International GCSE Physics · Pearson Edexcel 4PH1 · Topic 3
Waves
Original exam-style questions with mark schemes. Questions marked “Paper 2 only” use content assessed only on Paper 2. Notes 3(b) →Notes 3(c) →Notes 3(d) →
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Original questions in the style of each exam boardEvery answer re-solved independently before publishingCreated by Evan Kotronis with AI assistanceUpdated 13 October 2026 · new questions added regularlyHow every question is checked →
1IGCSE-PHY-3-001Order of 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-3-003Uses and dangers of electromagnetic radiationPapers 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-3-005Ocean waves at a buoy: period, frequency and speedPapers 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-3-021Reading amplitude and wavelength from a displacement–distance graphPapers 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-3-022Comparing red and violet light in spacePapers 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-3-023Electromagnetic radiation in a food factoryPapers 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-3-024The stadium 'Mexican wave' as a transverse wavePapers 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-3-025Reflection of light in a periscopePapers 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
11IGCSE-PHY-3-006Total internal reflection at a water surfacePapers 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
12IGCSE-PHY-3-007Radio and satellite communication with EM wavesPapers 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;
13IGCSE-PHY-3-008Practical: refractive index of a rectangular glass blockPapers 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)
14IGCSE-PHY-3-009Total internal reflection in a reflector prismPapers 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;
15IGCSE-PHY-3-010Compressions and rarefactions in front of a loudspeakerPapers 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)
16IGCSE-PHY-3-011Refraction of water waves in a ripple tankPapers 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)
17IGCSE-PHY-3-026Angle of incidence needed for a given angle of refraction in waterPapers 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
18IGCSE-PHY-3-027Doppler effect: frequency recorded as a drone flies pastPapers 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
19IGCSE-PHY-3-028Practical: tracing a ray through a triangular prismPapers 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
20IGCSE-PHY-3-029Law of reflection in a laser security alarmPapers 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;
21IGCSE-PHY-3-030Identifying electromagnetic radiations in a smart home from wavelength dataPapers 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
22IGCSE-PHY-3-031How an endoscope uses total internal reflectionPapers 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.
23IGCSE-PHY-3-012Practical: critical angle with a semicircular blockPapers 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.
24IGCSE-PHY-3-013Locating an earthquake from P-wave and S-wave arrival timesPapers 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);
25IGCSE-PHY-3-014Evaluating whether light stays inside an unclad plastic fibrePapers 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;
26IGCSE-PHY-3-015Doppler effect for water waves from a moving toy boatPapers 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;
27IGCSE-PHY-3-032Identifying a gemstone and explaining why diamonds sparklePapers 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, 2.42
(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)
28IGCSE-PHY-3-033Refraction of sound explains louder traffic noise at nightPapers 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.
29IGCSE-PHY-3-034Wave machine in a leisure-centre poolPapers 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
30IGCSE-PHY-3-035Practical: refractive index of a sugar solution from a graph of sin i against sin rPapers 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;
32IGCSE-PHY-3-017Practical: comparing two tuning forks with an oscilloscopePaper 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
33IGCSE-PHY-3-018High-frequency sound deterrent and the range of human hearingPaper 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);
34IGCSE-PHY-3-036Time period of a sound outside the range of human hearingPaper 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)
35IGCSE-PHY-3-037Practical: speed of sound by clapping in time with echoesPaper 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, 10.5
(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;
36IGCSE-PHY-3-038Practical: tuning a ukulele string with an oscilloscopePaper 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;
37IGCSE-PHY-3-019Practical: speed of sound from a time-delay graphPaper 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;
38IGCSE-PHY-3-020Bat echolocation: echo timing and pulse ratePaper 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
39IGCSE-PHY-3-039Practical: speed of sound using two microphones and a dual-trace oscilloscopePaper 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).
40IGCSE-PHY-3-040Evaluating a claim about two identical-looking oscilloscope tracesPaper 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
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