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IGCSE Physics Paper 4 (Theory / Extended)

Topic 3: Waves -- Cambridge Challenge Level
1 hour 15 minutes
80
7
75:00
0625 / 0972

⚡ Cambridge Challenge Level

These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!

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Question 1 -- Fibre Optic Communication and Total Internal Reflection
Total: 12 marks
An engineer is designing a fibre optic cable for high-speed broadband in Cambridge. The cable has a glass core with a refractive index of 1.52, surrounded by a glass cladding of lower refractive index.

The diagram below shows a ray of light entering the end of the fibre at point P and striking the core-cladding boundary at point Q.
Cladding (n = ?) Core (n = 1.52) Cladding (n = ?) Light in normal θ P Q
(a) [2]
State the two conditions that must be met for total internal reflection to occur at point Q.
Model Answer -- 1(a)
Light must be travelling from a more optically dense medium to a less optically dense medium / from a medium of higher refractive index to lower refractive index [1]
The angle of incidence must be greater than the critical angle [1]
⚠ If you missed marks here: Most students remember "angle of incidence > critical angle" but forget the OTHER condition: the light must be going from the MORE optically dense medium to the LESS dense one (higher n → lower n, i.e. core → cladding). Both conditions are separate marks — "the light reflects inside" describes the result, not a condition.
Mark 1 -- light travelling from higher refractive index to lower refractive index medium
Mark 2 -- angle of incidence greater than critical angle
(b) [3]
The critical angle at the core-cladding boundary is 62°.
Calculate the refractive index of the cladding material.
Model Answer -- 1(b)
Using n1 sin c = n2 sin 90°, or equivalently sin c = ncladding / ncore [1]
sin 62° = ncladding / 1.52
Correct substitution [1]
ncladding = 1.52 × sin 62° = 1.52 × 0.8829 = 1.34
Correct answer: ncladding = 1.34 (accept 1.34 to 1.35) [1]
⚠ If you missed marks here: The classic slip is inverting the ratio: sin c = ncladding/ncore, so ncladding = 1.52 × sin 62° = 1.34. If you got 1.72 you divided 1.52 by sin 62° — sanity check: the cladding MUST have a lower n than the core (1.52) or total internal reflection couldn't happen. Also check your calculator is in degrees mode.
Mark 1 -- correct relationship stated (sin c = n2/n1)
Mark 2 -- correct substitution of values
Mark 3 -- correct answer 1.34 or 1.35
(c) [3]
The angle θ in the diagram is 55°.
Determine whether total internal reflection occurs at point Q. Show your reasoning.
Model Answer -- 1(c)
The angle θ = 55° is measured between the ray and the normal at the boundary [1]
The critical angle is 62° [1]
Since 55° < 62°, the angle of incidence is less than the critical angle, so total internal reflection does NOT occur. The light is refracted into the cladding and the signal is lost. [1]
⚠ If you missed marks here: The trap is assuming TIR always happens in optical fibres — here 55° < 62°, so it does NOT occur and the light escapes into the cladding. A bare "no" doesn't score: you need the explicit comparison (55° is less than the critical angle 62°) and to note that θ is measured from the NORMAL, not from the boundary surface.
Mark 1 -- identifies theta as the angle of incidence at the boundary
Mark 2 -- compares angle of incidence to critical angle of 62 degrees
Mark 3 -- concludes TIR does not occur because angle is less than critical angle
(d) [4]
Explain two advantages of using optical fibres instead of copper cables for transmitting internet data over long distances.

Suggest why optical fibres are particularly useful in medical endoscopes.
Model Answer -- 1(d)
Advantage 1: Optical fibres carry signals as light, which travels much faster / higher bandwidth / can carry more data per second than electrical signals in copper [1]
Advantage 2: Optical fibres do not suffer from electromagnetic interference / are more secure / less signal loss over long distances / lighter and thinner [1]
Endoscope use: Optical fibres can be made very thin and flexible, so they can be inserted into the body through small openings [1]
Light is transmitted along the fibre by total internal reflection, allowing doctors to see images of internal organs without surgery / one bundle carries light in, another carries the image back [1]
⚠ If you missed marks here: Your two advantages must be DISTINCT — "faster" and "more data" often count as the same point; pair speed/bandwidth with something different (no electromagnetic interference, less signal loss, more secure). For the endoscope, "doctors can see inside you" alone misses both marks: you need thin/flexible for insertion AND that light travels along the fibre by total internal reflection.
Mark 1 -- advantage: higher bandwidth or faster data or no electromagnetic interference
Mark 2 -- second distinct advantage stated
Mark 3 -- thin and flexible so can be inserted into body
Mark 4 -- TIR allows light to travel along fibre to view internal organs
Question 2 -- Earthquake Seismic Waves
Total: 12 marks
During an earthquake, two types of seismic wave are produced: P-waves (primary) and S-waves (secondary). A seismometer station records the arrival of both waves from a distant earthquake.
(a) [3]
P-waves are longitudinal and S-waves are transverse.

(i) Describe the difference between longitudinal and transverse waves in terms of the direction of vibration relative to the direction of energy transfer. [2]

(ii) State one consequence of S-waves being transverse that is important in seismology. [1]
Model Answer -- 2(a)
(i) In longitudinal waves (P-waves), the vibrations are parallel to the direction of energy transfer / wave travel [1]
In transverse waves (S-waves), the vibrations are perpendicular to the direction of energy transfer / wave travel [1]
(ii) S-waves cannot travel through liquids (so they cannot pass through Earth's liquid outer core), which helps scientists determine the internal structure of the Earth [1]
⚠ If you missed marks here: Both definitions must be anchored to the DIRECTION OF ENERGY TRANSFER (or wave travel): longitudinal = vibrations PARALLEL to it, transverse = PERPENDICULAR to it — "up and down" or "side to side" without that reference doesn't score. In (ii), the expected fact is that S-waves cannot travel through liquids (that's how we know the outer core is liquid).
Mark 1 -- longitudinal vibrations parallel to direction of energy transfer
Mark 2 -- transverse vibrations perpendicular to direction of energy transfer
Mark 3 -- S-waves cannot travel through liquids
(b) [4]
The P-waves travel through the Earth's crust at a speed of 6.0 km/s and the S-waves travel at 3.5 km/s.

The seismometer detects the P-waves arriving 45 seconds before the S-waves.

Calculate the distance from the seismometer to the earthquake epicentre.
Model Answer -- 2(b)
Let d = distance. Time for P-waves: tP = d / 6.0. Time for S-waves: tS = d / 3.5 [1]
tS − tP = 45 s
d / 3.5 − d / 6.0 = 45
Correct equation set up [1]
d(1/3.5 − 1/6.0) = 45
d(0.2857 − 0.1667) = 45
d × 0.1190 = 45
Correct algebraic manipulation [1]
d = 45 / 0.1190 = 378 km
Correct answer: d = 378 km (accept 375 to 380 km) with unit [1]
⚠ If you missed marks here: The 45 s is NOT the travel time of either wave — it's the DIFFERENCE between their arrival times, so d = 6.0 × 45 = 270 km or d = 3.5 × 45 = 158 km are the classic wrong answers. You must set up d/3.5 − d/6.0 = 45 (same distance, different times) and solve for d ≈ 378 km.
Mark 1 -- correct expressions for time of P-waves and S-waves
Mark 2 -- correct equation linking time difference to distance
Mark 3 -- correct algebraic rearrangement
Mark 4 -- correct answer 375 to 380 km with unit
(c) [3]
The P-waves from this earthquake have a frequency of 0.50 Hz as they pass through the crust.

(i) Calculate the wavelength of these P-waves. [2]

(ii) As the P-waves pass from the crust into a denser layer of rock, their speed increases to 8.0 km/s. State what happens to the wavelength and the frequency. [1]
Model Answer -- 2(c)
v = f × λ
λ = v / f = 6.0 / 0.50 = 12 km
(i) Correct formula and substitution [1]
Correct answer: λ = 12 km with unit [1]
(ii) The frequency stays the same (frequency is determined by the source). The wavelength increases (since v = fλ and v increases while f is constant) [1]
⚠ If you missed marks here: In (i) keep the units consistent: λ = v/f = 6.0 km/s ÷ 0.50 Hz = 12 km — and the "km" (or 12 000 m) is part of the mark. In (ii), the rule to memorise: when a wave changes medium the FREQUENCY never changes (it's set by the source); only speed and wavelength change together.
Mark 1 -- correct formula v = f lambda and correct substitution
Mark 2 -- correct answer wavelength 12 km with unit
Mark 3 -- frequency stays same and wavelength increases
(d) [2]
Explain why at least three seismometer stations are needed to locate the exact position of an earthquake epicentre.
Model Answer -- 2(d)
Each station can only determine the distance to the epicentre, not the direction. This gives a circle of possible locations around each station [1]
Three circles from three different stations intersect at a single unique point, which is the epicentre. Two stations would give two possible points where circles intersect [1]
⚠ If you missed marks here: "For accuracy" or "to double-check" scores nothing — the key idea is that one station gives only a DISTANCE (a circle of possible positions), not a direction. Two circles still cross at TWO points; only a third circle picks out the single unique point.
Mark 1 -- each station gives distance only, forming a circle of possible locations
Mark 2 -- three circles intersect at one unique point to locate epicentre
Question 3 -- Refraction and the Sparkle of Diamonds
Total: 11 marks
A narrow beam of white light passes from air into a glass block at an angle of incidence of 40°. The angle of refraction inside the glass is measured to be 25°.
Glass Air normal 40° 25°
(a) [3]
Calculate the refractive index of the glass.
Model Answer -- 3(a)
n = sin i / sin r
Correct formula stated [1]
n = sin 40° / sin 25° = 0.6428 / 0.4226
Correct substitution [1]
n = 1.52
Correct answer: n = 1.52 (accept 1.52 to 1.53) [1]
⚠ If you missed marks here: n = sin i / sin r — the SINES of the angles, not the angles themselves (40/25 = 1.6 is the classic wrong answer). Also check: i is in the AIR (40°) and r in the glass (25°) — inverting gives 0.66, impossible since n of glass must be > 1 — and make sure your calculator is in degrees, not radians.
Mark 1 -- correct formula n = sin i / sin r
Mark 2 -- correct substitution of sin 40 and sin 25
Mark 3 -- correct answer 1.52
(b) [3]
Using your answer from (a), calculate the critical angle of this glass.
Model Answer -- 3(b)
n = 1 / sin c
Correct formula relating refractive index and critical angle [1]
sin c = 1 / 1.52 = 0.6579
Correct substitution [1]
c = sin−1(0.6579) = 41.1°
Correct answer: c = 41° (accept 41° to 42°) [1]
⚠ If you missed marks here: Two common slips: stopping at sin c = 1/1.52 = 0.658 and forgetting to take sin−1 (the answer is an ANGLE, 41°, not 0.66), or writing c = 1/1.52 without the sine at all. The formula is n = 1/sin c, so sin c = 1/n.
Mark 1 -- correct formula n = 1 / sin c or sin c = 1/n
Mark 2 -- correct substitution sin c = 1/1.52
Mark 3 -- correct answer critical angle 41 degrees
(c) [3]
Diamond has a refractive index of 2.42 and glass has a refractive index of 1.52.

Calculate the critical angle for diamond.

Suggest why a diamond sparkles much more than a piece of glass of the same shape.
Model Answer -- 3(c)
sin c = 1 / 2.42 = 0.4132
c = sin−1(0.4132) = 24.4°
Critical angle of diamond = 24° (accept 24° to 25°) [1]
Diamond has a much smaller critical angle (24°) compared to glass (41°) [1]
This means light entering the diamond undergoes total internal reflection at a wider range of angles / more light is reflected internally and eventually exits through the top. In glass, light escapes more easily through the sides because the critical angle is larger [1]
⚠ If you missed marks here: "Diamond is shinier / reflects more" without the mechanism scores only the calculation mark. The chain the examiner wants: diamond's critical angle (24°) is much SMALLER than glass's (41°) → far more rays hit internal faces above the critical angle → more total internal reflection, so more light bounces around inside and leaves through the top as sparkle.
Mark 1 -- correct critical angle for diamond approximately 24 degrees
Mark 2 -- diamond has smaller critical angle than glass
Mark 3 -- more total internal reflection occurs so more light exits from top creating sparkle
(d) [2]
When white light enters a glass prism, it splits into a spectrum of colours. This is called dispersion.

Explain why white light disperses as it enters glass.
Model Answer -- 3(d)
White light is a mixture of different colours / wavelengths / frequencies [1]
Each colour has a slightly different speed in glass / each colour is refracted by a slightly different amount. Violet light is refracted the most and red the least (because violet has the shortest wavelength / highest frequency) [1]
⚠ If you missed marks here: "The prism splits the light" restates the question. The two marks are: white light is a MIXTURE of colours/wavelengths, and each colour travels at a slightly different SPEED in glass so each is refracted by a different amount (violet bent most, red least). If you swapped red and violet, that's the classic reversal.
Mark 1 -- white light contains different colours or wavelengths
Mark 2 -- different colours refracted by different amounts because they travel at different speeds in glass
Question 4 -- Submarine Sonar and Sea Floor Mapping
Total: 12 marks
A research submarine uses sonar to map the sea floor. It emits short pulses of ultrasound vertically downward and detects the echoes reflected from the sea bed. The speed of sound in seawater is 1500 m/s.
SUB A B C D E
(a) [2]
(i) State what is meant by ultrasound. [1]

(ii) Explain why ultrasound is used for sonar rather than audible sound. [1]
Model Answer -- 4(a)
(i) Ultrasound is sound with a frequency above the upper limit of human hearing / above 20 000 Hz (20 kHz) [1]
(ii) Ultrasound has a shorter wavelength, so it diffracts less, giving a narrower beam and producing sharper / more detailed echoes / better resolution [1]
⚠ If you missed marks here: The definition needs the NUMBER or the boundary: frequency above 20 000 Hz (20 kHz) / above the upper limit of human hearing — "very high-pitched sound" is too vague. In (ii), "humans can't hear it" is not the physics reason: the mark is for shorter wavelength → less diffraction → narrower beam / sharper echoes.
Mark 1 -- sound above 20000 Hz or above human hearing range
Mark 2 -- shorter wavelength gives better resolution or less diffraction or sharper echoes
(b) [5]
As the submarine moves horizontally, it records the following echo times at five positions A to E:

Position A B C D E
Echo time / s 0.320 0.240 0.160 0.280 0.360

(i) Show that the depth of the sea floor below the submarine at position C is 120 m. [2]

(ii) Calculate the depth at each of the other four positions and complete the table below. [3]
Position A B C D E
Depth / m     120    
Model Answer -- 4(b)
(i) The echo time is the time for the sound to travel to the sea floor and back, so the one-way time = 0.160 / 2 = 0.080 s [1]
depth = speed × time = 1500 × 0.080 = 120 m ✔
Correct calculation showing depth = 120 m [1]
(ii) Using depth = speed × (echo time / 2) = 1500 × (echo time / 2): [1]
A: 1500 × 0.160 = 240 m
B: 1500 × 0.120 = 180 m
D: 1500 × 0.140 = 210 m
E: 1500 × 0.180 = 270 m
All four depths correct (A=240, B=180, D=210, E=270) [1]
Correct method clearly shown (dividing echo time by 2 before multiplying) [1]
⚠ If you missed marks here: The number one sonar error: forgetting the echo travels DOWN AND BACK, so halve the echo time before multiplying — 1500 × 0.160 = 240 m for C (double the true 120 m) means you skipped the ÷2. If all your depths came out exactly twice the correct values (480, 360, 420, 540), that single slip cost you several marks.
Mark 1 -- echo time divided by 2 for one-way distance
Mark 2 -- correct calculation showing depth at C is 120 m
Mark 3 -- correct method for remaining positions
Mark 4 -- all four depths correct A=240 B=180 D=210 E=270
Mark 5 -- correct units metres shown
(c) [3]
The submarine moves into warmer water where the speed of sound increases to 1540 m/s. The sonar system is not recalibrated and still uses 1500 m/s for its calculations.

(i) Suggest whether the sonar will overestimate or underestimate the depth. [1]

(ii) Calculate the percentage error in the depth measurement if the actual depth is 200 m. [2]
Model Answer -- 4(c)
(i) The sonar will underestimate the depth. The sound actually travels faster (1540 m/s) so the echo returns sooner than expected, but the system calculates using the slower speed (1500 m/s), giving a shorter calculated distance [1]
(ii) Actual depth = 200 m. True echo time = 2 × 200 / 1540 = 0.2597 s [1]
Calculated depth = 1500 × 0.2597 / 2 = 194.8 m
Percentage error = (200 − 194.8) / 200 × 100% = 2.6%
Correct answer: 2.6% (accept also (1540−1500)/1540 × 100 = 2.6%). Accept 2.5% to 2.7% [1]
⚠ If you missed marks here: The intuition trap in (i): faster sound means the echo returns SOONER, and a shorter time processed with the old slower speed gives a SMALLER depth — so the sonar UNDERESTIMATES; many students guess "overestimate" because the speed went up. In (ii), the shortcut (1540 − 1500)/1540 × 100 ≈ 2.6% works; just make sure you divided by the TRUE value.
Mark 1 -- underestimates depth with correct reasoning
Mark 2 -- correct method for percentage error calculation
Mark 3 -- correct answer approximately 2.6 percent
(d) [2]
Explain why the submarine uses short pulses of sound rather than a continuous signal.
Model Answer -- 4(d)
Short pulses ensure the emitted sound has stopped before the echo returns / prevents the outgoing signal from overlapping with the returning echo [1]
This allows the echo to be detected clearly and the time interval to be measured accurately [1]
⚠ If you missed marks here: "To save energy" or "so it's quieter" scores nothing. The point is OVERLAP: with a continuous signal the outgoing sound and the returning echo would mix and you couldn't tell when the echo arrives — short pulses leave a silent gap so the echo is detected cleanly and its time measured accurately.
Mark 1 -- prevents overlap between emitted pulse and returning echo
Mark 2 -- allows clear detection and accurate time measurement
Question 5 -- Ripple Tank Experiments and Diffraction
Total: 11 marks
A student uses a ripple tank to investigate the behaviour of waves. The ripple tank has a motor-driven dipper that produces straight waves at a frequency of 10 Hz.
Ripple Tank (top view) gap wave direction Diffracted waves (semicircular) Plane waves
(a) [2]
The student measures the distance across 5 complete waves to be 10.0 cm.

Calculate the speed of the water waves.
Model Answer -- 5(a)
λ = 10.0 / 5 = 2.0 cm = 0.020 m
v = f × λ = 10 × 0.020 = 0.20 m/s
Correct wavelength calculation [1]
Correct speed: v = 0.20 m/s (or 20 cm/s) [1]
⚠ If you missed marks here: The 10.0 cm spans FIVE waves, so one wavelength = 10.0/5 = 2.0 cm — using λ = 10 cm gives v = 1.0 m/s, five times too big. Then v = fλ = 10 × 0.020 m = 0.20 m/s; if you kept centimetres, state the unit as cm/s (20 cm/s), don't mix them.
Mark 1 -- correct wavelength 2.0 cm or 0.020 m
Mark 2 -- correct speed 0.20 m/s or 20 cm/s
(b) [3]
The diagram above shows the waves passing through a gap in a barrier. The gap width is approximately 5 cm (about 2.5 wavelengths).

(i) State the name of this wave phenomenon. [1]

(ii) Describe what happens to the pattern if the gap width is reduced to approximately 2.0 cm (about 1 wavelength). [1]

(iii) Explain why FM radio signals (wavelength about 3 m) can be received behind a building, but the building casts a shadow for visible light (wavelength about 5 × 10−7 m). [1]
Model Answer -- 5(b)
(i) Diffraction [1]
(ii) The waves spread out more / the diffraction is greater / the wavefronts become more nearly semicircular. Maximum diffraction occurs when the gap width is approximately equal to the wavelength [1]
(iii) The wavelength of FM radio waves (~3 m) is comparable to the size of buildings, so significant diffraction occurs and the radio waves spread around the building. The wavelength of visible light (~5 × 10−7 m) is much smaller than the building, so negligible diffraction occurs and a shadow is formed [1]
⚠ If you missed marks here: Don't confuse diffraction with refraction (no change of medium here). The scoring idea throughout is the COMPARISON of wavelength to gap/obstacle size: maximum spreading when they're similar — that's why narrowing the gap to ~1 wavelength increases spreading, and why 3 m radio waves bend around a building but 5 × 10−7 m light cannot. Answers that never compare wavelength to size lose the marks.
Mark 1 -- names the phenomenon as diffraction
Mark 2 -- waves spread out more when gap is reduced to about one wavelength
Mark 3 -- radio wavelength comparable to building size so diffracts, light wavelength much smaller so does not diffract around building
(c) [3]
The student now places a glass plate in the ripple tank to create a shallow region. The waves slow down from 0.20 m/s to 0.14 m/s as they enter the shallow region at an angle.

(i) Calculate the new wavelength in the shallow region. [2]

(ii) Describe what happens to the direction of the waves as they enter the shallow region and explain why. [1]
Model Answer -- 5(c)
(i) Frequency remains the same at 10 Hz (frequency is set by the source) [1]
λ = v / f = 0.14 / 10 = 0.014 m = 1.4 cm
Correct answer: λ = 1.4 cm (or 0.014 m) [1]
(ii) The waves change direction (refract) towards the normal as they enter the shallower (slower) region. This is because the part of the wavefront that enters the shallow region first slows down while the rest continues at the faster speed, causing the wavefront to turn [1]
⚠ If you missed marks here: In (i) the first mark is literally for saying the FREQUENCY STAYS 10 Hz (set by the dipper) — then λ = v/f = 0.14/10 = 1.4 cm; using the old speed 0.20 keeps the old wavelength and scores nothing. In (ii), slowing down means the waves bend TOWARDS the normal — "away from the normal" is the standard reversal.
Mark 1 -- frequency stays the same at 10 Hz
Mark 2 -- correct new wavelength 1.4 cm or 0.014 m
Mark 3 -- waves refract towards normal because they slow down
(d) [3]
Describe how you would use the ripple tank to demonstrate reflection of waves. Include:
• what you would place in the tank
• what you would observe
• one measurement you could take to verify the law of reflection
Model Answer -- 5(d)
Place a straight barrier (flat reflector) at an angle in the ripple tank [1]
Observe that the straight wavefronts reflect off the barrier and travel in a different direction. The reflected wavefronts are straight and have the same wavelength and speed as the incident wavefronts [1]
Measure the angle of incidence and angle of reflection (between the wavefront/ray and the normal to the barrier). The angle of incidence should equal the angle of reflection [1]
⚠ If you missed marks here: The question lists three bullets, and each is a mark — most lost marks come from skipping the last one: MEASURE the angle of incidence and angle of reflection (from the normal) and show they are equal. Also place the barrier at an ANGLE to the waves; head-on, the reflection just goes straight back and there's nothing to measure.
Mark 1 -- place a straight barrier or flat reflector in the tank
Mark 2 -- observe reflected wavefronts with same wavelength and speed
Mark 3 -- measure angle of incidence equals angle of reflection
Question 6 -- Electromagnetic Spectrum and Satellite Communication
Total: 10 marks
(a) [3]
The electromagnetic spectrum consists of a family of waves that all travel at the same speed in a vacuum.

(i) State the speed of electromagnetic waves in a vacuum. [1]

(ii) The table lists three regions of the EM spectrum with some information missing. Complete the blanks.
EM Region One use One danger to humans
Ultraviolet ___________ ___________
[2]
Model Answer -- 6(a)
(i) 3.0 × 108 m/s [1]
(ii) UV use: detecting forged banknotes / sterilising water / fluorescent lighting / security marking [1]
(ii) UV danger: can cause skin cancer / sunburn / damage to eyes [1]
⚠ If you missed marks here: The speed must be exact: 3.0 × 108 m/s — check the power (108, not 106) and the unit (m/s, not km/s). For UV, "it burns you" is weak — give an exam-standard use (detecting forged banknotes, sterilising) and danger (skin cancer, eye damage); don't mix up UV's uses with infrared's (heating) or X-rays' (imaging bones).
Mark 1 -- speed of EM waves 3.0 x 10^8 m/s
Mark 2 -- correct use of ultraviolet
Mark 3 -- correct danger of ultraviolet such as skin cancer
(b) [3]
A geostationary communications satellite orbits at a height of 36 000 km above the Earth's surface. A ground station sends a microwave signal up to the satellite, which amplifies and retransmits it back to Earth.

(i) Calculate the minimum time delay for a signal to travel from the ground station to the satellite and back. [2]

(ii) Suggest why this time delay is a problem for live telephone conversations via satellite. [1]
Model Answer -- 6(b)
distance = 2 × 36 000 km = 72 000 km = 7.2 × 107 m
time = distance / speed = 7.2 × 107 / 3.0 × 108 = 0.24 s
(i) Correct calculation: 0.24 s (accept 0.24 s; full round trip from person A to satellite to person B and back = 0.48 s is also acceptable if stated) [1]
Correct working shown with unit conversion [1]
(ii) The delay means there is a noticeable pause between one person speaking and the other hearing it, making conversation unnatural / speakers may talk over each other / it feels like the other person is slow to respond [1]
⚠ If you missed marks here: Two slips dominate: forgetting to DOUBLE the 36 000 km (the signal goes up AND comes back down: 72 000 km), and botching the km → m conversion (7.2 × 107 m). Then t = 7.2 × 107 / 3.0 × 108 = 0.24 s — an answer of 0.12 s means you forgot the return trip.
Mark 1 -- correct distance (72000 km or 7.2 x 10^7 m round trip)
Mark 2 -- correct time delay 0.24 s
Mark 3 -- noticeable pause causes unnatural conversation or talking over each other
(c) [4]
Television signals can be transmitted as either analogue or digital signals.

The diagram shows the effect of noise on both types of signal during transmission.
Analogue Signal Original: With noise: Digital Signal Original: With noise: Regenerated:
(i) Describe what happens to the analogue signal when noise is added during transmission. [1]

(ii) Explain why the digital signal can be regenerated to its original form but the analogue signal cannot. [2]

(iii) State one other advantage of digital signals over analogue signals. [1]
Model Answer -- 6(c)
(i) The noise adds random variations to the signal / the shape of the signal is distorted / the signal becomes corrupted and the noise cannot be separated from the original signal [1]
(ii) A digital signal only has two states: on (1) and off (0). Even when noise is added, the signal can still be identified as either a 1 or a 0, because the noise is not large enough to change a high voltage to a low voltage or vice versa [1]
A repeater/regenerator can therefore reconstruct the original clean digital signal perfectly. For an analogue signal, the noise has the same continuous nature as the signal itself, so it is impossible to separate the noise from the original signal [1]
(iii) Digital signals can be encrypted more easily / can carry more information / can be processed by computers / can be compressed / multiple signals can be sent simultaneously (multiplexing) [1]
⚠ If you missed marks here: "Digital is better quality" is the classic non-answer. The key mechanism: digital has only TWO states, so even a noisy pulse is still clearly a 1 or a 0 and a regenerator can rebuild the signal perfectly; analogue noise has the same continuous form as the signal itself, so it can never be separated out. In (iii), don't repeat noise resistance — give a NEW advantage (encryption, compression, multiplexing).
Mark 1 -- noise distorts or corrupts the analogue signal and cannot be separated
Mark 2 -- digital has only two states so noisy signal can still be identified as 1 or 0
Mark 3 -- regenerator reconstructs clean signal; analogue noise cannot be separated
Mark 4 -- one other advantage such as encryption or compression or multiplexing
Question 7 -- Sound, Echo Experiments and Concert Hall Design
Total: 12 marks
(a) [3]
A student wants to measure the speed of sound in air using the echo method. She stands 80 m from a large flat wall and claps her hands.

Describe a procedure she should follow to obtain an accurate value for the speed of sound. Include at least two precautions to improve accuracy.
Model Answer -- 7(a)
Clap hands and listen for the echo. Adjust the rate of clapping so that each clap coincides with the return of the previous echo. Time a large number of claps (e.g. 20 or more) using a stopwatch, then divide total time by number of intervals to find the time for one echo [1]
Precaution 1: Time many claps (e.g. 20+) and divide, rather than timing a single echo, to reduce the percentage error in timing / reduce the effect of reaction time [1]
Precaution 2: Measure the distance to the wall carefully and use a large distance (at least 50 m) so the echo time is long enough to measure. Use speed = 2 × distance / time (factor of 2 because sound travels there and back) [1]
⚠ If you missed marks here: Timing ONE clap-echo (~0.5 s) with a stopwatch is swamped by ~0.2 s reaction time — the expected method is timing MANY claps (20+) and dividing, which is also the precaution most students fail to justify ("repeat for accuracy" isn't enough; say it reduces the percentage timing error). And the speed formula needs the factor of 2: sound travels 80 m there AND 80 m back.
Mark 1 -- clap repeatedly synchronising with echoes and time many claps
Mark 2 -- precaution: time many claps to reduce percentage error from reaction time
Mark 3 -- precaution: use large distance and account for return journey (2x distance)
(b) [3]
In her experiment, the student times 20 clap-echo intervals and records a total time of 9.4 s.

(i) Calculate the time for one clap-echo interval. [1]

(ii) Calculate the speed of sound using her data. [2]
Model Answer -- 7(b)
time for one interval = 9.4 / 20 = 0.47 s
(i) 0.47 s [1]
speed = 2 × distance / time = 2 × 80 / 0.47
(ii) Correct formula with factor of 2 for return trip [1]
speed = 160 / 0.47 = 340 m/s
Correct answer: 340 m/s (accept 340 to 342 m/s) [1]
⚠ If you missed marks here: If your speed came out near 170 m/s, you forgot the echo's return journey — the sound covers 2 × 80 = 160 m per clap-echo interval, so v = 160/0.47 = 340 m/s. And use 0.47 s (one interval = 9.4/20), not the full 9.4 s.
Mark 1 -- correct time for one interval 0.47 s
Mark 2 -- correct formula with 2 x distance for return journey
Mark 3 -- correct answer 340 m/s
(c) [3]
An architect is designing a concert hall in Manchester. The hall is 45 m long from the stage to the back wall. Sound from a speaker on the stage reflects off the back wall.

(i) Calculate the time delay between the direct sound reaching a listener in the front row and the echo from the back wall reaching the same listener. Assume the listener sits 5 m from the stage. [2]

(ii) For good acoustics, the time delay between the direct sound and any reflection should be less than 0.05 s. Suggest one feature the architect could add to the back wall to solve any echo problems. [1]
Model Answer -- 7(c)
(i) Direct sound distance = 5 m. Direct sound time = 5 / 340 = 0.0147 s [1]
Echo distance: sound travels from stage to back wall (45 m) then back to listener (45 − 5 = 40 m) = 85 m total
Echo time = 85 / 340 = 0.250 s
Time delay = 0.250 − 0.015 = 0.235 s
Correct answer: time delay approximately 0.24 s (accept 0.23 to 0.25 s). This is much greater than 0.05 s, so there is an echo problem [1]
(ii) Add sound-absorbing material / soft panels / acoustic tiles / curtains / irregular/diffusing surfaces to the back wall to reduce reflections and prevent distinct echoes [1]
⚠ If you missed marks here: The echo path is the tricky bit: stage → back wall (45 m) → back to the listener (45 − 5 = 40 m) = 85 m, NOT 90 m and not 2 × 45 — and the answer is the DIFFERENCE between echo time (85/340 = 0.25 s) and direct time (5/340 = 0.015 s), about 0.24 s. In (ii), "make the wall thicker" doesn't help; the wall must ABSORB or diffuse the sound, not reflect it better.
Mark 1 -- correct calculation of direct and echo distances
Mark 2 -- correct time delay approximately 0.24 s
Mark 3 -- sound absorbing material or diffusing surfaces on back wall
(d) [3]
A noise pollution survey in a neighbourhood near an airport measures sound levels at different distances from the runway. The results are shown below.

Distance from runway / m 500 1000 2000 4000
Sound level / dB 95 89 83 77

(i) Describe the pattern shown by the data. [1]

(ii) A school is located 2000 m from the runway. Prolonged exposure to sound levels above 85 dB can cause hearing damage. A student suggests building a 5 m high concrete wall between the runway and the school to block the sound.

Explain whether this barrier would be effective at reducing the sound reaching the school. Refer to a specific wave property in your answer. [2]
Model Answer -- 7(d)
(i) As the distance from the runway doubles, the sound level decreases by 6 dB (each time) / the sound level decreases as distance increases [1]
(ii) The barrier would have limited effectiveness. Sound waves have wavelengths of the order of metres (e.g. at 340 Hz, λ = 340/340 = 1 m), which is comparable to or smaller than the barrier height [1]
Sound waves will diffract around and over the barrier because the wavelength is comparable to the size of the gap/obstacle. The lower-frequency (longer-wavelength) sounds will diffract the most and still reach the school. The barrier may reduce high-frequency sounds but will not block the lower-frequency rumble of aircraft engines effectively [1]
⚠ If you missed marks here: In (i), "it decreases" is weak — quote the pattern from the data: every time the distance DOUBLES the level drops by 6 dB (95 → 89 → 83 → 77). In (ii), the question demands a named wave property: sound DIFFRACTS over the 5 m wall because its wavelength (~1 m or more) is comparable to the wall's size — "sound goes over the top" without the word diffraction and the wavelength comparison loses the marks.
Mark 1 -- sound level decreases by 6 dB each time distance doubles
Mark 2 -- sound diffracts over or around barrier because wavelength is comparable to barrier size
Mark 3 -- lower frequency sounds diffract more so barrier is only partially effective

Self-Assessment

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IGCSE Physics 0625/0972 -- Paper 4 Theory (Extended) -- Topic 3: Waves -- Cambridge Challenge Level
For examination preparation only. Not an official Cambridge Assessment paper.