Reflection, sound waves, ultrasound and seismic waves (physics only, RP9) – GCSE Physics exam questions with mark scheme

1Grade 6–7HigherTriple only5 marks

A ship uses pulses of ultrasound to measure the depth of the sea. The echo of a pulse is detected 0.40 s after the pulse is sent. The speed of sound in sea water is 1500 m/s.

(a) Calculate the depth of the sea. [3]

(b) The ultrasound has a frequency of 30 kHz. Calculate its wavelength in sea water. [2]

(a) distance travelled by pulse = 1500 × 0.40 = 600 m (allow time for one-way journey = 0.40 ÷ 2 = 0.20 s)(1)
(a) pulse travels down and back, so depth = 600 ÷ 2 (or depth = 1500 × 0.20)(1)
(a) 300 m (a correct answer with no working scores 3 marks; 600 m scores 1 mark)(1)
(b) $1500 = 30\,000 \times \lambda$ or λ = 1500 ÷ 30 000 (30 kHz correctly converted to 30 000 Hz)(1)
(b) 0.050 m (allow 0.05 m or 5.0 cm; a correct answer with no working scores 2 marks)(1)
Answer: (a) 300 m (b) 0.050 m
Examiner tip: The echo time is for the journey down AND back — halve the distance.

2Grade 4–5Triple only3 marks

The diagram shows a ray of light hitting a plane mirror. The ray makes an angle of 35° with the surface of the mirror. Line X is drawn at 90° to the mirror where the ray hits it.

(a) Give the name of line X. [1]

(b) Determine the angle of reflection. [1]

(a) (the) normal(1)
(b) 55° (do not accept 35°)(1)
(c) 110° (allow ecf: 2 × their (b))(1)
Answer: (a) normal (b) 55° (c) 110°
Examiner tip: Angles of incidence and reflection are measured from the normal, not from the mirror, so 35° is the most common wrong answer.

3Grade 4–5Triple only4 marks

A student investigated the reflection of light by a plane mirror. She used a ray box to shine a narrow ray of light at the mirror and measured the angle of incidence and the angle of reflection with a protractor.

Angle of incidence in degrees	20	30	40	50	60
Angle of reflection in degrees	20	31	48	50	59

(a) Identify the anomalous result. [1]

(b) Write a conclusion for this investigation. [1]

(d) The student replaced the mirror with a sheet of white paper. She could not see a single reflected ray. Explain why. [1]

(a) angle of reflection of 48° / the result for an angle of incidence of 40°(1)
(b) the angle of reflection is equal to the angle of incidence (allow: i = r; do not accept 'they are proportional' alone)(1)
(c) the centre / direction of a narrow ray can be marked more accurately so the angles can be measured more accurately (allow: a wide beam makes it hard to tell where to measure from)(1)
(d) the rough surface reflects the light in many different directions (allow: diffuse reflection / the light is scattered; ignore 'the paper absorbs the light')(1)
Answer: (a) 48° (at i = 40°) (b) angle of reflection = angle of incidence (c) more accurate marking/measuring of the ray (d) diffuse reflection: rough surface scatters light in many directions
Examiner tip: A conclusion must state the relationship (angle of reflection = angle of incidence), not just describe the numbers in the table.

4Grade 6–7Triple only5 marks

A student investigated how light is refracted by different substances. She shone a ray of light into a rectangular glass block and then into a clear rectangular box filled with water. She measured the angle of refraction for different angles of incidence.

Angle of incidence in degrees	20	40	60
Angle of refraction in glass in degrees	13	25	35
Angle of refraction in water in degrees	15	29	41

(a) State the independent variable in this investigation. [1]

(b) Which substance changes the direction of the light ray more? Use data from the table to explain your answer. [2]

(c) Describe how the student could find the path of the ray inside the block, so that she can measure the angle of refraction. [2]

(a) angle of incidence (allow: the substance / type of material)(1)
(b) glass(1)
(b) for the same angle of incidence the angle of refraction is smaller in glass, e.g. at 40° glass 25° and water 29° (so the ray bends more towards the normal in glass); reference to data must be a matched pair(1)
(c) draw around the block and mark the path of the ray entering and leaving the block with crosses / dots (spaced well apart)(1)
(c) remove the block and join the point where the ray entered to the point where it left, then draw the normal at the entry point and measure the angle between the ray and the normal with a protractor(1)
Answer: (a) angle of incidence (b) glass: smaller angle of refraction for the same angle of incidence, e.g. 25° vs 29° at 40° (c) trace the block, mark incident and emerging rays, join entry and exit points, measure from the normal
Examiner tip: Comparing angles is only valid at the same angle of incidence, so quote one matched pair from the table.

5Grade 6–7HigherTriple only5 marks

An ultrasound probe is placed on a patient's leg. The probe sends a short pulse of ultrasound into the leg and detects the echoes.

The diagram shows the pulse sent and two echoes. Echo A comes from the boundary between fat and muscle. Echo B comes from the boundary between muscle and bone.

(a) Explain why echoes are detected from both boundaries. [2]

(b) The speed of ultrasound in muscle is 1500 m/s. Calculate the thickness of the muscle layer. [3]

(a) ultrasound is partially reflected at a boundary between two different media / tissues (allow: some of the ultrasound is reflected)(1)
(a) the rest of the ultrasound is transmitted (through the first boundary) so it can reach the second boundary and be reflected there(1)
(b) time between echoes = 80 − 20 = 60 µs (allow: 6.0 × 10⁻⁵ s)(1)
(b) distance = 1500 × 30 × 10⁻⁶ (i.e. using half the time) or 1500 × 60 × 10⁻⁶ = 0.090 m then ÷ 2(1)
(b) 0.045 m (allow: 4.5 cm; correct answer with no working scores 3; 0.090 m scores 2)(1)
Answer: (a) partial reflection at each boundary, rest transmitted to the next (b) 0.045 m (4.5 cm)
Examiner tip: The echo time is for the journey there and back, so halve the time (or the distance) before giving the thickness.

6Grade 6–7HigherTriple only6 marks

When an earthquake happens, P-waves and S-waves travel through the Earth. Seismometers all over the Earth's surface record which waves arrive.

The diagram shows where P-waves and S-waves are not detected after one earthquake.

Explain how these observations give evidence about the structure of the Earth's core. [6]

P-waves are longitudinal; S-waves are transverse(1)
P-waves can travel through solids and liquids(1)
S-waves can travel through solids but not through liquids(1)
S-waves are detected only up to about 103° from the earthquake(1)
no S-waves are detected on the opposite side of the Earth (a large S-wave shadow zone)(1)
so S-waves cannot pass through the core, which shows (part of) the core is liquid(1)
P-waves are detected on the far side, so they pass through the core(1)
P-waves change speed / direction (refract) at the boundary of the core, so there are regions where no P-waves are detected(1)
the angle at which the shadow zones start can be used to work out the size of the core(1)
waves follow curved paths because their speed changes gradually with depth (density changes)(1)

Level	Marks	What it looks like
Level 3	5–6	Links the properties of both wave types to the observations in the diagram to give a full, logical explanation that the core is (partly) liquid, including how the shadow zones arise and what they reveal about the core.
Level 2	3–4	Gives relevant properties of P- and S-waves and links at least one of them to an observation, e.g. no S-waves on the far side so the core is liquid; the reasoning may have gaps.
Level 1	1–2	Makes simple relevant statements about P- or S-waves or the observations, with little or no link to the structure of the core.
No relevant content	0	Nothing worthy of credit.

Answer: S-waves (transverse) cannot travel through liquids and are absent beyond about 103°, so the core is (at least partly) liquid; P-waves (longitudinal) travel through the core but are refracted at its boundary, giving P shadow zones; the size of the shadow zones gives the size of the core.
Examiner tip: Link each observation to a wave property: 'S-waves can't travel through liquids' only scores when it is used to conclude that the core is liquid.

7Grade 8–9HigherTriple only5 marks

An earthquake produces P-waves and S-waves. A seismometer some distance away records both types of wave.

Average speed of P-waves = 8.0 km/s
Average speed of S-waves = 4.5 km/s

(a) Which type of wave is recorded first? Give a reason for your answer. [1]

(b) The S-waves arrive 70 s after the P-waves. Calculate the distance from the earthquake to the seismometer. Assume both waves travel the same distance. [4]

(a) P-waves, because they travel faster than S-waves (reason needed for the mark)(1)
(b) time for P-waves = d ÷ 8.0 and time for S-waves = d ÷ 4.5 (allow d in km, t in s)(1)
(b) d/4.5 − d/8.0 = 70(1)
(b) d × (1/4.5 − 1/8.0) = 70 or d = 70 ÷ 0.0972 (allow: correct trial with a distance giving a 70 s gap)(1)
(b) 720 km (allow: 7.2 × 10⁵ m)(1)
Answer: (a) P-waves (faster) (b) 720 km
Examiner tip: The 70 s is the DIFFERENCE between two travel times, so you need an equation in d, not 70 × (8.0 − 4.5).

8Grade 1–3FoundationTriple only3 marks

When light waves reach the boundary of a material, the waves can be absorbed, reflected or transmitted.

Choose the correct word from the box for each situation.

absorbed | reflected | transmitted

(a) Sunlight passes through a clear glass window into a room. [1]

(b) A person sees their own face in a bathroom mirror. [1]

(a) transmitted (allow: transmission)(1)
(b) reflected (allow: reflection)(1)
(c) absorbed (allow: absorption; ignore reference to infrared or heat)(1)
Answer: (a) transmitted (b) reflected (c) absorbed
Examiner tip: Transmitted means the wave passes through the material; do not confuse it with 'transferred' or 'refracted', which are not in the box.

9Grade 4–5HigherTriple only4 marks

A loudspeaker produces a sound wave in air. The sound reaches a person's ear.

(a) Give the range of frequencies of normal human hearing. [1]

(b) Describe how the sound wave makes the person's ear drum vibrate. [2]

(c) A dog-training whistle produces a sound of frequency 25 kHz. A dog reacts to the whistle but the person hears nothing. Suggest why the person cannot hear the whistle. [1]

(a) 20 Hz to 20 kHz (allow: 20 Hz to 20 000 Hz; both limits with units needed)(1)
(b) the sound wave is a series of compressions and rarefactions (allow: pressure changes / air particles vibrating) that reach the ear drum(1)
(b) these make the ear drum (a solid) vibrate back and forth at the same frequency as the sound(1)
(c) 25 kHz is above the upper limit of human hearing (20 kHz) / the whistle produces ultrasound (allow: the human ear can only convert sound into vibrations over a limited range of frequencies / the ear drum cannot vibrate at this frequency)(1)
Answer: (a) 20 Hz–20 kHz (b) compressions/rarefactions (pressure variations) push the ear drum back and forth at the frequency of the sound (c) 25 kHz is above the upper limit of human hearing (ultrasound)
Examiner tip: Give both limits of the hearing range with units, and say the ear drum vibrates at the SAME frequency as the sound.

10Grade 8–9HigherTriple only7 marks

Ultrasound is used to check a metal block for hidden cracks. The block is 0.24 m thick.

A pulse of ultrasound is sent into the top of the block. An echo from the back surface of the block is detected 80 µs after the pulse is sent. Another echo, from a crack, is detected 30 µs after the pulse is sent.

(a) Calculate the speed of ultrasound in the metal. [3]

(b) Calculate the depth of the crack below the top surface of the block. [2]

(c) A second block of the same metal has a crack closer to the top surface. State how the time at which the echo from this crack is detected would be different. Give a reason for your answer. [2]

(a) total distance travelled = 2 × 0.24 = 0.48 m(1)
(a) speed = 0.48 ÷ (80 × 10⁻⁶)(1)
(a) 6000 m/s (allow: 6.0 × 10³ m/s; 3000 m/s scores 1 for correct conversion of time)(1)
(b) depth = 6000 × 15 × 10⁻⁶ (allow ecf from (a); allow 6000 × 30 × 10⁻⁶ ÷ 2)(1)
(b) 0.090 m (allow: 0.09 m / 9 cm; allow ratio method 0.24 × 30/80)(1)
(c) the echo would be detected sooner / less than 30 µs after the pulse is sent (ignore references to the size of the echo)(1)
(c) because the ultrasound travels a shorter distance (to the crack and back) at the same speed(1)
Answer: (a) 6000 m/s (b) 0.090 m (c) echo arrives earlier, as the distance to the crack and back is shorter
Examiner tip: Double the thickness for the speed, then halve the crack echo time for its depth: missing either factor of 2 is where most marks go.

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