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!
Instructions
Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.
Question Navigation
Question 1 -- At the Opticians
Total: 12 marks
An optician in York uses a thin converging lens of focal length 5.0 cm to demonstrate image formation to a trainee. The lens is used first as a camera-style lens, then as a magnifying glass.
(a)[2]
Define the principal focus of a converging lens, and its focal length.
Model Answer -- 1(a)
Principal focus: the point where rays (arriving) parallel to the principal axis converge/are brought to a focus after passing through the lens [1]
Focal length: the distance from the (centre of the) lens to the principal focus [1]
⚠ If you missed marks here: The principal focus is defined by PARALLEL rays — only rays parallel to the axis converge there. The focal length is measured from the lens centre to that point, not from the object.
Mark 1 -- principal focus where parallel rays converge (1 mark)
Mark 2 -- focal length distance from lens to focus (1 mark)
(b)[3]
An object is placed 8.0 cm from the lens — between one and two focal lengths away. Describe the nature of the image formed (three properties).
Model Answer -- 1(b)
The image is REAL (light actually passes through it; it can be caught on a screen) [1]
It is INVERTED (upside down) [1]
It is ENLARGED (bigger than the object), and formed on the far side of the lens beyond 2F [1]
⚠ If you missed marks here: Object between F and 2F is the projector arrangement: real, inverted, magnified, beyond 2F on the other side. Give all three properties — "real and inverted" alone leaves a mark on the table.
Mark 1 -- image real stated (1 mark)
Mark 2 -- image inverted stated (1 mark)
Mark 3 -- image enlarged beyond 2F stated (1 mark)
(c)[3]
The trainee then uses the same lens as a magnifying glass to read small print. State where the object must be placed, and describe the image now formed.
Model Answer -- 1(c)
The object must be CLOSER to the lens than the principal focus (inside F, less than 5.0 cm away) [1]
The image is VIRTUAL (cannot be formed on a screen; rays only appear to come from it) and UPRIGHT [1]
It is ENLARGED, and on the SAME side of the lens as the object [1]
⚠ If you missed marks here: A magnifying glass only works with the object INSIDE the focal length. The image flips character completely: virtual, upright, enlarged, same side. If your ray diagram gives a real image, the object has drifted outside F.
Mark 1 -- object inside focal length (1 mark)
Mark 2 -- image virtual and upright (1 mark)
Mark 3 -- image enlarged on same side as object (1 mark)
(d)[4]
A patient is long-sighted. State what a long-sighted eye can and cannot see clearly, explain what goes wrong with the focusing, and state the type of lens used to correct it.
Model Answer -- 1(d)
A long-sighted person sees DISTANT objects clearly but cannot focus on NEAR objects [1]
Light from a near object is brought to a focus BEHIND the retina... [1]
...because the eye/lens system does not converge (bend) the light strongly enough [1]
Correction: spectacles with a CONVERGING (convex) lens, which adds converging power so the focus lands on the retina [1]
⚠ If you missed marks here: Long sight = near vision lost: the eye converges too weakly and the would-be focus falls behind the retina. The fix ADDS convergence — a converging lens. (Short sight is the mirror image: focus in front of the retina, corrected with a diverging lens.)
Mark 1 -- distant clear near blurred stated (1 mark)
Mark 2 -- focus falls behind the retina (1 mark)
Mark 3 -- eye converges light too weakly (1 mark)
Mark 4 -- converging lens corrects it (1 mark)
Question 2 -- The Swimming Pool Illusion
Total: 11 marks
At Ponds Forge pool in Sheffield, a coach on the poolside notices the pool looks shallower than its true 3.0 m depth. Light travelling from the pool floor refracts as it leaves the water. The refractive index of the pool water is 1.33.
(a)[2]
Explain, in terms of refraction, why the pool appears shallower than it really is.
Model Answer -- 2(a)
Light from the pool floor speeds up on leaving the water and bends AWAY from the normal at the surface [1]
The observer’s brain traces the rays straight back, so the floor appears to be where the backwards extensions meet — higher than it really is (an apparent depth less than the real depth) [1]
⚠ If you missed marks here: The bending happens at the SURFACE (away from the normal, since light speeds up leaving water), but the illusion happens in the BRAIN, which assumes light travels straight. Tracing the refracted rays back places the floor too high.
Mark 1 -- light bends away from normal leaving water (1 mark)
Mark 2 -- rays traced straight back make floor appear higher (1 mark)
(b)[3]
A ray of light inside the water strikes the underside of the surface at an angle of incidence of 40°. Calculate the angle of refraction in the air. (n = sin i₀/sin r where i₀ is the angle in air.)
Model Answer -- 2(b)
Going water → air: sin (angle in air) = n × sin 40° = 1.33 × 0.643 [1]
sin (angle in air) = 0.855 [1]
angle in air = 58.8° ≈ 59° [1]
⚠ If you missed marks here: The refractive index is defined with the AIR angle on top, so leaving the water you MULTIPLY: sin(air) = 1.33 × sin 40° = 0.855, giving 59°. Dividing instead gives 28.9° — the answer for light going the other way. Check the sense: leaving water, the ray bends AWAY from the normal, so the air angle must be BIGGER than 40°.
Mark 1 -- sin air angle equals n times sin 40 set up (1 mark)
Mark 2 -- sine value 0.855 computed (1 mark)
Mark 3 -- angle in air about 59 degrees (1 mark)
(c)[3]
Calculate the critical angle for the water–air surface.
Model Answer -- 2(c)
sin c = 1 / n [1]
sin c = 1 / 1.33 = 0.752 [1]
c = 48.8° ≈ 49° [1]
⚠ If you missed marks here: sin c = 1/n = 0.752, so c = 48.8°. This is the incidence angle (in the water) at which the refracted ray skims along the surface at 90°. If you calculated sin c = n you got a sine bigger than 1 — always a sign the formula is upside down.
Mark 1 -- sin c equals 1 over n stated (1 mark)
Mark 2 -- sine value 0.752 computed (1 mark)
Mark 3 -- critical angle about 49 degrees (1 mark)
(d)[3]
A diver on the pool floor shines a torch upwards at 55° to the vertical. State and explain what happens to the beam at the surface, and give one practical device that makes use of this effect.
Model Answer -- 2(d)
55° is GREATER than the critical angle (48.8°), so the beam undergoes total internal reflection [1]
ALL of the light is reflected back down into the water (the surface acts as a perfect mirror); none escapes into the air [1]
Used in e.g. prismatic binoculars/periscopes, bicycle/road reflectors (cat’s eyes), or reflecting prisms in cameras [1]
⚠ If you missed marks here: Compare with the critical angle FIRST: 55° > 48.8°, so no refraction at all — total internal reflection, with the surface acting as a perfect mirror. The two conditions (denser medium, angle above critical) plus one device carry the marks.
Mark 1 -- 55 exceeds critical angle so TIR occurs (1 mark)
Mark 2 -- all light reflected back into water (1 mark)
Mark 3 -- valid device using TIR named (1 mark)
Question 3 -- In the Recording Studio
Total: 12 marks
A sound engineer at a recording studio in London checks a singer’s note on an oscilloscope connected to a microphone. The trace is shown below. The speed of sound in air is 340 m/s.
(a)[3]
Use the trace and the time-base setting to determine the period and hence the frequency of the note.
Model Answer -- 3(a)
One complete cycle spans 5 divisions at 0.5 ms per division [1]
period T = 5 × 0.5 = 2.5 ms = 0.0025 s [1]
f = 1 / T = 1 / 0.0025 = 400 Hz [1]
⚠ If you missed marks here: Count divisions for ONE full cycle (peak to peak the long way): 5 divisions × 0.5 ms = 2.5 ms. Then f = 1/T = 400 Hz — converting ms to s first. Reading half a cycle (2.5 divisions) doubles your frequency; the trace shows exactly two full cycles across the screen.
Mark 1 -- one cycle read as 5 divisions (1 mark)
Mark 2 -- period 2.5 ms converted to seconds (1 mark)
Mark 3 -- frequency 400 Hz calculated (1 mark)
(b)[2]
Calculate the wavelength of this note in air.
Model Answer -- 3(b)
λ = v / f = 340 / 400 [1]
λ = 0.85 m [1]
⚠ If you missed marks here: λ = v/f = 340/400 = 0.85 m — nearly a metre of air per cycle. Multiplying (340 × 400 = 136 000) produces a wavelength longer than the M25; divide, and check the plausibility.
Mark 1 -- wavelength v over f substituted (1 mark)
Mark 2 -- wavelength 0.85 m calculated (1 mark)
(c)[3]
The singer sings the SAME note louder, and then a HIGHER note at the original loudness. Describe how the trace changes in each case.
Model Answer -- 3(c)
Louder, same note: the amplitude (height) of the trace increases... [1]
...but the period/spacing of the cycles is unchanged (same frequency) [1]
Higher note, same loudness: more cycles fit on the screen — shorter period/wavelength on the trace — with the SAME amplitude [1]
⚠ If you missed marks here: Loudness lives in the AMPLITUDE, pitch in the FREQUENCY — and each control changes one without the other. Say both halves each time: "taller, same spacing" then "closer together, same height".
Mark 1 -- louder gives larger amplitude (1 mark)
Mark 2 -- period unchanged for louder note (1 mark)
Mark 3 -- higher pitch more cycles same amplitude (1 mark)
(d)[2]
Sound is a longitudinal wave. Describe how the air particles move as the note travels across the studio, and name the regions formed.
Model Answer -- 3(d)
The air particles vibrate/oscillate back and forth PARALLEL to the direction the sound travels (they do not travel with the wave) [1]
This produces regions of squashed-together air (COMPRESSIONS) and stretched-out air (RAREFACTIONS) [1]
⚠ If you missed marks here: Longitudinal means the vibration is along the travel direction — the particles shuttle on the spot while the pattern of compressions and rarefactions moves. Both technical names are expected.
Mark 1 -- particles vibrate parallel to wave direction (1 mark)
Mark 2 -- compressions and rarefactions named (1 mark)
(e)[2]
State the approximate range of frequencies a healthy young person can hear, and state what the 400 Hz note lies within it.
Model Answer -- 3(e)
(About) 20 Hz to 20 000 Hz (20 kHz) [1]
400 Hz is comfortably inside the audible range (towards the lower-middle) [1]
⚠ If you missed marks here: The examiners want the standard limits: 20 Hz to 20 kHz. Both ends are needed; and the upper limit falls with age — which is why the 400 Hz note is safe for every listener in the room.
Mark 1 -- range 20 Hz to 20 kHz stated (1 mark)
Mark 2 -- 400 Hz identified as within range (1 mark)
Question 4 -- Security and Signals at the Airport
Total: 11 marks
Gatwick Airport uses several parts of the electromagnetic spectrum: X-rays scan luggage, infrared cameras check passenger temperatures, microwaves carry signals to a communications satellite 3.6 × 10⁷ m above the equator, and visible light floods the terminal. All electromagnetic waves travel at 3.0 × 10⁸ m/s in a vacuum.
(a)[3]
Arrange the four named radiations (X-rays, infrared, microwaves, visible light) in order of INCREASING wavelength, and state one property all electromagnetic waves share.
Model Answer -- 4(a)
X-rays (shortest) → visible light [1]
→ infrared → microwaves (longest) [1]
All travel at the same (high) speed in a vacuum (3.0 × 10⁸ m/s) / all are transverse waves [1]
⚠ If you missed marks here: Increasing wavelength runs X-rays → visible → infrared → microwaves. The order reverses for frequency — read which the question asks. The shared property examiners expect: same speed in vacuum.
Mark 1 -- X-rays then visible placed correctly (1 mark)
Mark 2 -- infrared then microwaves complete order (1 mark)
Mark 3 -- common vacuum speed or transverse nature stated (1 mark)
(b)[3]
Calculate the time for a microwave signal to travel from the ground station up to the satellite.
Model Answer -- 4(b)
t = d / v [1]
t = 3.6 × 10⁷ / 3.0 × 10⁸ [1]
t = 0.12 s [1]
⚠ If you missed marks here: 3.6 × 10⁷ ÷ 3.0 × 10⁸ = 0.12 s each way — which is why satellite phone conversations have that awkward quarter-second round-trip lag. Powers of ten do all the work here; keep them explicit.
Mark 1 -- time distance over speed stated (1 mark)
Mark 2 -- powers of ten substituted correctly (1 mark)
Mark 3 -- travel time 0.12 s calculated (1 mark)
(c)[3]
X-rays are ionising radiation. Explain what harm over-exposure to X-rays can cause, why X-rays are dangerous in this way while visible light is not, and one precaution taken to protect the security staff who operate the scanners all day.
Model Answer -- 4(c)
Ionising radiation can damage/kill living cells and damage DNA, leading to mutation / cancer risk [1]
X-ray photons carry far more energy (much higher frequency) than visible light, enough to ionise atoms in the body — visible light cannot [1]
Precaution: the scanner is shielded (lead/metal casing and curtains), staff keep their distance / exposure time is limited (or wear dose badges to monitor exposure) [1]
⚠ If you missed marks here: The danger word is IONISING: X-rays carry enough energy per photon to knock electrons off atoms, damaging DNA. Visible light, same wave family, simply lacks the photon energy. Precautions follow the usual trio — shielding, distance, time (plus monitoring badges).
Mark 1 -- cell and DNA damage cancer risk stated (1 mark)
Mark 2 -- X-rays ionise but visible light cannot (1 mark)
Mark 3 -- shielding distance or monitoring precaution given (1 mark)
(d)[2]
Explain why an infrared camera can pick out a feverish passenger in a crowd.
Model Answer -- 4(d)
Every warm object emits infrared radiation, and HOTTER surfaces emit more (more intensely) [1]
The camera detects the infrared and displays skin temperature; a feverish face shows up brighter/hotter than the people around it [1]
⚠ If you missed marks here: All bodies emit infrared all the time; the camera is just an eye for it. The fever shows because emission rises with temperature — a degree or two of extra skin heat is a visible step in infrared brightness.
Mark 1 -- warm objects emit infrared more when hotter (1 mark)
Mark 2 -- camera maps emission showing feverish face hotter (1 mark)
Question 5 -- Waves at Woolacombe
Total: 12 marks
A surf instructor at Woolacombe beach in Devon watches regular waves rolling in. Standing on the pier, she counts 10 wave crests passing a post in 80 s. The distance between neighbouring crests is 12 m.
(a)[3]
Define wavelength, frequency and amplitude for these water waves.
Model Answer -- 5(a)
Wavelength: the distance between two successive crests (or any two corresponding points on successive waves) — here 12 m [1]
Frequency: the number of waves passing a point per second (measured in hertz) [1]
Amplitude: the maximum displacement of the water surface from its undisturbed (rest) position — crest height above the mean level, not crest-to-trough [1]
⚠ If you missed marks here: Amplitude is measured from the REST position to the crest — half the crest-to-trough distance. Frequency must say "per second" (or Hz), and wavelength needs "successive" or "corresponding points", not just "between two waves".
Mark 1 -- wavelength crest to crest defined (1 mark)
Mark 2 -- frequency waves per second defined (1 mark)
Mark 3 -- amplitude from rest position to crest (1 mark)
(b)[4]
Calculate the frequency of the waves, and hence their speed.
Model Answer -- 5(b)
f = 10 / 80 [1]
f = 0.125 Hz [1]
v = f λ = 0.125 × 12 [1]
v = 1.5 m/s [1]
⚠ If you missed marks here: 10 crests in 80 s means f = 10/80 = 0.125 Hz (a frequency below 1 Hz is fine — one wave every 8 s). Then v = fλ = 0.125 × 12 = 1.5 m/s. Dividing 80/10 gives the PERIOD (8 s); usable, but only if you then take 1/T.
Mark 1 -- frequency 10 over 80 set up (1 mark)
Mark 2 -- frequency 0.125 Hz calculated (1 mark)
Mark 3 -- speed f lambda substituted (1 mark)
Mark 4 -- speed 1.5 m per s calculated (1 mark)
(c)[3]
As the waves run from deep water into the shallows, they slow down. State what happens to their frequency and wavelength, and explain why waves approaching the beach at an angle bend to arrive nearly parallel to the shore.
Model Answer -- 5(c)
The frequency stays the SAME (the source sets it; waves cannot pile up or vanish) [1]
Since v = fλ and v falls, the wavelength SHORTENS — crests bunch together near the beach [1]
A wavefront arriving at an angle has its shallow end slow down first while the deep end keeps speed, swinging the front round — refraction — until it lies nearly parallel to the shore [1]
⚠ If you missed marks here: Frequency is fixed by the source — in any refraction it is wavelength and speed that change together. The bending argument is one end of the wavefront braking before the other, pivoting the front — the same reason light bends entering glass.
Mark 1 -- frequency unchanged in shallow water (1 mark)
Mark 2 -- wavelength shortens as speed falls (1 mark)
Mark 3 -- one end slows first turning wavefront parallel (1 mark)
(d)[2]
State the difference between a transverse and a longitudinal wave, and classify these surface water waves.
Model Answer -- 5(d)
Transverse: vibrations at RIGHT ANGLES to the direction of travel; longitudinal: vibrations PARALLEL to it [1]
The water surface moves (mainly) up and down as the wave travels horizontally, so these are (treated as) TRANSVERSE waves [1]
⚠ If you missed marks here: The classification follows the definition: the surface bobs vertically while the wave travels horizontally — vibration perpendicular to travel, so transverse. Sound is the standard longitudinal contrast.
Mark 1 -- transverse perpendicular longitudinal parallel stated (1 mark)
Mark 2 -- water surface waves classified transverse (1 mark)
Question 6 -- The Ultrasound Scan
Total: 12 marks
At a hospital in Leicester, a sonographer performs a prenatal scan. The probe emits pulses of ultrasound at a frequency of 3.0 MHz; the pulses travel through body tissue at 1540 m/s and partially reflect at boundaries between different tissues. One pulse returns from a boundary 52 μs after it was emitted.
(a)[2]
State what ultrasound is.
Model Answer -- 6(a)
Sound (a longitudinal pressure wave)... [1]
...with a frequency ABOVE 20 kHz — higher than the upper limit of human hearing [1]
⚠ If you missed marks here: Ultrasound is ordinary sound in every mechanical respect — just pitched above 20 kHz, past the top of human hearing. Give the number; "sound we cannot hear" without 20 kHz is half an answer.
Mark 1 -- ultrasound is sound wave stated (1 mark)
Mark 2 -- frequency above 20 kHz beyond hearing (1 mark)
(b)[4]
Calculate the depth of the reflecting tissue boundary below the probe.
Model Answer -- 6(b)
t = 52 μs = 52 × 10⁻⁶ s [1]
Total path (there and back) = v t = 1540 × 52 × 10⁻⁶ = 0.080 m [1]
The pulse travels DOWN and BACK, so depth = half the path [1]
depth = 0.040 m = 4.0 cm [1]
⚠ If you missed marks here: Two traps in one question: the microsecond conversion (52 μs = 5.2 × 10⁻⁵ s) and the FACTOR OF TWO — the 52 μs covers the round trip, so the boundary is at half of 8.0 cm, i.e. 4.0 cm. Forgetting the halving is the most common echo-calculation error in every exam session.
Mark 1 -- time converted from microseconds (1 mark)
Mark 2 -- total path 0.080 m from v times t (1 mark)
Mark 3 -- halved for there-and-back journey (1 mark)
Mark 4 -- depth 4.0 cm calculated (1 mark)
(c)[2]
Explain why ultrasound, rather than X-rays, is used to image an unborn baby.
Model Answer -- 6(c)
X-rays are IONISING and could damage the cells/DNA of the developing baby [1]
Ultrasound is not ionising (it is a sound wave), so at diagnostic intensities it causes no such damage — and it still shows soft-tissue boundaries well [1]
⚠ If you missed marks here: The contrast is ionising vs not: X-ray photons damage dividing cells — and no tissue divides faster than a fetus — while ultrasound is a pressure wave with no ionising power. Bonus truth: soft tissue barely shows on X-rays anyway.
Mark 1 -- X-rays ionising harm developing cells (1 mark)
Mark 2 -- ultrasound non-ionising and images soft tissue (1 mark)
(d)[2]
State two OTHER uses of ultrasound (not medical scanning and not sonar depth-finding).
Model Answer -- 6(d)
Cleaning delicate objects (jewellery, surgical instruments, watch parts) in an ultrasonic bath [1]
Detecting flaws/cracks inside metal (non-destructive testing of rails, welds, aircraft parts); also physiotherapy treatment, cutting, or pest deterrents [1]
⚠ If you missed marks here: The two standard answers are ultrasonic cleaning baths and flaw detection in metals (the echo technique of the scan, pointed at a railway rail). The question excludes scanning and sonar — re-offering them scores zero.
Mark 1 -- ultrasonic cleaning use given (1 mark)
Mark 2 -- metal flaw detection or other valid use (1 mark)
(e)[2]
Calculate the wavelength of the 3.0 MHz ultrasound in body tissue.
Model Answer -- 6(e)
λ = v / f = 1540 / 3.0 × 10⁶ [1]
λ = 5.1 × 10⁻⁴ m (≈ 0.51 mm) [1]
⚠ If you missed marks here: 1540 ÷ 3.0 × 10⁶ = 5.1 × 10⁻⁴ m — about half a millimetre. The tiny wavelength is the point: detail smaller than the wavelength cannot be resolved, so megahertz frequencies are what make sub-millimetre imaging possible.
Mark 1 -- wavelength v over f with 3 MHz converted (1 mark)
Mark 2 -- wavelength about 0.51 mm calculated (1 mark)
Question 7 -- Colour at the Theatre
Total: 10 marks
A lighting designer at a West End theatre uses a glass prism to split a white spotlight beam into a spectrum across the backdrop, and a red laser (wavelength 650 nm in air) for a special effect. Speed of light in air: 3.0 × 10⁸ m/s.
(a)[3]
Describe what happens to the white light as it passes through the prism, name the effect, and give the order of colours produced from least deviated to most deviated.
Model Answer -- 7(a)
The white light is split/separated into its component colours (a continuous spectrum) — the effect is called DISPERSION [1]
Red is deviated (bent) LEAST; violet is deviated MOST [1]
⚠ If you missed marks here: Name the effect (dispersion), state the extremes (red least bent, violet most), and give the full running order. White light is a MIXTURE — the prism does not colour the light, it sorts what was already there.
Mark 1 -- white light dispersed into spectrum named (1 mark)
Mark 2 -- red least violet most deviated (1 mark)
Mark 3 -- full colour order given (1 mark)
(b)[1]
The red laser light is described as monochromatic. State what this means.
Model Answer -- 7(b)
Light of a single frequency (single wavelength / one pure colour) [1]
⚠ If you missed marks here: Monochromatic = ONE frequency. "One colour" is acceptable shorthand, but frequency is the precise word — a laser’s light is a single frequency to extraordinary purity.
Mark 1 -- single frequency or wavelength stated (1 mark)
(c)[3]
Explain why the prism is able to separate the colours, in terms of the speed of light in glass.
Model Answer -- 7(c)
In glass, different frequencies/colours travel at (slightly) DIFFERENT speeds [1]
Violet travels slowest in glass, so it is refracted (bent) the most; red travels fastest and bends least [1]
Each colour therefore leaves the prism at a slightly different angle, spreading the beam into a spectrum (in a vacuum all colours travel at the same speed, so no separation happens there) [1]
⚠ If you missed marks here: Dispersion exists because glass is speed-selective: each colour has its own speed in glass, hence its own refractive index, hence its own bending angle. The vacuum contrast (all colours at c, no dispersion) shows you understand the cause lives in the glass, not the light.
Mark 1 -- colours travel at different speeds in glass (1 mark)
Mark 2 -- violet slowest bent most red least (1 mark)
Mark 3 -- different angles spread beam into spectrum (1 mark)
(d)[3]
Calculate the frequency of the red laser light in air.
Model Answer -- 7(d)
f = v / λ [1]
f = 3.0 × 10⁸ / 650 × 10⁻⁹ [1]
f = 4.6 × 10¹⁴ Hz [1]
⚠ If you missed marks here: Convert nanometres first: 650 nm = 6.5 × 10⁻⁷ m. Then f = 3.0 × 10⁸ ÷ 6.5 × 10⁻⁷ = 4.6 × 10¹⁴ Hz — hundreds of trillions of cycles per second. An answer near 10⁶ Hz means the nano prefix was dropped.
Mark 1 -- frequency v over lambda stated (1 mark)
Mark 2 -- nanometre conversion in substitution (1 mark)
Mark 3 -- frequency 4.6 times 10 to 14 Hz (1 mark)
When you have finished answering all questions, click Submit to see the model answers.
Self-Assessment
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