Edexcel International A Level Physics · Unit 2 · Waves and Electricity
Intensity, refraction and polarisation
40 free, high-quality past-paper-style questions on Intensity, refraction and polarisation for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 144 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U2-3-A-01Unit 2 · ASEasy1 mark
Which of these waves cannot be plane polarised?
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Answer
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Notes
Polarisation restricts the oscillations to one plane containing the direction of travel, so it is possible only for transverse waves. Infrared, microwaves and radio waves are all transverse electromagnetic waves. Sound in air is longitudinal: its oscillations are along the direction of travel, so there is no plane to select.
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Answer D
Answer: D
Exam tip: Link polarisation straight to the type of wave: transverse waves can be polarised, longitudinal waves cannot.
A type of flint glass has a refractive index of 1.65. c = 3.00 × 108 m s−1. What is the speed of light in this glass?
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Answer
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Notes
n = c/v, so v = c/n = 3.00 × 108 ÷ 1.65 = 1.82 × 108 m s−1.
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Answer B
Answer: B
Exam tip: Light is always slower in a material than in a vacuum, so v must come out less than 3.00 × 108 m s−1. An answer above c means the equation was inverted.
A ray of light in air strikes the flat surface of an acrylic block at an angle of incidence of 35°. The refractive index of the acrylic is 1.49. c = 3.00 × 108 m s−1.
Figure: ray entering the acrylic block (not to scale)
(a)
State what is meant by the refractive index of a material.
(1)
(b)
Calculate the angle of refraction in the acrylic.
(2)
(c)
Calculate the speed of light in the acrylic.
(1)
(d)
State what happens to the frequency and to the wavelength of the light as it enters the acrylic.
(1)
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Answer
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Notes
Part (a)
the speed of light in a vacuum divided by the speed of light in the material
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Part (b)
sin r = sin 35° ÷ 1.49 = 0.385
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r = 22.6°
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Part (c)
3.00 × 108 ÷ 1.49 = 2.01 × 108 m s−1
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Part (d)
frequency unchanged and wavelength decreases (both needed)
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Exam tip: Measure both angles from the normal. Going into a material of higher refractive index the ray bends towards the normal, so the angle of refraction must be smaller than 35°.
A laser produces a parallel beam of light that strikes a screen at right angles. The figure shows the power of the beam and the diameter of the circular spot it makes on the screen. Assume the power is spread evenly over the spot. What is the intensity of the light on the screen?
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Answer
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Notes
The radius of the spot is 1.0 mm = 1.0 × 10−3 m, so its area is πr2 = 3.14 × 10−6 m2. I = P/A = 5.0 × 10−3 W ÷ 3.14 × 10−6 m2 = 1.6 × 103 W m−2 (1590).
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Answer C
Answer: C
Common error: Putting the diameter into πr2, or leaving mm and mW unconverted. Halve the diameter first, then convert to metres and watts before dividing.
A ray of light passes from air into a liquid. The figure shows the angle the incident ray makes with the surface and the angle the refracted ray makes with the normal. What is the refractive index of the liquid?
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Answer
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Notes
The angle of incidence is measured from the normal: θ1 = 90° − 30° = 60°. Then n = sin 60° ÷ sin 40° = 0.866 ÷ 0.643 = 1.35.
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Answer B
Answer: B
Common error: Using an angle measured from the surface. Snell's law uses angles from the normal, so turn 30° to the surface into 60° to the normal first.
The figure shows three waves, P, Q and R, each viewed looking along its direction of travel. The symbols show the directions in which each wave oscillates. Which of the waves is plane polarised?
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Answer
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Notes
A plane-polarised wave oscillates in one plane only, a plane that contains the direction of travel. P oscillates along a single line, so it is plane polarised. Q is transverse but oscillates in many planes, so it is unpolarised. R is longitudinal: it oscillates along its direction of travel, so it cannot be polarised at all.
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Answer A
Answer: A
Exam tip: Polarised means one plane of oscillation. Only transverse waves can be polarised; a longitudinal wave never can, whatever it passes through.
An ultrasonic distance sensor on a robot sends a pulse of sound towards a wall and detects the echo. The oscilloscope trace shows the transmitted pulse and the echo. The speed of sound in air is 340 m s−1. How far is the sensor from the wall?
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Answer
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Notes
From the centre of the transmitted pulse to the centre of the echo is 5.0 − 1.0 = 4.0 divisions, so t = 4.0 × 2.0 ms = 8.0 ms. The sound goes there and back: d = vt/2 = 340 × 8.0 × 10−3 ÷ 2 = 1.4 m (1.36 m).
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Answer B
Answer: B
Exam tip: Time the gap between matching points on the two pulses (centre to centre), not from the edge of the screen, and halve vt because the sound travels there and back.
The figure shows the Sun and the orbits of Earth and Mars, with the data you need. Assume that the Sun radiates equally in all directions and that no radiation is absorbed between the Sun and the planets.
(a)
Calculate the intensity of the Sun's radiation at the distance of Earth's orbit.
(2)
(b)
Calculate the intensity of the Sun's radiation at the distance of Mars's orbit.
(1)
(c)
A rover on Mars carries a solar panel of area 1.8 m2, facing the Sun. Calculate the power of the Sun's radiation reaching the panel. Ignore the Martian atmosphere.
The figure shows a ray of light from a coin on the bottom of a swimming pool meeting the water surface. The refractive index of the water is 1.33. c = 3.00 × 108 m s−1.
(a)
Calculate the angle that the refracted ray makes with the normal in the air.
(2)
(b)
Calculate the speed of light in the water.
(1)
(c)
Not all of the light is refracted at the surface. State what happens to the rest of it, and the angle it makes with the normal.
(1)
(d)
The coin is moved so that the ray from it meets the surface at 52° to the normal. Calculate the critical angle for water and hence state what happens to this ray at the surface.
(1)
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Answer
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Notes
Part (a)
1.33 × sin 32° = 1.00 × sin θ
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sin θ = 0.705, so θ = 44.8°
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Part (b)
v = c/n = 3.00 × 108 ÷ 1.33 = 2.26 × 108 m s−1
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Part (c)
it is reflected back into the water, at 32° to the normal (both needed)
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Part (d)
sin C = 1/1.33 gives C = 48.8°; 52° is greater than 48.8°, so the ray is totally internally reflected
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Exam tip: Light leaving water bends away from the normal, so the angle in air must be bigger than 32°. If your answer is smaller, the refractive indices are the wrong way round.
A student uses a long rope to model polarisation. The figure shows the rope passing through two slits. To the left of slit 1 she shakes the end of the rope in all directions at right angles to it.
(a)
State the effect of slit 1 on the wave.
(1)
(b)
Describe and explain what happens to the wave at slit 2.
(2)
(c)
The rope is replaced by a long spring carrying a longitudinal wave. Explain why this wave passes through both slits.
(1)
(d)
State one practical application of polarisation of light, other than sunglasses.
(1)
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Answer
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Notes
Part (a)
it plane polarises the wave: beyond slit 1 the rope oscillates in a vertical plane only
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Part (b)
the wave is stopped: no wave travels beyond slit 2
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slit 2 only allows horizontal oscillations, and the wave reaching it oscillates only vertically
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Part (c)
its oscillations are along the spring, in the direction of travel, so neither slit blocks them / longitudinal waves cannot be polarised
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Part (d)
stress analysis: a transparent plastic model between crossed polarising filters shows coloured fringes where the stress is high / liquid-crystal displays / camera filters to cut reflections (any one)
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Common error: Writing that slit 2 “polarises the wave again”. A slit or filter can only pass oscillations along its own direction, so two crossed slits pass nothing.
A radioactive source emits gamma radiation equally in all directions. A detector 3.0 m from the source measures an intensity I. Absorption in air is negligible. What intensity is measured when the detector is moved to 7.5 m from the source?
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Answer
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Notes
For a point source I ∝ 1/r2, so the new intensity is I × (3.0/7.5)2 = I × 0.402 = 0.16I.
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Answer A
Answer: A
Exam tip: With the inverse-square law you do not need the power: multiply by (old distance ÷ new distance)2, and check that moving away gives a smaller intensity.
Light travelling in water strikes the flat glass base of a water-filled dish. The angle of incidence in the water is 50°. The refractive index of water is 1.33 and that of the glass is 1.52. What is the angle of refraction in the glass?
Figure: ray passing from water into glass (not to scale)
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Answer
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Notes
n1 sin θ1 = n2 sin θ2: sin θ2 = 1.33 × sin 50° ÷ 1.52 = 0.670, so θ2 = 42.1°. It is smaller than 50° because the light enters a medium of higher refractive index.
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Answer B
Answer: B
Common error: Putting n1 = 1 out of habit. Only use 1.00 when the first medium really is air.
An optical fibre has a core of refractive index 1.47 and cladding of refractive index 1.43. A ray of light in the core meets the core–cladding boundary at an angle of incidence of 70°. What happens to the ray at the boundary?
Figure: ray in the core meeting the core–cladding boundary (not to scale)
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Answer
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Notes
At the core–cladding boundary sin C = 1.43 ÷ 1.47 = 0.973, so C = 76.6°. The angle of incidence, 70°, is less than C, so there is no total internal reflection: part of the light is refracted into the cladding and part is reflected, as at any boundary.
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Answer D
Answer: D
Common error: Using sin C = 1/n for a fibre. That is the critical angle for core to air; the light actually meets the cladding, so use sin C = ncladding/ncore.
An air-traffic radar sends out a pulse of microwaves. The echo from an aircraft is received 0.24 ms after the pulse is sent. c = 3.00 × 108 m s−1. How far away is the aircraft?
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Answer
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Notes
The echo time is for the journey there and back, so d = ct/2 = 3.00 × 108 × 0.24 × 10−3 ÷ 2 = 3.6 × 104 m = 36 km.
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Answer C
Answer: C
Exam tip: Write “time there and back, so divide by 2” as a line of working in every pulse-echo calculation, and convert ms or μs to s first.
A microwave transmitter and receiver face each other 0.50 m apart. The receiver is connected to a meter. The receiver is rotated about the line joining it to the transmitter, and the meter reading is recorded: angle of rotation / °: 0, 90, 180, 270 meter reading / μA: 80, 2, 79, 3
Figure: microwave transmitter and receiver; the receiver is rotated about the line joining them (not to scale)
(a)
Describe how the reading varies and deduce what this shows about the microwaves from the transmitter.
(2)
(b)
Explain why this result shows that microwaves are transverse waves.
(1)
(c)
Suggest why the reading at 90° is not exactly zero.
(1)
(d)
Suggest why a television aerial on a roof should be lined up in a particular orientation.
(1)
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Answer
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Notes
Part (a)
maximum at 0° and 180°, almost zero at 90° and 270°
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the microwaves are plane polarised
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Part (b)
only transverse waves can be polarised, since their oscillations are perpendicular to the direction of travel
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Part (c)
microwaves reflected from the bench or nearby objects reach the receiver with a different plane of oscillation / the receiver is not exactly at 90°
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Part (d)
the transmitted waves are plane polarised, so the aerial must be aligned with their plane of oscillation for the strongest signal
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Exam tip: When a question asks you to deduce from data, quote the data: “near zero at 90° and 270°” earns more than “it goes up and down”.
A fishing boat uses an echo sounder that sends pulses of ultrasound vertically downwards. The speed of sound in sea water is 1520 m s−1. The frequency of the ultrasound is 38 kHz and each pulse lasts 1.0 ms.
Figure: echo sounder sending a pulse vertically down to the sea bed (not to scale)
(a)
The echo from the sea bed is received 0.84 s after a pulse is sent. Calculate the depth of the sea bed.
(2)
(b)
Calculate the wavelength of the ultrasound in sea water.
(1)
(c)
Calculate the smallest vertical separation of two fish whose echoes can be told apart.
(2)
(d)
Explain why a higher-frequency echo sounder can show more detail.
(2)
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Answer
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Notes
Part (a)
time there and back, so d = vt/2 = 1520 × 0.84 ÷ 2
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= 6.4 × 102 m (638 m)
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Part (b)
λ = v/f = 1520 ÷ 38 000 = 0.040 m
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Part (c)
pulse length = 1520 × 1.0 × 10−3 = 1.52 m
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echoes overlap unless the extra path 2Δd ≥ 1.52 m, so Δd = 0.76 m
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Part (d)
higher frequency gives a shorter wavelength
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smaller objects / features reflect clearly when they are not much smaller than the wavelength (less diffraction round them) / shorter pulses can be made, so closer objects can be separated
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Exam tip: Two different things limit the detail: the wavelength (size of the smallest object that reflects clearly) and the pulse duration (how close two objects can be along the beam). Name the one the question asks about.
A student measures the intensity I of the radiation from a small heater lamp at several distances r from it. The lamp radiates equally in all directions. The graph shows I plotted against 1/r2. What is the power emitted by the lamp?
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Answer
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Notes
I = P/4πr2 = (P/4π) × (1/r2), so the gradient of the line is P/4π. The line passes through (0.40, 2.0), so the gradient = 2.0 ÷ 0.40 = 5.0 W. P = 4π × 5.0 = 63 W (62.8 W).
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Answer D
Answer: D
Exam tip: When a graph is a straight line through the origin, write the equation as y = (gradient) × x and match the terms. Here the gradient is P/4π, not P itself.
Water waves in a ripple tank travel from deep water into shallow water. The figure shows a top view of the wavefronts, with a scale beneath. The waves travel at 0.24 m s−1 in the deep water. What is the speed of the waves in the shallow water?
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Answer
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Notes
From the scale, the wavelength is 12 mm in the deep water and 8 mm in the shallow water. The frequency does not change at the boundary, so v = fλ changes in the same ratio as λ: v = 0.24 × 8 ÷ 12 = 0.16 m s−1.
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Answer B
Answer: B
Exam tip: The frequency is set by the source and stays the same across a boundary, so speed changes in proportion to wavelength. Measure across several wavelengths and divide, for a more precise value.
The figure shows a right-angled glass prism ABC. A ray of light enters face AB at right angles and travels parallel to BC until it meets face AC at F. What is the minimum refractive index of the glass for the ray to be totally internally reflected at F?
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Answer
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Notes
The ray is parallel to BC, so it meets face AC at 47° to the face. Its angle of incidence, measured from the normal, is 90° − 47° = 43°. Total internal reflection needs the critical angle to be no more than 43°: sin C = 1/n ≤ sin 43°, so n ≥ 1/sin 43° = 1.47 (1.466).
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Answer D
Answer: D
Exam tip: In prism problems, find the angle between the ray and the face from the geometry first, then subtract it from 90° to get the angle of incidence from the normal.
A ray of light in air strikes a flat glass surface at P. Part of the light is reflected and part is refracted. The figure shows the angle of incidence and the angle between the reflected ray and the refracted ray. What is the refractive index of the glass?
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Answer
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Notes
The angle of reflection is 50°, so the reflected ray makes 40° with the surface. The refracted ray therefore makes 100° − 40° = 60° with the surface, which is 30° to the normal. n = sin 50° ÷ sin 30° = 0.766 ÷ 0.500 = 1.53.
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Answer C
Answer: C
Common error: Taking the angle with the surface as the angle of refraction. Work round the point P: the reflected ray makes 90° − 50° with the surface, and both angles of refraction and incidence are measured from the normal.
A student investigates how the intensity of light from a small lamp varies with distance. She uses a light meter in a darkened room and plots the graph shown. The lamp emits light equally in all directions.
(a)
Use the graph to find the intensity at r = 0.25 m and at r = 0.50 m.
(1)
(b)
Show that these two readings are consistent with an inverse-square law.
(2)
(c)
Calculate the power of the light emitted by the lamp.
(2)
(d)
Suggest why readings taken only a few centimetres from the lamp would not fit the same curve.
(1)
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Answer
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Notes
Part (a)
1.6 W m−2 and 0.40 W m−2 (both; allow ±0.05)
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Part (b)
doubling r from 0.25 m to 0.50 m divides I by 1.6 ÷ 0.40 = 4.0
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4 = 22, as required if I ∝ 1/r2 / Ir2 = 0.10 W at both distances
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Part (c)
P = 4πr2I = 4π × 0.502 × 0.40
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= 1.3 W (1.26 W)
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Part (d)
the lamp (or the meter's sensor) is no longer small compared with the distance, so the lamp cannot be treated as a point source
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Common error: Testing an inverse-square law by checking that I halves when r doubles. For I ∝ 1/r2, doubling r divides I by four; equivalently, I × r2 stays constant.
A thin layer of water rests on a flat glass plate, as shown in the figure. A ray of light in air strikes the water surface at 55° to the normal and passes through the water and the glass into the air below.
(a)
Calculate the angle θw that the ray makes with the normal in the water.
(2)
(b)
Calculate the angle θg that the ray makes with the normal in the glass.
(2)
(c)
Explain why the ray leaves the bottom of the glass at 55° to the normal.
(1)
(d)
Calculate the ratio speed of light in water ÷ speed of light in glass.
(1)
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Answer
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Notes
Part (a)
sin θw = sin 55° ÷ 1.33 = 0.616
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θw = 38.0°
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Part (b)
1.33 × sin 38.0° = 1.60 × sin θg, so sin θg = 0.512
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θg = 30.8°
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Part (c)
the boundaries are parallel, so n sin θ has the same value in every layer: 1.00 × sin θ = 1.00 × sin 55° in the air below
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Part (d)
v = c/n, so the ratio = 1.60 ÷ 1.33 = 1.20
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Exam tip: For a stack of parallel layers, n sin θ stays the same, so you can go straight from the air to any layer: nlayer sin θ = 1.00 × sin 55°.
A photographer points a light meter at the bright glare reflected from the surface of a lake and slowly rotates a polarising filter in front of the meter. The graph shows the reading against the angle θ between the filter's transmission axis and the horizontal.
(a)
State the values of θ, between 0° and 360°, at which the reading is a minimum.
(1)
(b)
Explain what the graph shows about the light reflected from the lake.
(2)
(c)
State the angle θ at which the photographer should set the filter to cut out as much of the glare as possible.
(1)
(d)
The meter is pointed instead at a filament lamp, which gives unpolarised light. Describe the graph that would now be obtained.
(1)
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Answer
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Notes
Part (a)
90° and 270° (both)
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Part (b)
the reading changes as the filter turns, so the reflected light is (partly) polarised, oscillating mainly horizontally since the reading is greatest when the axis is horizontal
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the minimum reading is not zero, so the light is not completely plane polarised: some of it is unpolarised
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Part (c)
θ = 90° (or 270°): transmission axis vertical
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Part (d)
a horizontal straight line: the reading is the same at every angle
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Exam tip: A minimum that is not zero means the light is only partly polarised. Only completely plane-polarised light can be cut almost to zero by a single filter.
An ultrasound probe is placed against the closed eyelid to measure the eye. The figure shows the echo trace. The speed of ultrasound is 1640 m s−1 in the lens of the eye and 1530 m s−1 in the fluid between the back of the lens and the retina.
(a)
Explain why the echoes become smaller from deeper boundaries.
(1)
(b)
Calculate the thickness of the lens.
(2)
(c)
Calculate the distance from the back of the lens to the retina.
(2)
(d)
Each pulse lasts 0.50 μs. Calculate the smallest separation of two boundaries inside the lens whose echoes do not overlap.
(2)
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Answer
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Notes
Part (a)
at each boundary only part of the pulse is reflected and the rest is transmitted (and some energy is absorbed), so less energy reaches and returns from deeper boundaries
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Part (b)
time there and back through the lens = 9.0 − 4.0 = 5.0 μs
length of pulse in the lens = 1640 × 0.50 × 10−6 = 8.2 × 10−4 m
✓ 1
the extra path is twice the separation, so separation = 4.1 × 10−4 m (0.41 mm)
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Exam tip: For the thickness of a layer, use the time between the echoes from its two faces and the speed in that layer, then halve for the there-and-back path.
An ultrasound scanner sends pulses that each last 2.0 μs into soft tissue, where the speed of ultrasound is 1540 m s−1. What is the smallest separation, along the beam, of two boundaries whose echoes do not overlap?
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Answer
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Notes
Each pulse is vτ = 1540 × 2.0 × 10−6 = 3.08 mm long. The echo from the deeper boundary travels an extra 2Δd, so the echoes separate when 2Δd ≥ vτ: Δd = 3.08 ÷ 2 = 1.54 mm (exactly; 1.5 mm would be too small).
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Answer B
Answer: B
Exam tip: The extra path for the deeper echo is twice the separation, because the pulse goes there and back. Shorter pulses give finer depth detail.
A pulse of ultrasound passes from soft tissue into a layer of bone. At each tissue–bone boundary, 40% of the incident intensity is reflected and 60% is transmitted. The pulse crosses the near boundary, reflects from the far boundary of the bone, and crosses the near boundary again on its way back. Ignore absorption. What fraction of the original intensity returns to the tissue as the echo from the far boundary?
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Answer
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Notes
Transmitted at the near boundary: 0.60. Reflected at the far boundary: × 0.40. Transmitted back through the near boundary: × 0.60. Fraction = 0.60 × 0.40 × 0.60 = 0.144 = 14%.
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Answer A
Answer: A
Exam tip: Follow the pulse boundary by boundary and multiply the fractions. Every crossing in either direction transmits only part of the intensity.
A cut diamond has a refractive index of 2.42. Much of its sparkle comes from light being totally internally reflected inside it.
Figure: ray inside the diamond meeting a face (not to scale)
(a)
Calculate the critical angle for light travelling from diamond into air.
(2)
(b)
Inside the diamond, a ray meets a face at an angle of incidence of 30°, with air outside. State and explain what happens to the ray.
(2)
(c)
The diamond is dropped into water of refractive index 1.33. Calculate the new critical angle at a diamond–water boundary.
(2)
(d)
Deduce what now happens to the ray in (b), and hence why a diamond sparkles less in water.
(1)
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Answer
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Notes
Part (a)
sin C = 1 ÷ 2.42 = 0.413
✓ 1
C = 24.4°
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Part (b)
total internal reflection
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light is travelling from a higher to a lower refractive index and 30° is greater than the critical angle of 24.4°
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Part (c)
sin C = nwater/ndiamond = 1.33 ÷ 2.42 = 0.550
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C = 33.3°
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Part (d)
30° is less than 33.3°, so the ray is now partly refracted out into the water (and only partly reflected), so less light is reflected back out of the top of the diamond
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Exam tip: When the outside medium is not air, put θ2 = 90° into n1 sin θ1 = n2 sin θ2: sin C = n2/n1. A smaller difference in refractive index gives a larger critical angle.
A student investigates refraction in a rectangular glass block. She records these angles of incidence i and angles of refraction r, each read from a protractor to the nearest degree: i / °: 25, 35, 45, 55, 65 r / °: 16, 22, 28, 32, 36
Figure: ray passing through the rectangular glass block (not to scale)
(a)
Describe how she could obtain one pair of readings of i and r. You should name the apparatus she uses.
(3)
(b)
Explain why a graph of sin i against sin r can be used to find the refractive index n of the glass.
(2)
(c)
Use the reading for i = 65° to calculate a value for n.
(2)
(d)
Calculate the percentage uncertainty in the reading r = 16°, and explain why the readings at large angles are more reliable.
(2)
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Answer
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Notes
Part (a)
place the block on paper, draw round it, and direct a narrow ray from a ray box (with a single slit) at one face
✓ 1
mark the incident and emergent rays with crosses, remove the block and join the points where the ray entered and left to give the refracted ray
✓ 1
draw the normal at the point of entry and measure i and r from the normal with a protractor
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Part (b)
with air as the first medium, Snell's law gives sin i = n sin r
✓ 1
so the graph is a straight line through the origin with gradient equal to n
the same ±0.5° is a smaller percentage of a larger angle
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Exam tip: In a method description, say what is measured, with what, and from where: angles are measured from the normal, which must be drawn with the protractor at the point where the ray enters.
An ultrasound probe is used to measure the depth of an organ boundary inside a patient's abdomen. The speed of ultrasound in soft tissue is 1540 m s−1.
(a)
Explain how the pulse-echo technique gives the depth of a boundary inside the body. Your answer should include why a gel is used between the probe and the skin, and what limits the amount of detail in the scan.
(6)
(b)
The echo from the organ boundary is received 65 μs after the pulse is sent. Calculate the depth of the boundary.
(2)
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Answer
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Notes
Part (a)
Indicative content:
—
a short pulse of ultrasound is sent into the body from the probe
✓ 1
at each boundary between different tissues part of the pulse is reflected and part is transmitted
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the time between sending the pulse and detecting the echo is measured, and depth = speed × time ÷ 2 because the pulse travels there and back
✓ 1
without gel, an air layer between probe and skin would reflect almost all of the ultrasound; the gel removes the air so most is transmitted into the body
✓ 1
features smaller than about one wavelength cannot be resolved, so a higher frequency (shorter wavelength) gives finer detail
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boundaries closer together than half the pulse length give overlapping echoes, so shorter pulses give better depth detail
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Part (b)
d = 1540 × 65 × 10−6 ÷ 2
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= 0.050 m (5.0 cm)
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Exam tip: In a six-mark explanation, make each point a separate sentence and cover every part of the question: here the depth calculation, the gel and both limits on detail.
The graph shows how the angle of refraction in air, θ2, depends on the angle of incidence, θ1, for light travelling from a transparent plastic into air. c = 3.00 × 108 m s−1. What is the speed of light in the plastic?
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Answer
Mark
Notes
The curve reaches θ2 = 90° at θ1 = 40°, so the critical angle is C = 40°. Then n = 1/sin C = 1.56 and v = c/n = c sin C = 3.00 × 108 × 0.643 = 1.93 × 108 m s−1.
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Answer A
Answer: A
Exam tip: The critical angle is the angle of incidence that makes the angle of refraction 90°. Read it where the curve reaches 90°, then use sin C = 1/n.
A radar transmitter sends out short pulses of microwaves. The graph shows the transmitted pulses against time. The echo of each pulse must return before the next pulse is sent, so that each echo can be matched to its own pulse. c = 3.00 × 108 m s−1. What is the greatest distance of an aircraft that this radar can locate without confusion?
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Answer
Mark
Notes
The pulses are sent every 400 μs (at 100, 500, 900 and 1300 μs). The longest allowed echo time is 400 μs, and the microwaves travel there and back: d = ct/2 = 3.00 × 108 × 400 × 10−6 ÷ 2 = 6.0 × 104 m = 60 km.
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Answer C
Answer: C
Exam tip: Read the time between neighbouring pulses, not the length of one pulse. The time between pulses limits the range; the pulse duration limits how close together two objects can be and still give separate echoes.
The figure shows a ray of light entering the flat end of an optical fibre and then meeting the boundary between the core and the cladding.
(a)
Calculate the critical angle at the core–cladding boundary.
(2)
(b)
The ray in the figure enters the end of the fibre at 20° to the normal. Show that it is totally internally reflected at the core–cladding boundary.
(3)
(c)
Calculate the largest angle to the normal at which a ray can enter the end of the fibre and still be totally internally reflected at the core–cladding boundary.
(2)
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Answer
Mark
Notes
Part (a)
sin C = 1.40 ÷ 1.50 = 0.933
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C = 69.0°
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Part (b)
sin r = sin 20° ÷ 1.50 = 0.228, so r = 13.2°
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angle of incidence at the boundary θ = 90° − 13.2° = 76.8°
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76.8° is greater than the critical angle of 69.0° (and the light is going into a lower refractive index), so it is totally internally reflected
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Part (c)
largest r = 90° − 69.0° = 21.0°
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sin α = 1.50 × sin 21.0° = 0.539, so α = 32.6° (allow 32.5°)
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Common error: Comparing the angle of refraction at the end face with the critical angle. The ray meets the side boundary at 90° − r, so convert before comparing.
A student shines rays of light from a ray box into a rectangular block of transparent resin and measures the angle of incidence i and the angle of refraction r for seven rays. She plots sin r against sin i and draws the best-fit line shown. Point X lies on the line.
(a)
Identify the anomalous point and suggest one reason for it.
(2)
(b)
Use point X to determine the refractive index of the resin.
(3)
(c)
Explain why the best-fit line should pass through the origin.
(1)
(d)
Suggest two ways in which the student could make her angle measurements more accurate.
(2)
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Answer
Mark
Notes
Part (a)
the point at sin i ≈ 0.77 (i = 50°), which lies above the line
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the angle r was misread / the normal was drawn in the wrong place / the refracted ray was traced inaccurately (any one)
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Part (b)
gradient = 0.50 ÷ 0.80 = 0.625
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sin i = n sin r, so sin r = (1/n) sin i and the gradient is 1/n
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n = 1 ÷ 0.625 = 1.60
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Part (c)
a ray along the normal (i = 0) is not bent, so r = 0; sin r is directly proportional to sin i
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Part (d)
use a narrow ray (single slit) and mark it with crosses placed far apart
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any one other: use a larger block so the refracted ray is longer / draw the normal with a set square or protractor / repeat each reading and average
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Exam tip: Check which quantity is on which axis before using a gradient. Here sin r is on the y-axis, so the gradient is 1/n, not n.
A survey ship sails along a straight line above a shipwreck, sending pulses of sound vertically downwards. The graph shows the time between sending a pulse and receiving its echo from below, measured every 10 m along the line, against the position of the ship. The speed of sound in sea water is 1500 m s−1.
(a)
Calculate the depth of the flat seabed.
(2)
(b)
Calculate the height of the wreck above the seabed.
(2)
(c)
The ship sends one pulse every 0.50 s. Explain why the echo from each pulse must be received before the next pulse is sent, and calculate the greatest depth that can be surveyed at this rate.
(2)
(d)
The sound has a frequency of 20 kHz and the ship moves at 2.0 m s−1. A thin mast, 4 cm in diameter, sticks up from the wreck. Explain why the survey may not detect the mast.
(3)
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Answer
Mark
Notes
Part (a)
d = vt/2 = 1500 × 0.080 ÷ 2
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= 60 m
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Part (b)
depth of the top of the wreck = 1500 × 0.064 ÷ 2 = 48 m
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height = 60 − 48 = 12 m
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Part (c)
otherwise an echo could be matched with the wrong pulse, giving a false depth
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greatest depth = 1500 × 0.50 ÷ 2 = 375 m (≈ 380 m)
the mast is narrower than the wavelength, so the sound diffracts round it and very little is reflected
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the ship moves 2.0 × 0.50 = 1.0 m between pulses, so a narrow mast may lie between the measured positions
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Exam tip: Read the echo times from the graph and convert ms to s before using d = vt/2. On this graph a shorter echo time means a shallower reflector, so the wreck shows as a raised section.
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