air.
(2)
(c)
Calculate the speed of sound in the steel rail.
(3)
(d)
Explain how the wavelength of the sound in the steel compares with its wavelength in the air.
(2)
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Part (a)
The sound travels to the colleague through both the steel rail and the air . Sound travels faster in steel, so the sound through the rail arrives first .
✓ 2
Part (b)
t = distance / speed = 510 / 340 . t = 1.5 s .
✓ 2
Part (c)
Time in steel = 1.5 − 1.4 = 0.10 s . v = distance / time = 510 / 0.10 . v = 5100 m/s .
✓ 3
Part (d)
The frequency is the same in both, because it is set by the source / the hammer blow . Since v = f × λ and the speed is greater in steel, the wavelength in steel is longer (15 times longer) .
✓ 2
Exam tip: The 1.4 s is the difference between the two travel times, not a travel time itself. Find the slower time first, then subtract.
190IGPE2-056-03-083(d) Light and SoundPaper 1PHard11 marks
A survey ship uses sonar to map the seabed. It sends a pulse of ultrasound with a frequency of 40 kHz straight down. The speed of sound in seawater is 1500 m/s.
(a)
Calculate the wavelength of the ultrasound in seawater.
(3)
(b)
The echo from the seabed is detected 0.24 s after the pulse is sent. Calculate the depth of the water.
(3)
(c)
Part of the pulse enters a layer of sediment on the seabed and reflects from the rock below it. This second echo arrives 0.064 s after the echo from the seabed. Sound travels at 1875 m/s in the sediment. Calculate the thickness of the sediment layer.
(3)
(d)
State what happens to the frequency and the wavelength of the ultrasound as it passes from seawater into the sediment.
(2)
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Part (a)
λ = v ÷ f
✓ 1
= 1500 ÷ 40 000
✓ 1
= 0.038 m (accept 0.0375 m)
✓ 1
Part (b)
distance travelled = 1500 × 0.24 = 360 m
✓ 1
sound travels down and back, so depth = 360 ÷ 2
✓ 1
= 180 m
✓ 1
Part (c)
the extra time is spent travelling down and back through the sediment
✓ 1
thickness = 1875 × 0.064 ÷ 2
✓ 1
= 60 m
✓ 1
Part (d)
Frequency stays the same . Wavelength increases (to about 0.047 m) because the speed increases .
✓ 2
Exam tip: For any echo calculation, the time measured covers the journey there and back. Halve the total distance.
191IGPE2-056-03-133(b) Properties of WavesPaper 1PHard6 marks
A doctor uses an ultrasound probe on a patient's leg to check blood flow in a vein. The probe sends out pulses of ultrasound and detects the reflected waves.
(a)
Explain how the reflected ultrasound can be used to find: • how deep the vein is below the skin • whether the blood is moving towards or away from the probe, and how fast it is moving.
(6)
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Part (a)
Depth:
—
The probe emits a pulse and some ultrasound reflects at the boundary of the vein . The time between sending the pulse and receiving the echo is measured . Depth = speed of ultrasound in tissue × time ÷ 2, because the pulse travels there and back .
✓ 3
Blood flow:
—
Ultrasound reflected from moving blood cells has a changed frequency because of the Doppler effect . Blood moving towards the probe gives reflected waves with a higher frequency (shorter wavelength). Blood moving away gives a lower frequency (longer wavelength) . A bigger change in frequency means the blood is moving faster .
✓ 3
Exam tip: In a 6-mark answer, cover both bullet points fully. Link each observation (time delay or frequency change) to the physics behind it (echo or Doppler effect).
192IGPE2-056-02-073(d) Light and SoundPaper 1PHard6 marks
Surgeons use an endoscope to look inside a patient's stomach. Light travels down thin, flexible glass optical fibres to light up the stomach, even when the fibres are curved.
(a)
Explain how light is carried along a curved optical fibre without escaping from its sides. Your answer should include the conditions that must be met.
(6)
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Part (a)
Light enters one end of the fibre and travels through the glass (core) until it meets the boundary with the material outside it (cladding or air)
✓ 1
The glass has a higher refractive index than the material outside it, so light is travelling from a higher to a lower refractive index (accept: optically denser to less dense)
✓ 1
The angle of incidence, measured from the normal, is greater than the critical angle
✓ 1
So total internal reflection occurs: all the light is reflected back into the glass and none is refracted out
✓ 1
The angle of reflection equals the angle of incidence, so the light meets the opposite wall at a large angle and is totally internally reflected again; repeated reflections zigzag the light along the fibre
✓ 1
At a gentle bend in a thin fibre, the angle of incidence stays above the critical angle; a very sharp bend makes it less than the critical angle, so some light refracts out and escapes
✓ 1
Exam tip: For extended answers on total internal reflection, state both conditions: light travelling from higher to lower refractive index, and an angle of incidence greater than the critical angle. Then link these to repeated reflections along the fibre.
193IGPE2-056-02-133(d) Light and SoundPaper 1PHard9 marks
Diamond has a refractive index of 2.42. Jewellers cut diamonds so that light entering the top is reflected around inside the stone before leaving, which makes it sparkle.
Diagram not to scale
(a)
A ray of light in air enters the top face of a diamond with an angle of incidence of 40°. Calculate the angle of refraction.
(3)
(b)
Calculate the critical angle for diamond.
(2)
(c)
Inside the diamond, the ray meets a sloping face with an angle of incidence of 35°. Explain what happens to the ray at this face. You may use your answer to part (b).
(2)
(d)
Glass has a refractive index of about 1.5. Explain why a diamond sparkles more than a glass copy of the same shape.
(2)
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Part (a)
sin r = sin 40° / 2.42
✓ 1
sin r = 0.266
✓ 1
r = 15.4°
✓ 1
Part (b)
sin c = 1 / n = 1 / 2.42 = 0.413
✓ 1
c = 24.4°
✓ 1
Part (c)
35° is greater than the critical angle of 24.4°
✓ 1
So the ray is totally internally reflected and no light leaves through this face
✓ 1
Part (d)
Diamond has a higher refractive index and so a smaller critical angle than glass (about 24° compared with about 42°)
✓ 1
So more of the rays inside the stone meet faces at angles greater than the critical angle and are totally internally reflected before leaving
✓ 1
Exam tip: A larger refractive index gives a smaller critical angle. Use this to compare materials without needing to do every calculation.
194IGPE-052-00-093(d) Light and SoundPaper 1PHard9 marks
A building uses a bundle of optical fibres to carry sunlight from a collector on the roof into a windowless storeroom. Each fibre has a core with a refractive index of 1.48. Assume the core is surrounded by air.
(a)
State the two conditions needed for total internal reflection to occur.
(2)
(b)
Calculate the critical angle for the core.
(3)
(c)
A ray inside the core meets the core boundary at an angle of incidence of 50°. State and explain what happens to this ray.
(2)
(d)
The installer is told not to bend the fibres sharply. Explain why a sharp bend reduces the amount of light reaching the storeroom.
(2)
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Notes
Part (a)
Light must be travelling from a material with a higher refractive index into one with a lower refractive index, e.g. from glass into air . The angle of incidence must be greater than the critical angle .
✓ 2
Part (b)
sin c = 1 ÷ n ; sin c = 1 ÷ 1.48 = 0.676 ; c = 42.5°
✓ 3
Part (c)
50° is greater than the critical angle of 42.5° , so the ray is totally internally reflected and stays inside the core .
✓ 2
Part (d)
At a sharp bend, the light meets the boundary at a smaller angle of incidence, which can be less than the critical angle . Some light then refracts out of the fibre instead of being reflected, so less light reaches the far end .
✓ 2
Exam tip: In comparison questions, quote both numbers, e.g. '50° is greater than 42.5°'. A bare 'yes' does not explain anything.
195IGPE3-3-14r22-143(d) Light and SoundPaper 1PHard6 marks
A band is playing in a hall. A person standing outside the building, away from the open doorway, hears the low-pitched bass notes clearly but the high-pitched notes only faintly. The doorway is 0.90 m wide. The speed of sound in air is 340 m/s.
(a)
State what is meant by diffraction.
(1)
(b)
Calculate the wavelength of a bass note of frequency 85 Hz and of a high note of frequency 3400 Hz.
(3)
(c)
Use your answers to explain why the bass notes are heard more clearly outside the building.
(2)
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Part (a)
B1 the spreading of waves as they pass through a gap or around an edge
✓ 1
Part (b)
M1 λ = v ÷ f
✓ 1
A1 85 Hz: 4.0 m
✓ 1
A1 3400 Hz: 0.10 m
✓ 1
Part (c)
B1 the bass wavelength (4.0 m) is much larger than the 0.90 m gap, so the diffraction is much greater
✓ 1
B1 the high note wavelength (0.10 m) is much smaller than the gap, so it diffracts little and spreads out less
✓ 1
Exam tip: Diffraction is greatest when the gap is about the same size as, or smaller than, the wavelength; compare the two sizes in your answer.
196IGPE3-3-15r22-123(d) Light and SoundPaper 1PHard8 marks
Storms far out in the Atlantic produce swell waves. At a monitoring buoy, the swell has a wavelength of 144 m and a time period of 9.0 s.
(a)
Calculate the speed of the swell.
(3)
(b)
Two buoys are 72 m apart along the direction in which the swell travels. State whether the two buoys are at the same point in their up-and-down cycles at any instant. Justify your answer.
(2)
(c)
A storm that produced the swell was 2400 km from the buoy. Use the speed from (a) to calculate how long the swell took to travel from the storm to the buoy. Give your answer in hours to 3 significant figures.
(3)
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Part (a)
use of v = fλ (with f = 1 ÷ T) or v = λ ÷ T
✓ 1
144 ÷ 9.0 (or f = 0.111 Hz and v = 0.111 × 144)
✓ 1
= 16 m/s
✓ 1
Part (b)
72 m is half a wavelength
✓ 1
so they are in antiphase: when one is at a crest the other is at a trough, so they are not at the same point
✓ 1
Part (c)
distance = 2.4×106 m
✓ 1
t = 2.4×106 ÷ 16 = 1.5×105 s (ECF from (a))
✓ 1
= 41.7 hours
✓ 1
Exam tip: Convert kilometres to metres before dividing, and seconds to hours at the end by dividing by 3600.
197IGPE3-3-16r28-113(d) Light and SoundPaper 1PHard8 marks
A bridge uses a glass optical fibre as a sensor. The fibre has refractive index 1.50.
Diagram not to scale
(a)
State the two conditions needed for total internal reflection at the inside surface of the fibre.
(2)
(b)
Calculate the critical angle for the glass-air boundary. Give your answer to 3 significant figures.
(3)
(c)
A ray of light in air hits the end of the fibre at an angle of incidence of 30°. Calculate the angle of refraction. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
Light is travelling from a denser material to a less dense material
✓ 1
The angle of incidence is greater than the critical angle
✓ 1
Part (b)
M1 sin c = 1 ÷ n
✓ 1
M1 c = sin−1(1 ÷ 1.50)
✓ 1
A1 41.8°
✓ 1
Part (c)
M1 n = sin i ÷ sin r
✓ 1
M1 sin r = sin 30° ÷ 1.50 = 0.3333
✓ 1
A1 19.5°
✓ 1
Exam tip: Take care that your calculator is in degree mode, and remember that the inverse sine (sin⁻¹) gives the angle.
198IGPE3-3-16r28-123(b) Properties of WavesPaper 1PHard9 marks
A wave machine in a leisure-centre pool makes water waves. A float on the surface completes 24 up-and-down oscillations in 60 s. The distance from the first crest to the seventh crest along the pool is 21 m. The float moves a vertical distance of 0.36 m from its highest to its lowest position.
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the wavelength of the waves.
(2)
(c)
Calculate the speed of the waves.
(2)
(d)
State the amplitude of the waves.
(1)
(e)
Near the shallow end of the pool the wave speed falls to 1.0 m/s. The frequency stays the same. Calculate the new wavelength.
(2)
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Part (a)
M1 24 ÷ 60
✓ 1
A1 0.40 Hz
✓ 1
Part (b)
M1 there are 6 wavelengths between the first and seventh crest
✓ 1
A1 21 ÷ 6 = 3.5 m
✓ 1
Part (c)
M1 v = f × λ = 0.40 × 3.5 (ECF from (a) and (b))
✓ 1
A1 1.4 m/s
✓ 1
Part (d)
0.18 m (half of 0.36 m)
✓ 1
Part (e)
M1 λ = v ÷ f = 1.0 ÷ 0.40 (ECF from (a))
✓ 1
A1 2.5 m
✓ 1
Exam tip: Count the gaps, not the crests: from the first to the seventh crest there are six wavelengths.
199IGPE3-3-16r32-123(b) Properties of WavesPaper 1PHard9 marks
A float bobs up and down 12 times in 20 s as waves from a wave machine in a leisure-centre pool pass it. The distance between neighbouring crests is 4.5 m.
(a)
Calculate the frequency of the waves.
(2)
(b)
Calculate the speed of the waves.
(2)
(c)
The operator changes the frequency to 0.45 Hz without changing the depth of water, so the wave speed stays the same. Calculate the new wavelength.
(2)
(d)
At the shallow end of the pool the speed falls to 1.8 m/s. The frequency is still 0.45 Hz. Calculate the wavelength at the shallow end.
(2)
(e)
Explain why the frequency of the waves is the same at the shallow end.
(1)
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Part (a)
M1 frequency = number of waves ÷ time = 12 ÷ 20
✓ 1
A1 0.60 Hz
✓ 1
Part (b)
M1 v = f × λ = 0.60 × 4.5 (ECF from (a))
✓ 1
A1 2.7 m/s
✓ 1
Part (c)
M1 λ = v ÷ f = 2.7 ÷ 0.45
✓ 1
A1 6.0 m
✓ 1
Part (d)
M1 λ = v ÷ f = 1.8 ÷ 0.45
✓ 1
A1 4.0 m
✓ 1
Part (e)
the same number of waves must pass each second as the machine produces (waves cannot pile up or be lost)
✓ 1
Exam tip: When a wave slows down, its frequency stays the same, so its wavelength shortens in proportion.
200IGPE3-3-16r32-133(d) Light and SoundPaper 1PHard9 marks
A cleaning bath in a pharmacy lab contains a sensor that sends out ultrasound pulses of frequency 40 kHz. Each pulse contains 6 complete waves. The speed of sound in water is 1500 m/s. The bath is filled to a depth of 0.45 m.
(a)
Calculate the time period of the ultrasound.
(2)
(b)
Calculate the wavelength of the ultrasound in water.
(2)
(c)
Calculate the duration of one pulse.
(2)
(d)
The pulse reflects from the bottom of the bath. Show that the echo starts to arrive back only after the pulse has finished being sent.
(3)
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Part (a)
M1 T = 1 ÷ f = 1 ÷ 40 000
✓ 1
A1 2.5 × 10−5 s
✓ 1
Part (b)
M1 λ = v ÷ f = 1500 ÷ 40 000
✓ 1
A1 0.0375 m
✓ 1
Part (c)
M1 6 × 2.5 × 10−5 (ECF from (a))
✓ 1
A1 1.5 × 10−4 s
✓ 1
Part (d)
distance travelled by the pulse = 2 × 0.45 = 0.90 m
✓ 1
M1 time for echo = 0.90 ÷ 1500 = 6.0 × 10−4 s
✓ 1
A1 6.0 × 10−4 s is greater than the pulse duration of 1.5 × 10−4 s
✓ 1
Exam tip: In 'show that' questions, write the final value and make the comparison in a clear statement.
201IGPE3-3-16r32-143(d) Light and SoundPaper 1PHard9 marks
A game-sound engineer displays a note from a loudspeaker on an oscilloscope. The time-base is set to 0.50 ms per division. Across 8.0 divisions the trace shows 4 complete waves.
Oscilloscope screen. Diagram not to scale
(a)
Calculate the time period of the note.
(2)
(b)
Calculate the frequency of the note.
(2)
(c)
A second note is displayed with the same settings. It shows 6 complete waves across the 8.0 divisions. Calculate its frequency.
(2)
(d)
State which note has the higher pitch. The speed of sound in air is 340 m/s. Calculate the wavelength of the second note, to 3 significant figures.
(3)
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Part (a)
M1 time for 4 waves = 8.0 × 0.50 = 4.0 ms
✓ 1
A1 period = 4.0 ÷ 4 = 1.0 ms
✓ 1
Part (b)
M1 f = 1 ÷ T = 1 ÷ 1.0 × 10−3 (ECF from (a))
✓ 1
A1 1000 Hz
✓ 1
Part (c)
M1 period = 4.0 ms ÷ 6, or f = 6 ÷ 4.0 × 10−3
✓ 1
A1 1500 Hz
✓ 1
Part (d)
the second note (higher frequency)
✓ 1
M1 λ = v ÷ f = 340 ÷ 1500
✓ 1
A1 0.227 m
✓ 1
Common error: Time-base readings are per division: multiply by the number of divisions, then divide by the number of waves to get one period.
202IGPE3-3-16r34-143(b) Properties of WavesPaper 1PHard8 marks
A research boat is anchored in the sea. Sound travels at 1500 m/s in sea water and at 340 m/s in air. The boat's engine produces a sound of frequency 500 Hz.
(a)
Calculate the wavelength of the sound in sea water.
(2)
(b)
The sound then travels into the air. State what happens to the frequency of the sound and calculate its wavelength in air.
(2)
(c)
A pulse of sound from an explosion travels 1530 m to the boat through the sea water. A second pulse travels the same distance through the air. Calculate the difference between the two arrival times.
(4)
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Part (a)
λ = v ÷ f = 1500 ÷ 500
✓ 1
3.0 m
✓ 1
Part (b)
the frequency stays at 500 Hz
✓ 1
λ = 340 ÷ 500 = 0.68 m
✓ 1
Part (c)
time = distance ÷ speed
✓ 1
time in water = 1530 ÷ 1500 = 1.02 s
✓ 1
time in air = 1530 ÷ 340 = 4.5 s
✓ 1
difference = 3.48 s
✓ 1
Exam tip: Sound travels faster in a liquid than in a gas. For a boundary crossing, the frequency stays the same while the speed and wavelength change.
203IGPE3-3-15r38-093(b) Properties of WavesPaper 1PHard6 marks
Two buoys, A and B, float in line with the direction in which regular water waves travel from A towards B. The waves have frequency 0.25 Hz and speed 3.0 m/s. The buoys are 90 m apart.
(a)
Calculate the wavelength of the waves.
(2)
(b)
Calculate the time taken for a wave crest to travel from A to B.
(2)
(c)
At one instant, buoy A is at its highest point. Determine whether buoy B is also at its highest point. Use the number of wavelengths between the buoys.
(2)
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Part (a)
M1 λ = v ÷ f = 3.0 ÷ 0.25
✓ 1
A1 12 m
✓ 1
Part (b)
M1 t = d ÷ v = 90 ÷ 3.0
✓ 1
A1 30 s
✓ 1
Part (c)
M1 90 ÷ 12 = 7.5 wavelengths (ECF from (a))
✓ 1
A1 the buoys are half a wavelength out of step, so B is at its lowest point, not its highest
✓ 1
Exam tip: A separation of a whole number of wavelengths means points vibrate in step; an extra half wavelength means they vibrate in opposite directions.
204IGPE3-3-15r38-133(d) Light and SoundPaper 1PHard6 marks
A robot in a swimming pool measures the depth of the water using pulses of ultrasound. The speed of ultrasound in water is 1500 m/s. A pulse sent downwards is detected back at the robot 0.0080 s later.
(a)
Calculate the depth of the water.
(3)
(b)
State what is meant by ultrasound.
(1)
(c)
The ultrasound has a frequency of 50 000 Hz. Calculate its wavelength in water.
(2)
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Part (a)
M1 distance travelled = speed × time = 1500 × 0.0080 = 12 m
✓ 1
M1 divides by 2 because the pulse travels down and back
✓ 1
A1 6.0 m
✓ 1
Part (b)
sound with a frequency above 20 000 Hz, too high for humans to hear
✓ 1
Part (c)
M1 λ = v ÷ f = 1500 ÷ 50 000
✓ 1
A1 0.030 m
✓ 1
Common error: The pulse travels to the bottom and back, so the distance found from speed × time is twice the depth.
205IGPE3-3-15r39-043(d) Light and SoundPaper 1PHard1 mark
Light passes from air into water of refractive index 1.33. The angle of refraction in the water is 30°. What is the angle of incidence in the air, to the nearest degree?
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Notes
n = sin i ÷ sin r, so sin i = 1.33 × sin 30° = 0.665 and i = 41.68°, which is 42°. 22° comes from dividing instead of multiplying (sin i = sin 30° ÷ 1.33). 30° assumes no bending. 40° comes from multiplying the angles (1.33 × 30°) instead of the sines.
✓ 1
Answer A
Answer: A
Common error: Refractive index links the sines of the angles, not the angles themselves.
206IGPE3-3-16r39-023(d) Light and SoundPaper 1PHard1 mark
A bat sends out pulses of ultrasound at a rate of 20 pulses per second. It waits for each echo before sending the next pulse. The speed of sound in air is 340 m/s. What is the greatest distance to an object that the bat can detect in this way?
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Mark
Notes
The time between pulses is 1 ÷ 20 = 0.050 s. In that time the sound travels 340 × 0.050 = 17 m, but this is the distance there and back, so the object is at most 17 ÷ 2 = 8.5 m away. 17 m forgets that the sound travels to the object and back. 4.25 m halves the distance twice. 0.029 m comes from dividing time by speed.
✓ 1
Answer B
Answer: B
Common error: An echo travels to the object and back, so the object's distance is half of speed × time.
207IGPE3-3-16r39-133(d) Light and SoundPaper 1PHard6 marks
A student stands between two flat parallel walls and claps once. Wall A is 60 m from the student. She hears two separate echoes, one from each wall. The echo from wall B arrives 0.50 s after the echo from wall A. The speed of sound in air is 340 m/s.
Diagram not to scale
(a)
Calculate the time between the clap and the echo from wall A. Give your answer to 2 significant figures.
(2)
(b)
Calculate the distance from the student to wall B. Use the unrounded value of the time from (a).
(3)
(c)
Explain why the student hears the two echoes as separate sounds.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 t = 2 × 60 ÷ 340
✓ 1
A1 0.35 s
✓ 1
Part (b)
M1 time for echo from wall B = 0.35294 + 0.50 = 0.853 s
✓ 1
M1 distance = 340 × 0.853 ÷ 2
✓ 1
A1 145 m
✓ 1
Part (c)
the echoes arrive more than about 0.1 s apart so the ear can tell them apart
✓ 1
Exam tip: Find each echo time from the total distance there and back, then compare the times.
208IGPE3-3-15r43-133(d) Light and SoundPaper 1PHard7 marks
A student tunes a ukulele string to 440 Hz using an oscilloscope with the time base set to 0.40 ms per division.
(a)
Calculate the period of a 440 Hz wave. Give your answer in ms to 3 significant figures.
(2)
(b)
Calculate how many divisions one complete wave of 440 Hz should occupy on the screen. Use the unrounded period from (a) and give your answer to 3 significant figures.
(2)
(c)
The string actually produces a wave 5.50 divisions wide. Calculate its frequency to 3 significant figures.
(2)
(d)
State whether the student should tighten or loosen the string, giving a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 T = 1/f = 1/440
✓ 1
A1 2.27 ms
✓ 1
Part (b)
M1 divisions = period ÷ time base = 2.2727 ms ÷ 0.40 ms
✓ 1
A1 5.68 divisions
✓ 1
Part (c)
M1 T = 5.50 × 0.40 ms = 2.20 ms and f = 1/T
✓ 1
A1 455 Hz
✓ 1
Part (d)
loosen, because the frequency is higher than 440 Hz and loosening lowers the frequency
✓ 1
Common error: A narrower wave on the screen means a shorter period and so a higher frequency.
209IGPE3-3-16r43-043(d) Light and SoundPaper 1PHard1 mark
A data centre uses an unclad plastic optical fibre of refractive index 1.50, surrounded by air. The critical angle is 41.8°. A ray inside the straight fibre travels at 55° to the axis (the centre line) of the fibre. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
The normal to the wall is at right angles to the axis, so the angle of incidence at the wall is 90° − 55° = 35°. This is less than 41.8°, so there is no total internal reflection and some light refracts out: C. A wrongly uses the angle to the axis as the angle of incidence. B ignores that some light is always reflected. D has the wrong condition for total internal reflection.
✓ 1
Answer C
Answer: C
Common error: Angles of incidence are measured from the normal, not from the surface or the fibre axis.
210IGPE3-3-16r43-123(d) Light and SoundPaper 1PHard6 marks
An archaeologist stands in a tomb corridor between two parallel walls and claps once. She hears the first echo 0.060 s after the clap and the second echo 0.100 s after the first. The speed of sound in air is 340 m/s. Do not round during your working.
(a)
Calculate the distance from the archaeologist to the nearer wall.
(2)
(b)
Calculate the time between the clap and the second echo.
(1)
(c)
Calculate the distance from the archaeologist to the farther wall.
(2)
(d)
Calculate the distance between the two walls.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the sound travels there and back, so distance = (340 × 0.060) ÷ 2
✓ 1
A1 10.2 m
✓ 1
Part (b)
A1 0.060 + 0.100 = 0.160 s
✓ 1
Part (c)
M1 distance = (340 × 0.160) ÷ 2 (ECF from (b))
✓ 1
A1 27.2 m
✓ 1
Part (d)
A1 10.2 + 27.2 = 37.4 m (ECF)
✓ 1
Common error: An echo travels to the wall and back, so use half of speed × time for the distance.
212IGCSE-PHY-3-0173(d) Light and SoundPaper 2 onlyGrade 6–710 marks
A student connects a microphone to an oscilloscope. She strikes tuning fork X and holds it near the microphone, then does the same with tuning fork Y. The oscilloscope settings are the same for both traces. The diagram shows the two traces.
(a)
Explain how the microphone and oscilloscope produce a trace of the sound wave.
(2)
(b)
Calculate the frequency of the sound from tuning fork X.
(3)
(c)
Compare the pitch and the loudness of the sound from fork Y with the sound from fork X. Give reasons for your answers.
(4)
(d)
Suggest how the student could make her measurement of the time period more accurate.
(1)
Show mark scheme
(a)
sound wave makes (diaphragm of) microphone vibrate;
producing an alternating / changing voltage (pd) which is displayed (against time) on the oscilloscope;
Allow
electrical signal for voltage
Do not accept
the microphone records the sound
(b)
period = 4.0 divisions;
T = 4.0 × 0.5 = 2.0 ms;
f = 1 ÷ 0.0020 = 500 (Hz);
Allow
ecf for period read from trace
0.5 kHz
0.5 (Hz) (ms not converted) scores 2
(c)
Y has a higher pitch;
because the period is shorter / frequency is higher / more waves on screen;
Y is quieter / less loud;
because amplitude (height of trace) is smaller;
Allow
reasons may refer to the trace or to the vibration of the fork
Do not accept
Y is louder because its waves are closer together
(d)
measure the time for several cycles and divide by the number of cycles;
Allow
adjust time-base so more cycles fill the screen and average
213IGCSE-PHY-3-0183(d) Light and SoundPaper 2 onlyGrade 6–78 marks
Some shops use a device that emits a high-pitched sound of frequency 17 000 Hz to discourage groups from gathering outside. The speed of sound in air is 340 m/s. The table shows the typical highest frequency that people of different ages can hear. Age / years: 15, 25, 35, 45, 55 Highest audible frequency / kHz: 19.0, 17.5, 16.0, 14.5, 13.0
(a)
Determine the time period of the sound wave from the device.
(1)
(b)
Calculate the wavelength of the sound from the device.
(2)
(c)
Use the table to explain which age groups are likely to hear the device.
(2)
(d)
The shopkeeper increases the amplitude of vibration of the loudspeaker in the device. State and explain the effect on the sound heard by a 15-year-old.
(2)
(e)
Explain why increasing the amplitude does not make the sound audible to a 45-year-old.
(1)
Show mark scheme
(a)
5.9 × 10⁻⁵ (s);
Allow
5.88 × 10⁻⁵ (s)
0.000059 (s)
0.059 ms
Do not accept
17 000 (s)
(b)
rearrangement and substitution e.g. λ = 340 ÷ 17 000;
evaluation: (λ =) 0.020 (m);
Allow
2.0 cm
Do not accept
5.8 × 10⁶ m (v × f)
(c)
people aged about 25 years or under / teenagers and young adults;
their highest audible frequency is above 17 kHz (and older people's is below);
Allow
up to about 28 years (interpolation)
(d)
(sound is) louder;
loudness depends on (increases with) amplitude (pitch / frequency unchanged);
Do not accept
higher pitch
(e)
frequency (pitch) is unchanged at 17 kHz, which is above their highest audible frequency (14.5 kHz);
214IGCSE-PHY-3-0363(d) Light and SoundPaper 2 onlyGrade 6–71 mark
A signal generator and loudspeaker produce sounds with different time periods. Which time period gives a sound that a person with normal hearing cannot hear?
Show mark scheme
D 0.045 ms;
A is incorrect: f = 1 ÷ 0.030 s = 33 Hz, which can be heard (treating ms as μs gives 33 000 Hz and wrongly suggests it is too high)
B is incorrect: f = 1 ÷ 0.000 25 s = 4000 Hz, which can be heard (chosen if the upper limit of hearing is misremembered as 2000 Hz)
C is incorrect: f = 1 ÷ 0.000 060 s = 16 700 Hz, which is below 20 000 Hz and can be heard (not converting ms gives 16.7 Hz, which wrongly appears to be below 20 Hz)
215IGCSE-PHY-3-0373(d) Light and SoundPaper 2 onlyGrade 6–79 marks
Two students measure the speed of sound in air using echoes from a large flat wall. Student A stands 85 m from the wall and claps. She keeps clapping so that each clap is made at the same moment as she hears the echo of the previous clap. Student B uses a stopwatch to measure the time for 20 intervals between claps. They repeat the measurement three times. Trial / number: 1, 2, 3 Time for 20 intervals / s: 10.6, 10.2, 10.5Diagram not to scale
(a)
Determine the mean time for 20 intervals.
(1)
(b)
Calculate the speed of sound in air.
(4)
(c)
Explain why the students time 20 intervals rather than a single interval.
(2)
(d)
Explain why the students stand a long distance from the wall.
(1)
(e)
The accepted value for the speed of sound in air is 340 m/s. Suggest one reason why the students' value is lower.
(1)
Show mark scheme
(a)
10.4 (s);
Allow
10.43 (s)
(b)
(time for one interval =) 10.43 ÷ 20 = 0.522 (s);
(distance =) 2 × 85 = 170 (m);
(v =) 170 ÷ 0.52;
326 (m/s);
Allow
325–330 (m/s)
327 (m/s) from rounded 0.52 s
ecf from (a)
163 (m/s) (distance not doubled) scores 3
16.3 (m/s) (time not divided by 20) scores 3
(c)
(single interval) too short to time accurately / reaction time is large compared with one interval;
(timing 20) reduces the percentage uncertainty / reaction time error shared between 20 intervals;
Do not accept
to make the experiment fair
so they can take an average (alone)
(d)
longer time for each echo so percentage (timing) uncertainty is smaller / echo can be heard separately from the clap;
Do not accept
so the sound is louder
(e)
any suitable reason, e.g. clapping slightly after hearing the echo (reaction time) makes each interval too long / stopwatch started early or stopped late / air temperature lower than that for 340 m/s;
216IGCSE-PHY-3-0383(d) Light and SoundPaper 2 onlyGrade 6–78 marks
A musician wants to tune the A string of a ukulele to a frequency of 440 Hz. She plucks the string near a microphone connected to an oscilloscope. The time-base of the oscilloscope is set to 0.50 ms per division. The diagram shows the trace produced by the string before tuning.
(a)
Calculate the time period of a sound of frequency 440 Hz.
(2)
(b)
Determine the number of horizontal divisions that one cycle of a 440 Hz sound would occupy on this oscilloscope.
(1)
(c)
Use the trace to explain whether the pitch of the string is too high or too low.
(3)
(d)
After tuning, the musician plucks the string harder. Describe how the trace changes and state the effect on the sound heard.
(2)
Show mark scheme
(a)
(T =) 1 ÷ 440;
0.00227 (s) / 2.27 (ms);
Allow
2.3 × 10⁻³ s
2.3 ms
Do not accept
2.27 (unit s)
(b)
4.5 (divisions);
Allow
4.54 / 4.55
ecf from (a)
Do not accept
0.22 (0.5 ÷ 2.27)
(c)
(trace) one cycle = 5.0 divisions / period = 2.5 ms;
period too long so frequency too low / frequency = 400 Hz (less than 440 Hz);
(so) pitch is too low;
Allow
MP3 dependent on a correct comparison of period or frequency
Do not accept
pitch too high because the waves are more spread out
(d)
(trace) taller / larger amplitude, with the same period / same number of waves on screen;
217IGPE2-056-00-063(d) Light and SoundPaper 2PMedium8 marks
A sports equipment company tests a referee's whistle. A microphone connected to an oscilloscope picks up the sound. The timebase is set to 0.1 ms per division. One complete wave on the screen covers 2.5 horizontal divisions. The speed of sound in air is 340 m/s.
Oscilloscope screen. Diagram not to scale
(a)
Calculate the period of the sound wave in seconds.
(2)
(b)
Calculate the frequency of the sound.
(2)
(c)
Calculate the wavelength of the sound in air.
(2)
(d)
The referee blows the same whistle harder, so the sound is louder but the pitch is unchanged. Describe how the trace on the oscilloscope changes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Period = number of divisions × timebase = 2.5 × 0.1 ms . T = 0.25 ms = 2.5 × 10−4 s .
✓ 2
Part (b)
f = 1 / T = 1 / (2.5 × 10−4) . f = 4000 Hz .
✓ 2
Part (c)
λ = v / f = 340 / 4000 . λ = 0.085 m .
✓ 2
Part (d)
The peaks are taller (larger amplitude / greater vertical height) . The horizontal spacing of the waves / number of waves on the screen stays the same .
✓ 2
Exam tip: Measure across several complete waves and divide by the number of waves. This reduces the uncertainty in the period.
218IGPE2-056-03-093(d) Light and SoundPaper 2PMedium7 marks
A microphone connected to an oscilloscope picks up the sound of a referee's whistle. The timebase is set to 0.20 ms per division. One complete wave on the screen spans 2.5 horizontal divisions.
Oscilloscope screen. Diagram not to scale
(a)
Calculate the time period of the sound wave.
(2)
(b)
Calculate the frequency of the sound.
(2)
(c)
State the range of frequencies that a typical human can hear.
(1)
(d)
The referee blows the whistle louder and at a higher pitch. The oscilloscope settings are not changed. Describe how the trace changes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 2.5 × 0.20 ms
✓ 1
= 0.50 ms (5.0 × 10−4 s)
✓ 1
Part (b)
f = 1 ÷ T = 1 ÷ 5.0 × 10−4
✓ 1
= 2000 Hz
✓ 1
Part (c)
20 Hz to 20 000 Hz
✓ 1
Part (d)
The waves are taller, showing a larger amplitude . More complete waves fit on the screen, showing a shorter period and higher frequency .
✓ 2
Exam tip: Convert ms to s before using f = 1/T, otherwise the frequency comes out 1000 times too small.
219IGPE2-056-02-033(b) Properties of WavesPaper 2PMedium1 mark
A delivery drone carries a buzzer that emits a sound of constant frequency. The drone flies directly away from a person standing on the ground. Compared with the sound emitted by the buzzer, what does the person detect?
Show mark scheme
Marking point
Mark
Notes
When the source moves away from the observer, the wavefronts reaching the observer are spread further apart, so the observed wavelength is longer. The speed of sound in air does not change, so from v = f × λ a longer wavelength means a lower observed frequency. This is the Doppler effect. B describes a source moving towards the observer. C wrongly assumes that the frequency heard must equal the source frequency. D is impossible, because at constant wave speed a lower frequency must go with a longer wavelength.
✓ 1
Answer A
Answer: A
Exam tip: Link wavelength and frequency through v = f × λ with the wave speed fixed. If one goes up, the other must go down.
220IGPE2-056-02-063(d) Light and SoundPaper 2PMedium7 marks
A music student tunes a violin string. She uses a microphone connected to an oscilloscope to study the sound the string produces.
Oscilloscope screen. Diagram not to scale
(a)
Describe how the microphone and oscilloscope together display the sound wave.
(2)
(b)
On the oscilloscope screen, one complete wave covers 4.0 horizontal divisions. The timebase is set to 0.50 ms per division. Calculate the frequency of the sound.
(3)
(c)
The student bows the same string more strongly, so the note is louder but has the same pitch. State how the trace changes.
(1)
(d)
She then tightens the string so that it plays a note of higher pitch, at the same loudness. State how the trace changes.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The microphone converts the sound (pressure variations) into an electrical signal / varying voltage
✓ 1
The oscilloscope displays this voltage against time as a wave trace
✓ 1
Part (b)
Time period T = 4.0 × 0.50 = 2.0 ms
✓ 1
f = 1 / T = 1 / 2.0×10−3
✓ 1
f = 500 Hz
✓ 1
Part (c)
The waves are taller / have a larger amplitude, with the same horizontal spacing
✓ 1
Part (d)
More complete waves appear on the screen / the waves are closer together horizontally
✓ 1
Exam tip: Find the time period first by multiplying divisions by the timebase, and convert ms to s before using f = 1/T.
221IGPE2-056-02-093(d) Light and SoundPaper 2PMedium8 marks
In elite sprint races, the starting signal is played through a small loudspeaker placed behind each athlete's starting blocks. Speed of sound in air = 340 m/s
Diagram not to scale
(a)
In an older system, a single starting gun was fired 12 m from the athlete in the outside lane. Calculate the time taken for the sound to reach this athlete.
(2)
(b)
Explain why a loudspeaker is now placed behind every athlete.
(2)
(c)
A student measures the speed of sound by standing 85 m from a large wall. She claps so that each clap happens at the same moment she hears the echo of the previous clap. A friend measures the time for 20 intervals between claps as 10.0 s. Calculate the speed of sound from these results.
(3)
(d)
Suggest why the friend timed 20 intervals rather than just one.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
t = d / v = 12 / 340
✓ 1
t = 0.035 s
✓ 1
Part (b)
Each athlete is then the same (short) distance from the sound source, so all hear the signal at the same time
✓ 1
Athletes further from a single gun would hear it later, which would be unfair / a disadvantage
✓ 1
Part (c)
Time for one interval = 10.0 / 20 = 0.50 s
✓ 1
Distance travelled by sound in one interval = 2 × 85 = 170 m
✓ 1
v = 170 / 0.50 = 340 m/s
✓ 1
Part (d)
Reduces the effect of reaction time on the measurement / gives a smaller percentage uncertainty
✓ 1
Exam tip: In echo methods, decide clearly how far the sound travels in the time measured. Here it is to the wall and back, so the distance is doubled.
222IGPE-052-00-073(b) Properties of WavesPaper 2PMedium5 marks
A delivery drone emits a warning beep at a constant frequency of 440 Hz. A person standing in a garden hears the drone fly towards them, pass overhead and then fly away.
(a)
Describe how the pitch of the beep heard by the person compares with 440 Hz as the drone approaches and as it moves away.
(2)
(b)
Explain why the frequency heard by the person is higher than 440 Hz while the drone is approaching.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
As the drone approaches, the pitch heard is higher than 440 Hz . As it moves away, the pitch heard is lower than 440 Hz .
✓ 2
Part (b)
The drone moves towards the person between emitting one wavefront and the next, so the wavefronts in front of it are bunched together . The wavelength of the sound reaching the person is shorter . The speed of sound is unchanged, so more waves reach the person each second, which means a higher frequency .
✓ 3
Common error: The drone itself still emits 440 Hz. Only the frequency and wavelength detected by the observer change. Do not write that the source 'changes its frequency'.
223IGPE-052-00-103(d) Light and SoundPaper 2PMedium8 marks
A microphone connected to an oscilloscope is used to study the note made by a metal wind chime. On the screen, one complete wave covers 4.0 horizontal divisions. The timebase is set to 0.50 ms per division.
Oscilloscope screen. Diagram not to scale
(a)
State the job of the microphone in this arrangement.
(1)
(b)
Calculate the time period of the sound wave.
(2)
(c)
Calculate the frequency of the note.
(2)
(d)
A second chime is tested with the same oscilloscope settings. Its trace is twice as tall and has more complete waves across the screen. Describe how its sound differs from the first chime.
(2)
(e)
A third chime produces a sound with a frequency of 25 kHz. State whether a person could hear this sound. Give a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
It converts the sound wave into an electrical signal (a varying voltage) that the oscilloscope can display .
✓ 1
Part (b)
T = 4.0 × 0.50 ms ; T = 2.0 ms, i.e. 0.0020 s
✓ 2
Part (c)
f = 1 ÷ T = 1 ÷ 0.0020 ; f = 500 Hz
✓ 2
Part (d)
It is louder, because the amplitude is larger . It has a higher pitch, because the frequency is greater .
✓ 2
Part (e)
No, because 25 kHz is above the upper limit of human hearing of about 20 kHz (20 000 Hz) .
✓ 1
Common error: Count the divisions for one complete cycle (peak to next peak), not for half a wave. Counting only half a wave halves the period and doubles your frequency.
224IGPE3-3-14r21-113(d) Light and SoundPaper 2PMedium6 marks
An archaeologist examines a polished crystal bead. A ray of light in air hits a flat face of the crystal at an angle of incidence of 40°. The angle of refraction in the crystal is 25°. The speed of light in a vacuum is 3.0 × 108 m/s.
Diagram not to scale
(a)
Calculate the refractive index of the crystal. Give your answer to 3 significant figures.
(2)
(b)
Calculate the speed of light in the crystal. Use n = 1.52 and give your answer to 3 significant figures.
(2)
(c)
Explain why the ray bends towards the normal as it enters the crystal.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 n = sin 40° ÷ sin 25°
✓ 1
A1 1.52
✓ 1
Part (b)
M1 v = c / n = 3.0 × 108 ÷ 1.52
✓ 1
A1 1.97 × 108 m/s
✓ 1
Part (c)
B1 light slows down in the crystal
✓ 1
B1 the wavefront slows on one side first, so the ray changes direction towards the normal
✓ 1
Common error: Put sin of the angle in air on top: n = sin i ÷ sin r, with both angles measured from the normal.
225IGPE3-3-14r22-033(d) Light and SoundPaper 2PMedium1 mark
A ray of light travels inside a glass reflector prism and meets the glass–air boundary at an angle of incidence of 45°. The critical angle of the glass is 41°. What happens to the ray at this boundary?
Show mark scheme
Marking point
Mark
Notes
The light is travelling from glass (denser) to air and the angle of incidence, 45°, is greater than the critical angle, 41°, so all of the light is reflected at the boundary. By the law of reflection the angle of reflection equals the angle of incidence, 45°. Refraction into the air would only happen if the angle of incidence were smaller than the critical angle. Bending towards the normal is what happens when light enters a denser medium, not leaves it. Partial reflection with a 41° reflected ray is wrong because the reflected angle must equal the incident angle and no light escapes above the critical angle.
✓ 1
Answer A
Answer: A
Exam tip: Compare the angle of incidence with the critical angle: total internal reflection needs it to be larger, and light going from a denser to a less dense medium.
227IGPE3-3-15r22-113(d) Light and SoundPaper 2PMedium6 marks
A jeweller identifies a clear gemstone by finding its refractive index. A narrow ray of light in air strikes a flat face of the gem. The angle of incidence is 60.0° and the angle of refraction is 23.7°. Refractive indices: diamond 2.42, cubic zirconia 2.15, quartz 1.54, glass 1.52.
Diagram not to scale
(a)
Calculate the refractive index of the gemstone. Give your answer to 3 significant figures.
(3)
(b)
Use the table to identify the gemstone.
(1)
(c)
Explain why the ray bends towards the normal as it enters the gem.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
n = sin i ÷ sin r
✓ 1
n = sin 60.0° ÷ sin 23.7°
✓ 1
= 2.15
✓ 1
Part (b)
cubic zirconia (ECF from (a))
✓ 1
Part (c)
light travels more slowly in the gem than in air
✓ 1
the change in speed makes the ray change direction, towards the normal when it slows down
✓ 1
Common error: Angles of incidence and refraction are measured from the normal, not from the surface of the gem.
228IGPE3-3-14r26-113(b) Properties of WavesPaper 2PMedium6 marks
A library visitor uses a converging lens as a magnifying glass to read small print. A letter of height 4.0 mm appears to have height 15 mm when viewed through the lens.
(a)
Calculate the magnification.
(2)
(b)
State two features of the image formed when the text is closer to the lens than its focal length.
(2)
(c)
The same lens is used with the same magnification to view a letter of height 2.8 mm. Calculate the height of its image.
229IGPE3-3-16r28-043(b) Properties of WavesPaper 2PMedium1 mark
Which statement about the waves produced by an earthquake is correct?
Show mark scheme
Marking point
Mark
Notes
P-waves are longitudinal and can travel through both solids and liquids, whereas S-waves are transverse and cannot travel through liquids. The two options that swap these properties are wrong, and the option with both longitudinal is wrong because S-waves are transverse and are slower than P-waves.
✓ 1
Answer B
Answer: B
Common error: A transverse wave needs a medium that resists shearing, so it cannot pass through a liquid.
231IGPE3-3-16r32-033(d) Light and SoundPaper 2PMedium1 mark
The glass core of a fibre-optic gaming cable has a refractive index of 1.60. What is the critical angle for light travelling in the glass towards a glass–air boundary?
Show mark scheme
Marking point
Mark
Notes
sin c = 1 ÷ n = 1 ÷ 1.60 = 0.625, so c = 38.7°. 51.3° is 90° − c, measured from the boundary instead of the normal. 32.0° comes from using tan instead of sin. 58.0° is 90° minus that tan-based angle.
✓ 1
Answer B
Answer: B
Common error: Critical angle is measured from the normal, and sin c = 1/n.
233IGPE3-3-16r32-093(d) Light and SoundPaper 2PMedium6 marks
A virtual-reality headset has lenses made of glass. A ray of light travels from air into the glass. The angle of incidence is 40° and the angle of refraction 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 3.00 × 108 m/s. Calculate the speed of light in the glass. Use n = 1.52.
(2)
(c)
A second ray has an angle of incidence of 60°. Calculate its angle of refraction. Use n = 1.52 and give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 n = sin i ÷ sin r = sin 40° ÷ sin 25°
✓ 1
A1 1.52
✓ 1
Part (b)
M1 speed in glass = 3.00 × 108 ÷ 1.52
✓ 1
A1 1.97 × 108 m/s
✓ 1
Part (c)
M1 sin r = sin 60° ÷ 1.52 = 0.570
✓ 1
A1 r = 34.7°
✓ 1
Common error: Use n = sin i ÷ sin r with both angles measured from the normal, and check your calculator is in degree mode.
234IGPE3-3-16r34-043(d) Light and SoundPaper 2PMedium1 mark
A transparent plastic has a refractive index of 1.25. What is the critical angle for light travelling from the plastic into air?
Show mark scheme
Marking point
Mark
Notes
sin c = 1 ÷ n = 1 ÷ 1.25 = 0.80, so c = 53°. 37° is found by using cos c = 0.80 instead of the sine. 39° comes from using tan c = 0.80. 72° comes from dividing 90° by 1.25, but the critical angle is found from a sine relationship, not a ratio of angles.
✓ 1
Answer B
Answer: B
Exam tip: Critical angle: sin c = 1 ÷ n for a boundary with air. Total internal reflection only occurs when the angle of incidence is greater than c.
235IGPE3-3-16r34-073(d) Light and SoundPaper 2PMedium7 marks
An endoscope uses a bundle of thin glass fibres to carry light into the body and to carry an image back out. Each fibre has a refractive index of 1.50. Light stays inside the fibre by total internal reflection at the boundary between the glass and the air.
(a)
State two conditions needed for total internal reflection to occur.
(2)
(b)
Calculate the critical angle for light going from the glass into air. Give your answer to 3 significant figures.
(3)
(c)
A ray reaches the glass–air boundary at an angle of incidence of 55°. Explain what happens to the ray.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the light travels from a more dense medium to a less dense medium (glass to air)
✓ 1
the angle of incidence is greater than the critical angle
✓ 1
Part (b)
sin c = 1 ÷ n
✓ 1
sin c = 1 ÷ 1.50 = 0.667
✓ 1
c = 41.8°
✓ 1
Part (c)
55° is greater than the critical angle (41.8°, ECF from (b))
✓ 1
so the ray is totally internally reflected: none of the light passes out into the air
✓ 1
Exam tip: To decide between refraction and total internal reflection, compare the angle of incidence with the critical angle calculated for that boundary.
236IGPE3-3-16r35-043(d) Light and SoundPaper 2PMedium1 mark
The refractive index of a glass is 1.25. What is the critical angle for light travelling from this glass into air?
Show mark scheme
Marking point
Mark
Notes
At the critical angle sin c = 1 ÷ n = 1 ÷ 1.25 = 0.80, so c = 53°. 37° comes from using cos instead of sin. 39° comes from using tan c = 1 ÷ n. 51° comes from using tan c = n.
✓ 1
Answer B
Answer: B
Exam tip: Use sin c = 1/n. A critical angle must always be less than 90°, so check your calculator is in degree mode.
237IGPE3-3-15r38-033(d) Light and SoundPaper 2PMedium1 mark
A ray of light travels inside a semicircular glass block along a radius, so it is not bent when it enters the curved surface. It reaches the centre of the flat face at an angle of incidence of 50°. The critical angle of the glass for a glass–air boundary is 42°. What happens to the ray at the flat face?
Show mark scheme
Marking point
Mark
Notes
The angle of incidence (50°) is greater than the critical angle (42°), so none of the light can leave the glass and it is totally internally reflected, which is the correct statement. Refraction away from the normal into air happens only when the angle of incidence is less than the critical angle. Emerging along the surface happens only at exactly the critical angle. Passing straight on happens only for a ray along the normal.
✓ 1
Answer D
Answer: D
Common error: Total internal reflection needs the angle of incidence to be greater than the critical angle, not just any angle in the denser material.
238IGPE3-3-15r38-103(d) Light and SoundPaper 2PMedium7 marks
A student finds the critical angle of a semicircular glass block using a ray box, a sheet of paper and a protractor. Use n = 1 ÷ sin c, where c is the critical angle.
(a)
Describe how the student can find the critical angle of the glass.
(4)
(b)
The student finds that the critical angle is 41°. Calculate the refractive index of the glass. Give your answer to 3 significant figures.
(2)
(c)
Suggest one way to improve the reliability of the result for the critical angle.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
place the block on the paper, draw round it and mark the centre of the flat face
✓ 1
aim a ray at the curved surface towards the centre so that it enters without bending and hits the flat face at the centre
✓ 1
increase the angle of incidence at the flat face, measuring it from the normal with the protractor, until the emerging ray just disappears along the flat face
✓ 1
the angle of incidence at that point is the critical angle
✓ 1
Part (b)
M1 n = 1 ÷ sin 41°
✓ 1
A1 1.52
✓ 1
Part (c)
repeat the measurement several times and calculate a mean
✓ 1
Exam tip: Measure all angles from the normal, not from the surface of the block.
239IGPE3-3-15r39-083(d) Light and SoundPaper 2PMedium7 marks
Light from a laser on a bridge deck is aimed at the surface of a river. The angle of incidence in the air is 50.0°. The refractive index of the water is 1.33. The refractive index is given by n = sin i ÷ sin r.
Diagram not to scale
(a)
State what happens to the speed of the light as it enters the water.
(1)
(b)
Calculate the angle of refraction in the water. Give your answer to 1 decimal place.
(3)
(c)
A second beam must travel through the water at an angle of refraction of 28.0°. Calculate the angle of incidence needed in the air. Give your answer to 1 decimal place.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
it decreases
✓ 1
Part (b)
M1 sin r = sin i ÷ n
✓ 1
M1 sin r = sin 50.0° ÷ 1.33 = 0.576
✓ 1
A1 35.2°
✓ 1
Part (c)
M1 sin i = n × sin r
✓ 1
M1 sin i = 1.33 × sin 28.0° = 0.624
✓ 1
A1 38.6°
✓ 1
Common error: When light enters a denser medium it bends towards the normal, so the angle of refraction must come out smaller than the angle of incidence.
240IGPE3-3-16r39-143(d) Light and SoundPaper 2PMedium6 marks
A robot arm aims a laser at a rectangular glass block of refractive index 1.52 so that the refracted ray inside the block makes an angle of 20.0° with the normal. Use n = sin i ÷ sin r.
Diagram not to scale
(a)
Calculate the angle of incidence at the first face. Give your answer to 1 decimal place.
(3)
(b)
The robot increases the angle of incidence. State what happens to the angle of refraction.
(1)
(c)
The ray leaves the block through the opposite parallel face. State the angle the emerging ray makes with the normal.
(1)
(d)
State how the direction of the emerging ray compares with the direction of the incident ray.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sin i = n × sin r
✓ 1
M1 sin i = 1.52 × sin 20.0° = 0.520
✓ 1
A1 31.3°
✓ 1
Part (b)
it increases
✓ 1
Part (c)
31.3° (the same as the angle of incidence) (ECF from (a))
✓ 1
Part (d)
parallel to it (but sideways displaced)
✓ 1
Exam tip: Light leaving a parallel-sided block emerges parallel to its original direction.
241IGPE3-3-15r43-063(d) Light and SoundPaper 2PMedium5 marks
A farm uses an unclad plastic optical fibre to carry light from a sensor in a grain store to a counter. The plastic has refractive index 1.48 and is surrounded by air.
(a)
Calculate the critical angle of the plastic. Give your answer to 3 significant figures.
(2)
(b)
Inside a straight section of the fibre, a ray meets the wall at an angle of incidence of 44°. Evaluate whether this ray stays inside the fibre.
(2)
(c)
Dust sticks to the outside wall of the unclad fibre. Suggest why this reduces the light reaching the counter.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of sin c = 1/n or sin c = 1/1.48
✓ 1
A1 42.5°
✓ 1
Part (b)
44° is greater than the critical angle (42.5°), ECF from (a)
✓ 1
so the ray is totally internally reflected and stays inside the fibre
✓ 1
Part (c)
total internal reflection no longer occurs where the dust touches the fibre, so light escapes or is absorbed by the dust (accept scattered)
✓ 1
Common error: Compare the angle of incidence with the critical angle before deciding; the ray stays in only if it is larger.
243IGCSE-PHY-3-0193(d) Light and SoundPaper 2 onlyGrade 8–98 marks
A student investigates the speed of sound in air. A hammer strikes a metal plate. Microphone A is next to the plate and microphone B is a distance d from microphone A. An electronic timer starts when the sound reaches microphone A and stops when it reaches microphone B. Distance d / m: 0.50, 1.00, 1.50, 2.00, 2.50 Time t / ms: 1.7, 3.1, 4.6, 6.0, 7.5 The graph shows the results with a line of best fit.
(a)
Explain why the student uses an electronic timer rather than a stopwatch.
(1)
(b)
Determine the gradient of the line of best fit, in ms per m.
(2)
(c)
Use your answer to (b) to calculate the speed of sound in air.
(2)
(d)
The line of best fit does not pass through the origin. This is because the timer has a small, constant delay. Explain why this delay does not affect the value of the speed of sound found from the gradient.
(2)
(e)
State one variable that should be kept constant during the investigation.
(1)
Show mark scheme
(a)
times are very short / a few ms, much shorter than human reaction time;
Do not accept
it is more accurate (no reason)
(b)
large triangle used / two points read from the line e.g. (0, 0.23) and (3.0, 8.9);
gradient in range 2.8 to 3.0 (ms/m);
Allow
use of data points on the line
(c)
speed = 1 ÷ gradient with conversion of ms to s, e.g. 1 ÷ 0.0029;
evaluation: 333 to 357 (m/s);
e.g. (v =) 345 (m/s)
Allow
ecf from (b)
0.34 (m/s) (ms not converted) scores 1
(d)
same (constant) extra time added to every reading / line shifted up (intercept changes);
so the gradient / change in t ÷ change in d is unchanged;
Do not accept
it is a random error
the delay is too small to matter
(e)
any one from: (air) temperature; wind speed / direction; same plate and hammer; humidity;
244IGCSE-PHY-3-0203(d) Light and SoundPaper 2 onlyGrade 8–99 marks
A bat finds insects by emitting short pulses of sound of frequency 45 kHz and listening for the echoes. The speed of sound in air is 340 m/s.
(a)
Explain why humans cannot hear the sound from the bat.
(1)
(b)
An echo from a moth returns to the bat 8.0 ms after the pulse is emitted. Calculate the distance between the bat and the moth.
(3)
(c)
When searching, the bat emits 25 pulses per second. Each echo must return before the next pulse is emitted. Calculate the greatest distance at which the bat can detect an object.
(3)
(d)
As the bat closes in on the moth, it increases its rate to 200 pulses per second. Explain why the bat can use this higher pulse rate at short range and why this helps it to catch the moth.
(2)
Show mark scheme
(a)
45 kHz / 45 000 Hz is above 20 000 Hz (upper limit of human hearing);
Do not accept
the sound is too quiet
(b)
substitution into distance = speed × time with t = 8.0 × 10⁻³ s;
division by 2 (sound travels there and back);
evaluation: 1.36 (m);
e.g. 340 × 0.0080 = 2.72 m, 2.72 ÷ 2
Allow
1.4 m
2.72 (m) (not halved) scores 2
Do not accept
2720 m (no ms conversion and no halving)
(c)
time between pulses T = 1 ÷ 25 = 0.040 s;
(maximum) total path = 340 × 0.040 = 13.6 m;
greatest distance = 13.6 ÷ 2 = 6.8 (m);
Allow
ecf for halving if already penalised in (b)
13.6 (m) as final answer scores 2
(d)
echo time is shorter (when the moth is close) so the next pulse can be sent sooner without overlapping the echo;
more frequent information about the moth's (changing) position;
Allow
at 200 pulses per second the maximum range is 0.85 m, which is enough at close range
245IGCSE-PHY-3-0393(d) Light and SoundPaper 2 onlyGrade 8–96 marks
A student is given the following apparatus: • a signal generator and a loudspeaker • two microphones • a dual-trace oscilloscope, which displays the signals from both microphones at the same time • a metre rule.
(a)
Describe how the student could use this apparatus to determine the speed of sound in air. You may draw a diagram to help your answer.
(6)
Show mark scheme
(a)
Indicative content:
MP1. connect loudspeaker to signal generator set to a known frequency (e.g. 1–4 kHz);
MP2. connect one microphone to each input of the oscilloscope;
MP3. place microphones side by side in front of / in line with the loudspeaker so the traces are in step (in phase);
MP4. move one microphone away from the loudspeaker (along the line) until the traces are in step again;
MP5. distance moved (measured with the metre rule) = one wavelength;
MP6. move through several wavelengths and divide by the number to reduce uncertainty;
MP7. calculate speed using v = f × λ;
MP8. frequency can be checked from the oscilloscope: period from time-base, f = 1 ÷ T;
MP9. repeat and calculate a mean / repeat at different frequencies;
MP10. avoid reflections (work away from walls / in open space);
Allow
award 1 mark per MP to a maximum of 6, but use the levels for the overall mark
a labelled diagram may credit MP1–MP3
Do not accept
timing the sound between the microphones with a stopwatch
measuring the wavelength from the oscilloscope screen in divisions
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated relevant steps, e.g. connecting the apparatus or stating v = fλ; no workable way of finding the wavelength.
Level 2 (3–4)
A workable method: microphones moved apart until the traces match again to find one wavelength, with frequency known and v = fλ; some detail missing.
Level 3 (5–6)
A detailed, logically sequenced method including the in-step condition, measuring the distance over several wavelengths, v = fλ and a valid improvement (repeats, mean, avoiding reflections).
Model sketch
Moving M2 until the two traces are in phase again (shift of one wavelength) gives the wavelength; then v = f × λ. Diagram not to scale
246IGCSE-PHY-3-0403(d) Light and SoundPaper 2 onlyGrade 8–99 marks
A student connects a microphone to an oscilloscope. A loudspeaker connected to a signal generator produces two sounds, P and Q, one after the other. For each sound she adjusts the oscilloscope settings so that the trace fills the screen. The two traces look exactly the same, as shown in the diagram. The table shows the oscilloscope settings used for each sound. Sound / name: P, Q Time-base / ms per division: 1.0, 0.020 Y-gain / mV per division: 5.0, 20 The student says: 'Sounds P and Q have the same pitch and the same loudness, because the traces look the same.'
(a)
Calculate the frequency of sound P and the frequency of sound Q.
(4)
(b)
Evaluate the student's statement about pitch. Use your answers to (a).
(3)
(c)
Evaluate the student's statement about loudness.
(2)
Show mark scheme
(a)
period = 2.0 divisions (from trace);
(T_P =) 2.0 ms and f_P = 500 (Hz);
(T_Q =) 0.040 ms / 4.0 × 10⁻⁵ s;
f_Q = 25 000 (Hz);
Allow
ecf from period read
25 kHz
0.5 (Hz) and 25 (Hz) (ms not converted) scores 2
Do not accept
1000 (Hz) for P (period read as one division)
(b)
statement is incorrect / pitch is not the same;
Q has a (much) higher frequency (so higher pitch) / 50 times the frequency of P;
25 000 Hz is above 20 000 Hz so Q cannot be heard (by humans);
Allow
ecf from (a)
Do not accept
Q has the same pitch because the waves have the same spacing
(c)
amplitude of P = 15 mV and amplitude of Q = 60 mV / Q has 4 times the amplitude;
(so) the sound wave reaching the microphone has a larger amplitude for Q / loudness not the same / Q louder (if it could be heard);
Allow
peak-to-peak values 30 mV and 120 mV
loudspeaker cone vibrates with a larger amplitude for Q for MP2
247IGPE3-3-14r21-053(d) Light and SoundPaper 2PHard8 marks
A plastic optical fibre has no cladding and is surrounded by air. The refractive index of the plastic is 1.50. A student says that a ray of light entering the flat end of the fibre at an angle of incidence of 70° will escape through the side wall. The ray enters in a plane containing the axis of the fibre.
Diagram not to scale
(a)
Calculate the critical angle for the plastic-to-air boundary. Give your answer to 1 decimal place.
(2)
(b)
Calculate the angle of refraction when the ray enters the end of the fibre. Give your answer to 1 decimal place.
(3)
(c)
Evaluate the student's statement. Use the value of r from part (b) rounded to 1 decimal place. Your answer should include the angle of incidence of the ray at the side wall.
(2)
(d)
Suggest one reason why real fibres have a cladding even though total internal reflection can occur at a plastic-to-air boundary.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sin c = 1/1.50
✓ 1
A1 41.8°
✓ 1
Part (b)
M1 uses n = sin i / sin r
✓ 1
M1 sin r = sin 70° ÷ 1.50 = 0.6265
✓ 1
A1 38.8°
✓ 1
Part (c)
A1 angle at the side wall = 90° − 38.8° = 51.2° (ECF from (b))
✓ 1
B1 51.2° is greater than the critical angle 41.8°, so the ray is totally internally reflected and stays in the fibre; the statement is wrong
✓ 1
Part (d)
B1 protects the surface from scratches, dirt or moisture (which would let light escape), or stops light leaking when fibres touch
✓ 1
Common error: The angle used for total internal reflection is measured from the normal to the side wall, so subtract the angle of refraction from 90°.
248IGPE3-3-14r21-063(b) Properties of WavesPaper 2PHard6 marks
A seismometer detects P-waves from an earthquake 30 s before it detects S-waves. P-waves travel through the crust at 8.0 km/s and S-waves at 5.0 km/s. Assume both travel in a straight line at constant speed.
(a)
State why S-waves cannot travel through the liquid outer core of the Earth.
(1)
(b)
Calculate the distance from the earthquake to the seismometer.
(3)
(c)
Explain why records from at least three seismometers are needed to find the position of the earthquake.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 S-waves are transverse and need a solid (liquids cannot support shear)
✓ 1
Part (b)
M1 time for P = d/8.0 and time for S = d/5.0
✓ 1
M1 d/5.0 − d/8.0 = 30
✓ 1
A1 400 km
✓ 1
Part (c)
B1 each record gives only a distance, so the epicentre lies somewhere on a circle around that station
✓ 1
B1 two circles meet at two points, so a third circle is needed to give a single position
✓ 1
Exam tip: Write the time for each wave in terms of the same distance, then use the given time difference to form an equation.
249IGPE3-3-14r22-083(d) Light and SoundPaper 2PHard8 marks
A bicycle reflector contains a glass prism with a right-angled corner. A ray of light enters one face along the normal, and then meets the sloping face of the glass at an angle of incidence of 45°. The refractive index of the glass is 1.52, and the refractive index n of a material at a boundary with air is related to the critical angle c by n = 1 ÷ sin c.
Diagram not to scale
(a)
State what is meant by the critical angle.
(1)
(b)
Calculate the critical angle for the glass–air boundary. Give your answer to 3 significant figures.
(2)
(c)
Explain why the ray is totally internally reflected at the sloping face.
(2)
(d)
The reflector is covered in water of refractive index 1.33. The critical angle at a glass–water boundary is given by sin c = 1.33 ÷ 1.52. Calculate this critical angle, and state whether the ray is still totally internally reflected.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 the angle of incidence in the denser medium above which all light is reflected (the angle at which the refracted ray travels along the boundary)
✓ 1
Part (b)
M1 sin c = 1 ÷ 1.52
✓ 1
A1 41.1° (41.14...)
✓ 1
Part (c)
B1 the light is travelling from glass to air (a denser to a less dense medium)
✓ 1
B1 the angle of incidence, 45°, is greater than the critical angle, 41.1°
✓ 1
Part (d)
M1 sin c = 1.33 ÷ 1.52
✓ 1
A1 61.0° (61.04...)
✓ 1
B1 no, because 45° is less than 61.0°, so the ray is partly refracted out
✓ 1
Exam tip: Always compare the angle of incidence with the critical angle in your answer; quoting both numbers earns the mark.
250IGPE3-3-15r22-043(d) Light and SoundPaper 2PHard1 mark
Gem A has refractive index 1.54 and gem B has refractive index 2.42. Inside each gem, a ray of light strikes a flat face that borders air. In both cases the angle of incidence is 30°. What happens to the rays?
Show mark scheme
Marking point
Mark
Notes
The critical angle is given by sin c = 1/n. For gem A, c is about 40.5°; for gem B, c is about 24.4°. A ray in gem A at 30° is below its critical angle, so some light leaves. A ray in gem B at 30° is above its critical angle, so it is totally internally reflected, and that makes the second statement correct. The first and fourth statements ignore one of the two critical angles. The third statement reverses the link between refractive index and critical angle: a higher refractive index gives a smaller critical angle.
✓ 1
Answer B
Answer: B
Common error: A larger refractive index gives a smaller critical angle, which is why high-index gems trap light more easily.
251IGPE3-3-15r22-103(d) Light and SoundPaper 2PHard10 marks
A diamond has a refractive index of 2.42. Jewellers cut diamonds so that light entering the top is reflected inside the stone and leaves through the top again.
(a)
State the two conditions needed for total internal reflection to occur.
(2)
(b)
Calculate the critical angle for light travelling from diamond to air.
(3)
(c)
A ray inside the diamond meets a surface with air at an angle of incidence of 35°. State and explain what happens to the ray.
(2)
(d)
Glass with a refractive index of 1.52 has a critical angle of 41.1°. Explain why a cut diamond sparkles more than a piece of glass cut in the same shape.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
light is travelling 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
sin c = 1 ÷ 2.42
✓ 1
c = 24.4°
✓ 1
Part (c)
35° is greater than the critical angle (24.4°) (ECF from (b))
✓ 1
so the ray is totally internally reflected
✓ 1
Part (d)
the critical angle for diamond is smaller
✓ 1
so more of the rays inside meet the surfaces at angles above the critical angle and are totally internally reflected
✓ 1
so more light is reflected several times inside and emerges through the top faces, instead of leaking out of the sides and bottom
✓ 1
Exam tip: Always compare the angle of incidence with the critical angle in words before stating that total internal reflection happens.
252IGPE3-3-14r26-073(d) Light and SoundPaper 2PHard9 marks
An endoscope is used to look inside a patient's stomach. It contains bundles of optical fibres. The glass core of each fibre has refractive index 1.50 and is surrounded by plastic cladding of lower refractive index.
(a)
State two conditions needed for total internal reflection at a boundary.
(2)
(b)
Calculate the critical angle for light travelling from this glass into air. Give your answer to 3 significant figures.
(3)
(c)
A ray strikes the boundary between the core and the cladding at an angle of incidence of 62°. The critical angle at this boundary is 73°. State whether the ray is totally internally reflected and explain your answer.
(2)
(d)
Describe how an endoscope uses two bundles of fibres to show the inside of the stomach.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The light travels from a higher refractive index (more optically dense) material towards a lower one
✓ 1
The angle of incidence is greater than the critical angle
✓ 1
Part (b)
M1 sin c = 1/n
✓ 1
M1 c = sin−1(1 ÷ 1.50)
✓ 1
A1 41.8°
✓ 1
Part (c)
No (the ray is not totally internally reflected)
✓ 1
The angle of incidence (62°) is smaller than the critical angle (73°), so some light refracts into the cladding
✓ 1
Part (d)
One bundle carries light into the stomach by total internal reflection to illuminate it
✓ 1
Light reflected from the inside of the stomach returns along a second bundle to form the image seen by the doctor
✓ 1
Common error: Total internal reflection can only happen when light tries to leave the denser material; it cannot happen going from a less dense to a denser material.
253IGPE3-3-14r26-103(d) Light and SoundPaper 2PHard8 marks
Engineers use ultrasound to check a wind-farm turbine blade. A transmitter sends a short pulse into the blade and a detector beside it records echoes. The speed of ultrasound in the blade material is 3200 m/s. An echo from a crack returns after 15.5 µs. A second echo, from the back surface of the blade, returns after 24.0 µs. (1 µs = 1 × 10−6 s)
Diagram not to scale
(a)
State what is meant by ultrasound.
(1)
(b)
Explain why an echo is produced by the crack.
(1)
(c)
Calculate the thickness of the blade. Give your answer in metres to 3 significant figures.
(3)
(d)
Calculate the depth of the crack below the front surface of the blade. Give your answer in millimetres to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Sound with a frequency above 20 000 Hz (above the range of human hearing)
✓ 1
Part (b)
Part of the pulse is reflected at a boundary between two different materials (blade material and air in the crack)
254IGPE3-3-16r28-103(d) Light and SoundPaper 2PHard7 marks
A surveyor stands on a river bank near a bridge and claps. The echo from a pier of the bridge returns after 0.50 s. The speed of sound in air is 340 m/s. The bridge engineer also uses ultrasound pulses to look for flaws in steel beams.
(a)
Calculate the distance from the surveyor to the pier.
(3)
(b)
State what is meant by ultrasound.
(1)
(c)
An ultrasound pulse sent into a steel beam of thickness 0.060 m reflects from the far side and returns after 24 μs. Calculate the speed of the pulse in steel.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 distance travelled by sound = 340 × 0.50 = 170 m
✓ 1
M1 halve, because the sound travels to the pier and back
✓ 1
A1 85 m
✓ 1
Part (b)
Sound with a frequency above 20000 Hz (above the range of human hearing)
✓ 1
Part (c)
M1 distance travelled = 2 × 0.060 = 0.12 m
✓ 1
M1 speed = 0.12 ÷ (24 × 10−6)
✓ 1
A1 5000 m/s
✓ 1
Exam tip: For echo questions, the wave covers the distance twice, so use twice the thickness or half the time.
255IGPE3-3-16r34-123(d) Light and SoundPaper 2PHard9 marks
A submarine periscope uses two right-angled glass prisms to turn light through 90° by total internal reflection. Light enters one face of a prism along the normal and strikes the long face at an angle of incidence of 45°. The glass has a refractive index of 1.52.
Diagram not to scale
(a)
State why the light is not refracted as it enters the prism through the first face.
(1)
(b)
Calculate the critical angle of the glass. Give your answer to 3 significant figures.
(3)
(c)
Explain why the light is totally internally reflected at the long face.
(2)
(d)
A designer suggests using a plastic with a refractive index of 1.35 instead. Calculate the critical angle of this plastic and explain why it would not work in the periscope. Give your angle to 3 significant figures.
(3)
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Notes
Part (a)
it enters along the normal, so the angle of incidence is 0° and there is no change of direction
✓ 1
Part (b)
sin c = 1 ÷ n
✓ 1
sin c = 1 ÷ 1.52 = 0.658
✓ 1
c = 41.1°
✓ 1
Part (c)
the angle of incidence (45°) is greater than the critical angle (41.1°, ECF from (b))
✓ 1
so all the light is reflected inside the prism
✓ 1
Part (d)
c = asin(1 ÷ 1.35) = 47.8°
✓ 1
45° is smaller than 47.8°, so the angle of incidence is less than the critical angle
✓ 1
some light is refracted out of the prism, so total internal reflection does not occur
✓ 1
Exam tip: A lower refractive index gives a larger critical angle, so total internal reflection needs a larger angle of incidence.
257IGPE3-3-16r35-113(d) Light and SoundPaper 2PHard9 marks
A lamp at the bottom of a swimming pool shines light upwards at the water surface. The refractive index of water is 1.33.
Diagram not to scale
(a)
A ray meets the surface at an angle of incidence of 30° in the water and emerges into air. State and explain what happens to the direction of the ray as it leaves the water.
(2)
(b)
Calculate the angle between the emerging ray and the normal. Give your answer to 3 significant figures.
(3)
(c)
Calculate the critical angle for the water-air boundary. Give your answer to the nearest degree.
(2)
(d)
A different ray reaches the surface at an angle of incidence of 60°. Explain why this ray does not emerge into the air.
(2)
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Mark
Notes
Part (a)
B1 it bends away from the normal
✓ 1
B1 light speeds up as it passes from water into air
✓ 1
Part (b)
M1 n = sin(angle in air) ÷ sin(angle in water)
✓ 1
M1 sin r = 1.33 × sin 30° = 0.665
✓ 1
A1 r = 41.7°
✓ 1
Part (c)
M1 sin c = 1 ÷ n
✓ 1
A1 c = sin−1(1 ÷ 1.33) = 49°
✓ 1
Part (d)
B1 60° is greater than the critical angle (ECF from (c))
✓ 1
B1 so the ray is totally internally reflected at the surface
✓ 1
Exam tip: Use sin c = 1/n only for light going from the denser medium to air. For a ray leaving the material, n = sin(angle in air) ÷ sin(angle in material).
258IGPE3-3-15r38-113(d) Light and SoundPaper 2PHard6 marks
A decorative sign in a library uses a plastic optical fibre. The core of the fibre has a refractive index of 1.50. Use n = 1 ÷ sin c, where c is the critical angle.
(a)
State two conditions needed for total internal reflection to take place.
(2)
(b)
Calculate the critical angle for the boundary between the core and air. Give your answer to 3 significant figures.
(2)
(c)
A ray inside the core meets the core–air boundary at an angle of incidence of 55°. State what happens to the ray and justify your answer.
(2)
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Notes
Part (a)
light travels from a more dense medium towards a less dense medium
✓ 1
the angle of incidence is greater than the critical angle
✓ 1
Part (b)
M1 sin c = 1 ÷ 1.50
✓ 1
A1 41.8°
✓ 1
Part (c)
55° is greater than the critical angle of 41.8° (ECF from (b))
✓ 1
so the ray is totally internally reflected and stays in the fibre
✓ 1
Common error: The critical angle is measured from the normal inside the denser material; check you use degrees mode when finding the inverse sine.
259IGPE3-3-15r39-143(d) Light and SoundPaper 2PHard7 marks
A sealed probe in a lake shines a beam of light upwards towards the surface of the water. In the water, the beam makes an angle of 20.0° with the normal to the surface. The refractive index of the water is 1.33. For light travelling from water into air, n = sin (angle in air) ÷ sin (angle in water).
Diagram not to scale
(a)
State why the beam bends away from the normal as it leaves the water.
(1)
(b)
Calculate the angle the emerging beam makes with the normal. Give your answer to 1 decimal place.
(3)
(c)
Calculate the critical angle for water. Give your answer to the nearest degree.
(2)
(d)
State what happens to a beam that meets the surface from below at an angle of 60°.
(1)
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Marking point
Mark
Notes
Part (a)
light speeds up in the air / goes from a more dense to a less dense medium
✓ 1
Part (b)
M1 sin (angle in air) = n × sin 20.0°
✓ 1
M1 = 1.33 × 0.342 = 0.455
✓ 1
A1 27.1°
✓ 1
Part (c)
M1 sin c = 1 ÷ 1.33
✓ 1
A1 48.75° which is 49°
✓ 1
Part (d)
total internal reflection
✓ 1
Exam tip: Total internal reflection needs the light to be in the denser medium and the angle of incidence to be greater than the critical angle.
260IGPE3-3-16r39-123(b) Properties of WavesPaper 2PHard7 marks
A bat flies directly towards a stationary moth at a constant speed of 6.0 m/s. At the moment the bat sends a pulse of ultrasound, the moth is 4.0 m away. The speed of sound in air is 340 m/s.
(a)
Calculate the time the pulse would take to reach the moth if the bat were stationary. Give your answer to 2 significant figures.
(2)
(b)
The echo reaches the bat while it is still moving. Together, the sound and the bat cover the distance there and back (2 × 4.0 m) in the time the echo takes to return. Calculate the time between the pulse being sent and the echo being received. Give your answer to 3 significant figures.
(3)
(c)
The bat sends 40 pulses each second. Determine whether the echo from this pulse is received before the next pulse is sent.
(2)
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Notes
Part (a)
M1 t = d ÷ v = 4.0 ÷ 340
✓ 1
A1 0.012 s
✓ 1
Part (b)
M1 sound distance + bat distance = 8.0 m, so (340 + 6.0) × t = 8.0
✓ 1
M1 t = 8.0 ÷ 346
✓ 1
A1 0.0231 s
✓ 1
Part (c)
time between pulses = 1 ÷ 40 = 0.025 s
✓ 1
0.0231 s is less than 0.025 s so the echo returns first (ECF from (b))
✓ 1
Common error: The bat moves while the sound is in flight, so the sound does not cover the full 8.0 m on its own.
261IGPE3-3-15r43-043(d) Light and SoundPaper 2PHard1 mark
A plastic fibre in an aircraft cabin has no cladding and is surrounded by air. The plastic has refractive index 1.60. A ray inside the fibre meets the side wall at an angle of incidence of 35°. Which statement describes what happens?
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Mark
Notes
The critical angle is given by sin c = 1/1.60, so c = 38.7°. The angle of incidence (35°) is less than this, so there is no total internal reflection: part of the light refracts out and part is reflected, making D correct. A needs an angle above 38.7°. B and C are wrong because at an ordinary boundary a refracted ray and a reflected ray both occur.
✓ 1
Answer D
Answer: D
Common error: Below the critical angle the ray is partly reflected and partly refracted; it is not all lost.
262IGPE3-3-15r43-093(d) Light and SoundPaper 2PHard6 marks
Light is carried along an unclad plastic fibre in an aircraft cabin. The fibre has a flat end face at right angles to its length, refractive index 1.50, and is surrounded by air. A ray in the air strikes the end face with an angle of incidence of 40°. Give all answers to 3 significant figures.
Diagram not to scale
(a)
Calculate the angle of refraction of the ray as it enters the fibre.
(2)
(b)
The side wall of the fibre is at right angles to the end face. Calculate the angle of incidence of the ray at the side wall.
(1)
(c)
Calculate the critical angle of the plastic.
(2)
(d)
Evaluate whether the ray stays inside the fibre at the side wall.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sin r = sin 40° ÷ 1.50
✓ 1
A1 25.4°
✓ 1
Part (b)
A1 90° − 25.4° = 64.6° (ECF from (a))
✓ 1
Part (c)
M1 sin c = 1/1.50
✓ 1
A1 41.8°
✓ 1
Part (d)
64.6° is greater than 41.8°, so it is totally internally reflected and stays inside (ECF)
✓ 1
Common error: The angle at the side wall is measured from the normal, so it is 90° minus the angle between the ray and the fibre axis.
263IGPE3-3-16r43-143(d) Light and SoundPaper 2PHard6 marks
An archaeologist's lamp rests on the seabed. The refractive index of the water is 1.33. Rays from the lamp meet the water surface at different angles of incidence.
(a)
Calculate the critical angle for light passing from the water to the air. Give your answer to the nearest degree.
(2)
(b)
A ray meets the surface at an angle of incidence of 40°. Calculate the angle of refraction in the air. Give your answer to the nearest degree.
(2)
(c)
Another ray meets the surface at an angle of incidence of 55°. Describe what happens to this ray.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sin c = 1/1.33
✓ 1
A1 49°
✓ 1
Part (b)
M1 sin r = 1.33 × sin 40°
✓ 1
A1 59°
✓ 1
Part (c)
it does not pass into the air, because 55° is greater than the critical angle (total internal reflection)
✓ 1
it is reflected back into the water at an angle of 55° to the normal
✓ 1
Common error: For light leaving water, sin r = n × sin i; a value above 1 would mean there is no refracted ray.
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