113 free, high-quality past-paper-style questions on Wave phenomena for IB Physics SL, each with a full mark scheme. Covers: Critical angle & TIR; Refractive index & speed; Diffraction; and more. For when you have finished the past papers, or want to practise one topic at a time.C.3 at Higher Level →Read the C.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.559 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1C-1A-19Critical angle & TIRPaper 1AEasy1 mark
Light meets the boundary between two transparent media.
Which row gives the conditions needed for total internal reflection to occur?
The light travels towards a medium withAngle of incidence
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Step 1Total internal reflection needs light travelling towards a medium of lower refractive index (where it would speed up) at an angle of incidence greater than the critical angle.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: only on passing into a medium of lower refractive index can the refracted angle reach 90°, and beyond the critical angle no light is transmitted.
BLight entering a medium of higher refractive index bends towards the normal, so a refracted ray always exists and total internal reflection cannot occur.
CThe direction is right, but below the critical angle a refracted ray emerges; there is only partial reflection.
DBoth conditions are reversed: light must travel towards the lower refractive index and meet the boundary beyond the critical angle.
Two identical lamps placed side by side illuminate a screen. No interference fringes are seen on the screen.
What is the reason?
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Step 1An observable interference pattern needs coherent sources — a constant phase difference; light from two separate lamps changes its relative phase randomly and very rapidly, so the pattern averages out.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AWaves from any sources superpose wherever they meet; the problem is that the resulting pattern is not steady.
BCorrect: the two lamps are not coherent, so the interference pattern shifts too rapidly to be seen.
CLight from a single source split by two slits does produce visible fringes, so a short wavelength does not prevent interference.
DUnequal amplitudes only reduce the contrast between bright and dark fringes; coherent waves of unequal amplitude still give fringes.
A mobile-phone network transmits signals at 700 MHz and at 28 GHz. In the region just behind tall buildings, the 700 MHz signal is received much better than the 28 GHz signal.
Which statement explains this in terms of diffraction?
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Step 1λ = c/f: about 0.43 m at 700 MHz and about 1.1 cm at 28 GHz. The amount of diffraction around an obstacle increases as the wavelength becomes more comparable with the size of the obstacle or edge, so the 700 MHz waves spread further into the region behind the buildings.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThe wavelengths are the wrong way round: the 28 GHz waves have the shorter wavelength (about 1 cm), because λ = c/f.
BAll electromagnetic waves travel through the air at (almost) the same speed, c. The difference is in wavelength, not speed.
CCorrect: the 700 MHz waves have a wavelength of about 0.43 m, about 40 times longer than the 28 GHz waves, so they diffract much more around the edges of buildings.
D700 MHz is the lower frequency. It is the longer wavelength, which goes with the lower frequency, that diffracts more.
6C-1A-73Reflection & transmissionPaper 1AEasy1 mark
Plane water waves in a ripple tank travel towards a straight barrier. The rays make an angle of 30° with the normal to the barrier, as shown.
Which diagram shows the reflected wavefronts and their direction of travel?
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Step 1On reflection the angle of reflection equals the angle of incidence (30° to the normal), on the other side of the normal; the reflected wavefronts are perpendicular to the reflected rays.
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Step 2The speed and frequency are unchanged, so the reflected wavefronts have the same spacing: diagram A.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the reflected rays make 30° with the normal, on the other side of it, and the wavefronts keep the same spacing because the wave stays in the same water.
BThe waves do not return along their own path unless they meet the barrier along the normal; here the angle of incidence is 30°.
CThe direction is right, but reflection does not change the speed or the frequency, so the wavelength — the spacing of the wavefronts — cannot change.
DHere the reflected rays make 60° with the normal: the 30° has been measured from the barrier instead of from the normal.
Plane water waves of wavelength 2.0 cm in a ripple tank meet two obstacles in turn: a thin rod of diameter 3 mm, and a solid block 25 cm wide.
Which statement describes the water just behind each obstacle?
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Step 1Waves diffract around an obstacle most when its size is comparable to or smaller than the wavelength: the 3 mm rod is much smaller than 2.0 cm, so the waves bend round it and close up; the 25 cm block is many wavelengths wide, so bending at its edges fills only a little of the region behind it.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis ignores diffraction: waves spread round the edges of an obstacle, and round a very small one they close up almost completely.
BCorrect: the rod is much narrower than one wavelength, so the waves diffract round it; the block is more than 10 wavelengths wide, so it casts a clear shadow.
CThis reverses the effect of size: diffraction is significant when the obstacle is comparable to or smaller than the wavelength, so it is the thin rod that casts almost no shadow.
DThe amount of diffraction depends on the size of the obstacle compared with the wavelength; behind the wide block the waves cannot close up.
A small dipper vibrates up and down at one point on the surface of the water in a ripple tank.
Which row describes the wavefronts and the rays of the waves that spread out from the dipper?
WavefrontsRays
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Step 1A point source sends out circular wavefronts (all points of one wavefront are the same distance from the source); the rays show the direction of travel and are perpendicular to the wavefronts, so they radiate outwards from the dipper.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: the wavefronts are circles centred on the point source and the rays, perpendicular to them, radiate outwards from it.
BRays show the direction in which the wave travels, so they are perpendicular to the wavefronts, not drawn along them.
CA point source produces circular wavefronts; straight, parallel wavefronts come from a straight bar, not from a single point. Straight wavefronts would also not match radial rays.
DThis describes plane waves from a straight vibrating bar. A dipper at a single point produces circular wavefronts and radial rays.
At a football stadium, two loudspeakers are driven in phase by the same amplifier. They emit sound of wavelength 0.28 m. A spectator sits 3.4 m from one loudspeaker and 4.1 m from the other. What does the spectator hear, and why?
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The path difference is 4.1 − 3.4 = 0.70 m, and 0.70 / 0.28 = 2.5. For in-phase sources, a path difference of a whole number plus a half of wavelengths means the waves arrive in antiphase and interfere destructively, so the spectator hears a minimum. Being larger than λ does not make a point a maximum, and in-phase sources still produce minima. A half-integer number of wavelengths is a minimum, not an intermediate point.
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Answer A
Answer: A
Exam tip: Always convert the path difference into a number of wavelengths. Whole number means a maximum; whole number + ½ means a minimum, provided the sources are in phase.
10IBP2S-022-02-02Coherent sourcesPaper 1AEasy1 mark
The two headlights of a car emit light of the same colour. No interference fringes are seen where their beams overlap on a wall. What is the main reason?
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A stable interference pattern needs coherent sources, meaning sources with a constant phase difference (and so the same frequency). Each headlight emits light in short, random bursts, so the phase difference between the two sources changes millions of times per second. The pattern shifts too fast to be seen. Unequal brightness only reduces the contrast of fringes. Waves from any sources superpose. A large separation changes the fringe spacing but would not remove the fringes.
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Answer A
Answer: A
Common error: Equal amplitude is not a condition for coherence. Coherence needs a constant phase difference; equal amplitudes only make the minima completely dark.
In a medical endoscope, light passes from air into the glass core of an optical fibre. What happens to the frequency, the wavelength and the speed of the light?
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The frequency is fixed by the source, so it does not change at the boundary. Glass has a higher refractive index than air, so the speed decreases (v = c/n). From v = fλ with f constant, the wavelength must decrease by the same factor as the speed.
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Answer A
Answer: A
Common error: Frequency never changes on refraction. If speed changes, it is the wavelength that adjusts.
A ray of light in air strikes the flat surface of a new transparent material for smartphone screens. The angle of incidence is 60°, and the angle of refraction is 30°. What is the refractive index of the material?
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From Snell's law, n = sin 60° / sin 30° = (√3/2)/(1/2) = √3 ≈ 1.73. The value 2 comes from dividing the angles themselves instead of their sines. The value 1/√3 inverts the ratio. The value √3/2 is just sin 60°.
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Answer A
Answer: A
Common error: Snell's law uses the ratio of the sines of the angles, never the ratio of the angles.
Ocean swell moves from deep water into the shallower water near a surfing beach. In the shallower water the waves travel more slowly. What happens to the frequency of the waves and to the separation of adjacent wavefronts?
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The frequency is set by the source, so it cannot change when the wave crosses into a new region. Each crest must arrive at the boundary at the same rate it leaves. From v = fλ with f constant, a lower speed means a shorter wavelength. The wavefronts are one wavelength apart, so they move closer together. Options A and B wrongly assume that slowing changes the frequency. Option D gets the direction of the wavelength change wrong.
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Answer C
Answer: C
Common error: When a wave changes medium, its frequency never changes. The speed and wavelength change together, in proportion.
14IBP2S-022-00-02Critical angle & TIRPaper 1AEasy1 mark
A medical endoscope carries light along glass fibres. Each fibre has a core surrounded by a cladding of lower refractive index. Which pair of conditions is necessary for total internal reflection at the core–cladding boundary?
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Total internal reflection needs two things. First, the light must be travelling in the medium of higher refractive index (the core) towards one of lower index (the cladding). Second, the angle of incidence must be greater than the critical angle. Light going from lower to higher index always refracts towards the normal and can never be totally internally reflected, which rules out A and D. Below the critical angle some light is refracted out of the core, which rules out C.
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Answer B
Answer: B
Exam tip: State both conditions when asked about TIR: the direction of travel (high n to low n) and the angle compared with the critical angle. Each is usually worth a mark.
A medical ultrasound transducer has a fixed diameter. The frequency of the ultrasound is increased, while its speed in tissue stays the same. What happens to the spreading of the beam, and why?
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From λ = v/f, a higher frequency at the same speed gives a shorter wavelength. The transducer face acts as an aperture. A smaller ratio of wavelength to aperture means less diffraction, so the beam spreads less and stays narrower. Option A reverses how diffraction depends on wavelength. Option B gets the wavelength change wrong. Amplitude has no effect on diffraction, so C is wrong.
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Answer D
Answer: D
Exam tip: Shorter wavelength means less diffraction for the same aperture. This is why high-frequency ultrasound gives narrower beams.
16IBP2S-022-01-01Critical angle & TIRPaper 1AEasy1 mark
A layered optical sensor contains two transparent materials. Material X has refractive index 1.2 and material Y has refractive index 1.6. What is the ratio \(\frac{\text{speed of light in X}}{\text{speed of light in Y}}\)?
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Since n = c/v, the speed is inversely proportional to refractive index. So vX/vY = nY/nX = 1.6/1.2 = 4/3. Light travels faster in the material with the lower refractive index. Option A assumes speed is proportional to n. Options C and D wrongly square the ratio.
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Answer B
Answer: B
Common error: A higher refractive index means a lower speed. Check that the answer to a ratio question makes physical sense.
17IBP2S-022-01-02Path differencePaper 1AEasy1 mark
During a sound check in a sports hall, two loudspeakers emit coherent sound in phase with a wavelength of 0.60 m. A microphone is 4.20 m from one speaker and 5.10 m from the other. What is detected at the microphone, and why?
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The path difference is 5.10 − 4.20 = 0.90 m, which equals 1.5λ. For sources in phase, a path difference of (n + ½)λ makes the waves arrive in antiphase, giving destructive interference: a minimum. Option B confuses multiples of λ/2 with multiples of λ. Option C ignores path difference. Option D is wrong because interference occurs at every point, not only at equidistant points.
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Answer A
Answer: A
Exam tip: Always divide the path difference by λ. A whole number gives a maximum and a half-integer gives a minimum, provided the sources are in phase.
Two pulses travel towards each other along a stretched guitar string. One pulse has a maximum displacement of 3.0 cm upwards and the other has a maximum displacement of 5.0 cm downwards. What is the displacement of the string at the instant the points of maximum displacement of the two pulses coincide?
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By the principle of superposition, the resultant displacement is the algebraic sum of the individual displacements. Taking upwards as positive: +3.0 + (−5.0) = −2.0 cm, which is 2.0 cm downwards. Option A adds the magnitudes and ignores direction. Option B gets the magnitude right but the direction wrong. Option D assumes that opposite pulses always cancel completely, which happens only when their amplitudes are equal.
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Answer C
Answer: C
Exam tip: Treat displacement as a signed quantity. Choose upwards as positive and add the values algebraically.
Ocean swell heading towards a harbour passes from deep water into the shallower water at the harbour entrance. How do the frequency, speed and wavelength of the waves in the shallow water compare with their values in the deep water?
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The frequency is set by the source and cannot change at a boundary, because crests cannot be created or lost there. Water waves travel more slowly in shallower water. Since v = fλ with f fixed, the wavelength falls in proportion to the speed. Option B wrongly lets the frequency change. Option C ignores the change of speed. Option D wrongly assumes the frequency rises when the waves are compressed.
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Answer A
Answer: A
Common error: At any boundary the frequency stays the same. Work out the speed change first, then use v = fλ to find what happens to the wavelength.
The cover glass of a smartphone screen has refractive index 1.50. The speed of light in a vacuum is 3.0×108 m s−1. What is the speed of light in the cover glass?
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The refractive index is defined by n = c/v, so v = c/n = 3.0×108/1.50 = 2.0×108 m s−1. Option A multiplies by n instead of dividing. Option C assumes light keeps the same speed in every medium. Option D divides by n twice.
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Answer B
Answer: B
Exam tip: A material's refractive index is always greater than 1, so light is always slower in it than in a vacuum. Any answer above c must be wrong.
A climber ties a light cord to the end of a heavy rope. Waves travel more slowly in the rope than in the cord. A single upward pulse is sent along the cord towards the junction. What are the orientations of the reflected pulse and the transmitted pulse?
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When a pulse reaches a medium in which it travels more slowly (here the heavier rope), the reflected pulse is inverted, a phase change of π, similar to reflection from a fixed end. The transmitted pulse is never inverted, because the junction moves upward and drives the rope upward. Option A describes reflection into a faster medium. Options B and C wrongly invert the transmitted pulse.
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Answer D
Answer: D
Common error: Only the reflected pulse can be inverted. The transmitted pulse always keeps the same orientation as the incident pulse.
22IBP2S-022-03-09Young's double slitPaper 1AEasy1 mark
A dental curing lamp emits blue light of wavelength 480 nm in a vacuum. Inside the polymer light guide of the lamp, the wavelength of this light is 300 nm. What is the refractive index of the polymer?
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The frequency is unchanged, so n = c/v = λvacuum/λmedium = 480/300 = 1.60. Option B inverts the ratio, giving an index below 1, which is impossible for a real material. Option C squares the ratio. Option D argues that n = 1 because the frequency is the same, confusing frequency with wavelength.
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Answer A
Answer: A
Exam tip: Since f is constant, n = λ₀/λ. The vacuum wavelength always goes on the top, so n comes out greater than 1.
Monochromatic light of wavelength λ passes through two narrow slits and forms bright and dark fringes on a screen. Point P is at the centre of the second bright fringe beyond the central bright fringe. Point Q is at the centre of the second dark fringe from the central bright fringe, on the same side.
Which row gives the path difference at P and at Q?
Path difference at PPath difference at Q
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Step 1Bright fringes have path difference nλ; the central fringe is n = 0, so the second bright fringe beyond it, P, has 2λ.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Dark fringes have path difference (n + ½)λ, with n = 0 for the first, so the second dark fringe, Q, has 1.5λ.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: P has path difference 2λ (constructive) and Q has 1.5λ (destructive).
BP is right, but the dark fringes have been numbered from n = 1: the first dark fringe has path difference 0.5λ, so the second has 1.5λ.
CBoth fringes have been miscounted: the central bright fringe counted as the first, and the dark fringes numbered from n = 1.
DQ is right, but the central bright fringe (path difference zero) has been counted as the first bright fringe.
27C-1A-45Refraction & Snell's lawPaper 1AMedium1 mark
A plane wavefront travelling in air meets the flat surface of a glass block at an angle to the normal, and passes into the glass.
What happens to the spacing of the wavefronts and to the direction of the rays?
Spacing of the wavefrontsDirection of the rays
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Step 1The frequency cannot change at the boundary and light is slower in glass, so λ = v/f falls: the wavefronts are closer together.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The edge of each wavefront that enters the glass first slows first, which swings the wavefront, and the ray perpendicular to it, towards the normal.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe direction is right but the spacing is not: a slower wave of the same frequency has a shorter wavelength, so the wavefronts move closer together.
BThe spacing is right but the direction is not: light bends away from the normal only when it speeds up, on entering a medium of lower refractive index.
CThe spacing cannot stay the same: only the frequency is fixed at the boundary, and the wavelength changes in step with the speed.
DCorrect: the frequency is unchanged and the speed falls, so the wavefronts bunch up, and the side of each wavefront that slows first turns it towards the normal.
Plane waves of wavelength λ1 in medium 1 meet a straight boundary with medium 2 at an angle of incidence θ1. In medium 2 the wavelength is λ2 and the angle of refraction is θ2.
Which equation is correct?
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Step 1The frequency does not change at the boundary, so v2/v1 = fλ2/fλ1 = λ2/λ1.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Snell's law gives sin θ2/sin θ1 = v2/v1 = λ2/λ1.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: at constant frequency the speed is proportional to the wavelength, and sin θ2/sin θ1 = v2/v1.
BThe ratio is inverted. A wave that slows down (shorter wavelength) bends towards the normal, so θ2 must be smaller when λ2 < λ1.
CSnell's law relates the sines of the angles, not the angles themselves.
DThis applies Snell's law to the angles between the rays and the boundary. In Snell's law θ is the angle between the ray and the normal.
Two dippers in a ripple tank vibrate in phase with the same frequency. At point X the two waves arrive in phase and the water oscillates with an amplitude of 7.0 mm. At a nearby point Y the two waves arrive in antiphase and the amplitude is 1.0 mm. Each wave has the same amplitude at X as at Y.
What are the amplitudes of the two waves?
Larger amplitude / mmSmaller amplitude / mm
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Step 1At X the displacements add: A1 + A2 = 7.0 mm. At Y they are opposite: A1 − A2 = 1.0 mm.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Adding and subtracting the two equations: A1 = 4.0 mm and A2 = 3.0 mm.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
ABoth resultant amplitudes have been halved, as if each were twice the amplitude of one wave. Only the in-phase resultant is a sum; the antiphase resultant is a difference.
BEqual amplitudes of 3.5 mm would cancel completely at Y, giving zero, not 1.0 mm. The waves must have different amplitudes.
CCorrect: A1 + A2 = 7.0 mm and A1 − A2 = 1.0 mm, so the amplitudes are 4.0 mm and 3.0 mm.
DThese are the two resultant amplitudes, not the amplitudes of the separate waves. A 7.0 mm wave and a 1.0 mm wave would give 8.0 mm in phase.
30C-1A-74Reflection & transmissionPaper 1AMedium1 mark
A ray of light travelling in glass of refractive index 1.50 meets a flat glass–air boundary at an angle of incidence of 30°.
Which row gives the angle of reflection and the angle of refraction?
Angle of reflectionAngle of refraction
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Step 1The critical angle is sin−1(1/1.50) = 41.8°; 30° is smaller, so some light is reflected (at 30°) and some is transmitted into the air.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 21.50 sin 30° = 1.00 sin θ, so sin θ = 0.75 and θ = 48.6°.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
ASnell's law has been applied as if the light were entering the glass: sin θ = sin 30°/1.50. Leaving the glass, the light speeds up and bends away from the normal.
BCorrect: the angle of reflection equals the angle of incidence, 30°, and 1.50 sin 30° = sin θ gives θ = 48.6°.
CTotal internal reflection needs an angle of incidence greater than the critical angle, 41.8°. At 30° the light is partly reflected and partly transmitted.
DThe angle of reflection has been measured from the boundary, 90° − 30°. Like the angle of incidence, it is measured from the normal.
In a double-slit experiment, the laser is replaced by a white-light filament lamp with a narrow single slit in front of it, so that the single slit illuminates the double slit.
Which statement describes the pattern seen on the screen?
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Step 1At the centre the path difference is zero for every wavelength, so every colour interferes constructively there and the central fringe is white.
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Step 2The other bright fringes are at distances nλD/d from the centre, which are proportional to λ; violet (shortest wavelength) has its maxima nearest to the centre and red farthest.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AEach wavelength forms its own fringe pattern, because the single slit makes the light from the two slits coherent; the patterns overlap to give coloured fringes.
BThe order is reversed. The fringe separation s = λD/d is largest for red, so in each bright fringe red is farthest from the centre.
CAt the centre the path difference is zero for all wavelengths, so all colours are bright there together and the central fringe is white.
DCorrect: zero path difference makes the centre white; elsewhere the maxima are at distances proportional to λ, so violet is nearest the centre and red farthest.
32C-1A-80Reflection & transmissionPaper 1AMedium1 mark
A radar reflector on a lifeboat contains two flat metal plates joined along one edge at 90°. A microwave ray, travelling in a plane perpendicular to the join, meets plate 1 at an angle of incidence of 25°. It is reflected on to plate 2 and is reflected again.
Which row gives the angle of incidence at plate 2 and the total angle through which the ray has been turned by the two reflections?
Diagram NOT to scale
Angle of incidence at plate 2Total angle turned
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Step 1The normals to the two plates are perpendicular, so the ray reflected from plate 1 at 25° to its normal meets plate 2 at 90° − 25° = 65° to the normal of plate 2.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each reflection turns the ray through 180° − 2 × (angle of incidence): 130° at plate 1 and 50° at plate 2, in the same sense, so 180° in total — the ray travels back parallel to its original path.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis assumes that the angle of incidence is 25° at both plates, so that each reflection turns the ray through 130°. The normals to the plates are perpendicular, so the angle at plate 2 is 90° − 25° = 65°.
B65° is right, but only the first reflection has been counted (180° − 2 × 25° = 130°). The reflection at plate 2 turns the ray through a further 180° − 2 × 65° = 50°.
CCorrect: the ray meets plate 2 at 90° − 25° = 65° to the normal; the two reflections turn it through 130° and 50°, 180° in total, so it returns parallel to its original path, back towards the radar.
D50° is the angle between the incident and the reflected ray at plate 1 (2 × 25°). The angle of incidence at plate 2 is measured between the ray and the normal to plate 2.
An optical fibre used in a broadband connection has a core of refractive index 1.50 and a cladding of refractive index 1.40. What is the critical angle at the core–cladding boundary?
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At the critical angle, the refracted ray travels along the boundary (90°). Applying Snell's law gives ncore sin c = ncladding sin 90°, so sin c = 1.40/1.50 and c ≈ 69°. The expression sin−1(1/1.50) is the critical angle for a core–air boundary, which ignores the cladding. The expression sin−1(1/1.40) uses the wrong medium. Subtracting from 90° gives the angle measured from the boundary rather than from the normal.
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Answer A
Answer: A
Exam tip: For a boundary between two media, sin c = n₂/n₁, where n₁ is the denser medium the light starts in. Only use 1/n when the second medium is air.
34IBP2S-022-02-05Young's double slitPaper 1AMedium1 mark
A double-slit experiment with a laser gives a fringe spacing s on a screen in air. The whole apparatus, including the slits and the screen, is then submerged in a tank of water of refractive index 1.33. The slit separation and the slit-to-screen distance stay the same. What is the new fringe spacing?
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The fringe spacing is s = λD/d. In water, the frequency is unchanged but the speed falls by a factor of 1.33, so the wavelength becomes λ/1.33. Since D and d are unchanged, the fringe spacing also becomes s/1.33. Multiplying by 1.33 wrongly assumes the wavelength increases. Leaving s unchanged ignores the change of wavelength. Dividing by 1.332 applies the factor twice.
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Answer A
Answer: A
Common error: In a denser medium the wavelength decreases, not the frequency, so the fringes move closer together.
A spear-fisher standing on a rock looks down at an angle at a fish in calm water. Which statement correctly describes what happens?
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Light travels from water (higher n) into air (lower n), so it speeds up and bends away from the normal. The eye traces the ray back in a straight line, so the image of the fish appears closer to the surface than the real fish. The spear-fisher must aim below the image. Bending towards the normal would happen for light entering a denser medium. Refraction occurs whenever light crosses the surface at a non-zero angle of incidence.
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Answer A
Answer: A
Exam tip: Light going from a denser to a less dense medium bends away from the normal, so objects under water appear shallower than they are.
36IBP2S-022-02-09Young's double slitPaper 1AMedium1 mark
A dental curing lamp emits blue light of wavelength 450 nm in air. In the resin filling, the wavelength of this light is 300 nm. The speed of light in air is 3.0 × 108 m s−1. What is the speed of the light in the resin?
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The frequency is unchanged, so speed is proportional to wavelength. The speed in the resin is 3.0 × 108 × 300/450 = 2.0 × 108 m s−1, which corresponds to n = 1.5. The answer 4.5 × 108 inverts the ratio, giving a speed greater than c. The answer 3.0 × 108 assumes the speed does not change. The answer 1.3 × 108 divides by n2.
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Answer A
Answer: A
Exam tip: Since f is constant across a boundary, v₁/v₂ = λ₁/λ₂ = n₂/n₁.
37IBP2S-022-00-03Critical angle & TIRPaper 1AMedium1 mark
A cut diamond (refractive index 2.4) sparkles because light is totally internally reflected inside it. The diamond is now placed in water (refractive index 1.3) instead of air. What happens to the critical angle at the diamond surface, and what is the effect?
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The critical angle obeys sin c = noutside/ndiamond. In air this is 1.0/2.4. In water it is 1.3/2.4, which is larger, so the critical angle increases. Fewer internal rays now exceed the critical angle, so more light escapes and the diamond sparkles less. Option C forgets that the surrounding medium appears in the ratio. Option D gets the effect of a larger critical angle the wrong way round.
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Answer A
Answer: A
Common error: The formula sin c = 1/n only applies when the outside medium is air. In general, sin c = n₂/n₁.
A person stands in a corridor, out of line of sight of an open doorway 0.9 m wide. They can clearly hear a band playing in the room but cannot see the band. Which statement best explains this?
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Diffraction is significant when the wavelength is similar to the width of the gap. Audible sound has wavelengths from a few centimetres to several metres, so it spreads widely through a 0.9 m doorway. Visible light has wavelengths around 5×10−7 m, so its diffraction through the doorway is negligible. All waves diffract, whether longitudinal or transverse, so A is wrong. Wave speed alone does not control spreading, so B is wrong. Sound has a much lower frequency than light, so C is wrong.
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Answer D
Answer: D
Exam tip: In diffraction explanations, always compare the wavelength with the aperture or obstacle size.
39IBP2S-022-00-06Critical angle & TIRPaper 1AMedium1 mark
A car dashboard uses an acrylic light pipe of refractive index n1, coated with a transparent polymer of refractive index n2, where n2 < n1. Light travels along the pipe by total internal reflection at the acrylic–polymer boundary. What is the sine of the critical angle at this boundary?
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At the critical angle, the refracted angle is 90°. Snell's law gives n1 sin c = n2 sin 90°, so sin c = n2/n1. Option A is the critical angle for an acrylic–air boundary, which ignores the coating. Option C is greater than 1, so it is impossible. Option D treats the coating as the denser medium.
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Answer B
Answer: B
Exam tip: Derive the critical angle from n₁ sin θ₁ = n₂ sin θ₂ with θ₂ = 90°. A sine greater than 1 shows the ratio is upside down.
40IBP2S-022-00-08Critical angle & TIRPaper 1AMedium1 mark
A diver looking straight up from below a calm water surface sees the sky through a bright circle. Outside this circle the surface looks like a mirror. The water is replaced by a transparent liquid of greater refractive index, and the diver stays at the same depth. What happens to the bright circle?
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The edge of the circle corresponds to rays meeting the surface at the critical angle c, so the circle's radius is r = h tan c. A larger refractive index gives a smaller value of sin c = 1/n, so c decreases. At the same depth, r therefore decreases. Beyond the edge, light from below is totally internally reflected, which is why the surface looks like a mirror there. Option A gets the effect on c the wrong way round. Option B ignores the dependence on c. Option D gives the right size change for the wrong reason.
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Answer C
Answer: C
Exam tip: Link geometry to the critical angle: the boundary between transmission and TIR marks the edge of the visible region.
41IBP2S-022-01-03Young's double slitPaper 1AMedium1 mark
A Young's double-slit apparatus produces fringes with spacing s in air. The entire apparatus, including the source region, slits and screen, is then submerged in water of refractive index 1.33. The slit separation and slit-to-screen distance are unchanged. What is the new fringe spacing?
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In water the frequency is unchanged but the speed falls by a factor of 1.33, so the wavelength becomes λ/1.33. Since s = λD/d, the fringe spacing also falls by a factor of 1.33. Option A assumes the wavelength increases. Option B assumes the wavelength is fixed. Option D applies the factor twice.
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Answer C
Answer: C
Common error: Frequency is set by the source and never changes on entering a new medium. It is the wavelength that changes.
42IBP2S-022-01-04Refractive index & speedPaper 1AMedium1 mark
A jeweller is told that light travels through a gemstone at 1.25 × 108 m s−1. The speed of light in a vacuum is 3.00 × 108 m s−1. What is the refractive index of the gemstone?
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The refractive index is the ratio of the speed of light in a vacuum to the speed in the material: n = c/v = (3.00 × 108) / (1.25 × 108) = 2.40. Option A inverts the ratio (v/c) and gives n < 1, which is impossible for a material. Option B quotes the speed in the gemstone without its power of ten. Option C subtracts the speeds (3.00 − 1.25) instead of dividing them.
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Answer D
Answer: D
Exam tip: Refractive indices of materials are always greater than 1. If your answer is below 1, you have inverted the ratio.
43IBP2S-022-01-05Young's double slitPaper 1AMedium1 mark
Laser light has wavelength 600 nm and frequency 5.0 × 1014 Hz in air. It enters a plastic spectacle lens of refractive index 1.5. Which row gives the wavelength and frequency of the light inside the lens?
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The frequency is fixed by the source, so it stays at 5.0 × 1014 Hz. The speed falls by a factor of 1.5, and since v = fλ the wavelength also falls by a factor of 1.5: 600/1.5 = 400 nm. Option B multiplies by n instead of dividing. Options C and D wrongly reduce the frequency.
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Answer A
Answer: A
Common error: Do not divide both the frequency and the wavelength by n. Only the wavelength and the speed change.
44IBP2S-022-01-07Young's double slitPaper 1AMedium1 mark
A double-slit arrangement is illuminated with white light instead of monochromatic light. Which statement correctly describes the pattern seen on the screen?
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At the centre the path difference is zero for every wavelength, so all colours interfere constructively and the central fringe is white. Away from the centre, fringe spacing s = λD/d is smaller for shorter wavelengths, so violet/blue maxima lie closer to the centre than red maxima. This gives coloured fringes with violet on the inner edge. Option B confuses diffraction spread with the zero-order fringe. Options C and D ignore the dependence of fringe spacing on wavelength.
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Answer A
Answer: A
Exam tip: The zero-order maximum is always at zero path difference, whatever the wavelength. This is why it stays white.
45IBP2S-022-01-10Young's double slitPaper 1AMedium1 mark
In a double-slit experiment, each slit on its own produces light of intensity I at the centre of the screen. Both slits are now opened, and the light from the slits is coherent. What is the intensity at the centre of the central bright fringe?
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At the centre the waves arrive in phase, so their amplitudes add: the resultant amplitude is 2A. Intensity is proportional to amplitude squared, so the intensity becomes (2)2 × I = 4I. Option B adds the intensities, which is correct only for incoherent sources. Option A ignores superposition. Option D confuses the centre with a minimum.
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Answer C
Answer: C
Common error: Superpose amplitudes, not intensities. Then square the resultant amplitude to find the intensity.
46IBP2S-022-03-03Critical angle & TIRPaper 1AMedium1 mark
A medical endoscope uses optical fibres with a core of refractive index 1.52 and a cladding of refractive index 1.48. What is the critical angle at the core–cladding boundary?
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At the critical angle the refracted ray travels along the boundary, so ncore sin θc = nclad sin 90°. This gives sin θc = 1.48/1.52 = 0.974 and θc = 76.8°. Option A uses 1/1.52, which treats the core as surrounded by air. Option D uses 1/1.48. Option B is the complement of the correct angle, measured from the boundary instead of from the normal.
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Answer C
Answer: C
Common error: sin θc = 1/n only applies when the second medium is air. Between two materials, use sin θc = n₂/n₁, with n₁ the denser medium.
Ultrasound travels at 1540 m s−1 in soft tissue and at 3080 m s−1 in a bone. A beam in the soft tissue strikes the tissue–bone boundary. Which statement is correct?
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For any wave, Snell's law gives sin θ1/sin θ2 = v1/v2. The sound speeds up on entering bone, so the beam bends away from the normal and a critical angle exists. Setting θ2 = 90° gives sin θc = 1540/3080 = 0.50, so θc = 30°. Option B measures the angle from the boundary. Option C confuses mass density with wave speed: what matters is moving into the faster medium. Option D is false because total internal reflection happens with all waves.
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Answer A
Answer: A
Exam tip: Total internal reflection needs the wave to travel towards a medium where it moves faster. For sound this is usually the 'denser' material, which is the opposite of light.
48IBP2S-022-03-06Critical angle & TIRPaper 1AMedium1 mark
A diver is 2.0 m below the flat, calm surface of a pool (refractive index of water 1.33). Looking upward, she sees the whole sky compressed into a bright circle on the surface, directly above her. The critical angle for water is 48.8° and tan 48.8° = 1.14. What is the diameter of the bright circle?
Diagram not to scale. Bright circle seen at the surface.
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Light from the sky arriving at grazing incidence is refracted into the water at the critical angle. So all skylight reaches the diver within a cone of half-angle 48.8° about the vertical. The radius of the circle is 2.0 × tan 48.8° = 2.28 m, so the diameter is 4.6 m. Option A gives only the radius. Option B uses sin instead of tan. Option C divides the depth by tan θ instead of multiplying.
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Answer D
Answer: D
Exam tip: Draw the right-angled triangle: the depth is the side adjacent to the angle and the radius is the opposite side. Check whether the question asks for the radius or the diameter.
In a test of materials for a laser projector, a ray of light passes from transparent plastic A into transparent plastic B. The angle of incidence in A is 30° and the angle of refraction in B is 45°. Which statement is correct?
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Snell's law gives nA sin 30° = nB sin 45°, so nA/nB = sin 45°/sin 30° = √2. The ray bends away from the normal, so it is faster in B: vB/vA = nA/nB = √2. The frequency does not change, so the wavelength scales with speed: λB/λA = √2. Option A inverts the index ratio. Option B wrongly lets the frequency change. Option D inverts the speed ratio.
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Answer C
Answer: C
Exam tip: Wave speed and wavelength are both inversely proportional to n. The medium with the larger angle to the normal has the larger speed and the larger wavelength.
In a double-slit experiment the slit separation is d and the screen is a distance D from the slits. The distance between the centres of the second dark fringes on either side of the central bright fringe is y.
What is the wavelength of the light?
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Step 1Dark fringes lie midway between bright fringes, at 0.5s, 1.5s, … from the centre of the central bright fringe.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The second dark fringes are 1.5s either side of the centre, so y = 3s and s = y/3.
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Step 3s = λD/d gives λ = sd/D = yd/(3D).
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis places the dark fringes at whole numbers of fringe spacings, giving y = 4s. The first dark fringe is only half a spacing from the centre.
BCorrect: y = 3s and λ = sd/D, so λ = yd/(3D).
CThis takes y as the distance from the centre to one second dark fringe (1.5s); y spans both sides of the centre.
DThe slit separation and the screen distance have been swapped: s = λD/d, so λ = sd/D.
Two loudspeakers emit sound of the same frequency and amplitude, but exactly in antiphase. Point X is the same distance from both loudspeakers. At point Y the path difference is half a wavelength.
Which row describes the sound detected at X and at Y?
At XAt Y
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Step 1At X there is no path difference, so the waves arrive with the phase difference of the sources, π: destructive interference, a minimum.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2At Y the half-wavelength path difference adds a further π, making 2π: the waves arrive in phase, a maximum.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThese are the results for loudspeakers in phase; the antiphase of the sources reverses both.
BCorrect: the sources' phase difference of π gives a minimum at X, and the extra π from the half-wavelength path difference gives a maximum at Y.
CX is right, but at Y the phase difference from the path (π) has not been added to that of the sources (π); together they make 2π.
DThe antiphase of the sources has been used at Y but ignored at X, where the equal paths leave the waves in antiphase.
A double slit and a screen are fixed inside a glass tank. Monochromatic laser light passes through the slits and forms fringes of separation s on the screen. The tank is then filled with water of refractive index 1.33, without moving the slits or the screen.
What is (new fringe separation)/s?
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Step 1The frequency of the light is unchanged in the water but its speed falls to c/1.33, so the wavelength becomes λ/1.33.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2s = λD/d with D and d unchanged, so s is multiplied by 1/1.33 = 0.75.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis divides by the refractive index twice (1/1.332). Only the wavelength changes; D and d are fixed.
BCorrect: the wavelength in water is λ/1.33, and s ∝ λ, so the ratio is 1/1.33 = 0.75.
CThe frequency is unchanged, but the wavelength is not: the light travels more slowly in water, so λ = v/f is shorter.
DThis is the inverse ratio. Light travels more slowly in water, so its wavelength, and the fringe separation, decrease.
Two small dippers in a ripple tank are 6.0 cm apart and vibrate in phase, producing waves of wavelength 2.5 cm.
How many points of maximum disturbance are there on the straight line between the two dippers?
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Step 1Moving a distance x from the midpoint towards one dipper increases one path and decreases the other, so the path difference changes by 2x: between the dippers it ranges from −6.0 cm to +6.0 cm, i.e. −2.4λ to +2.4λ.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Maxima occur where the path difference is 0, ±λ or ±2λ: 5 points (half a wavelength apart).
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis counts the maxima on one side of the midpoint only, and leaves out the central maximum: there are 5 in all.
BThis assumes adjacent maxima are a whole wavelength apart along the line. Moving x changes the path difference by 2x, so adjacent maxima are only λ/2 apart.
CThis is the number of minima, where the path difference is ±0.5λ and ±1.5λ.
DCorrect: the path difference runs from −2.4λ to +2.4λ, so it equals 0, ±λ and ±2λ at 5 points.
Light passes from air into glass of refractive index 1.50. The angle of incidence θ1 is increased from 0° to 90°, and the angle of refraction θ2 is measured.
Which graph shows how θ2 varies with θ1?
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Step 1sin θ2 = sin θ1/1.50, so θ2 < θ1 at every angle and the graph starts at the origin with gradient 1/1.50.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2As θ1 → 90°, sin θ2 → 1/1.50: the curve flattens and ends at θ2 = 42°, the critical angle — graph B.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis applies Snell's law to the angles themselves, θ2 = θ1/1.50, giving a straight line to 60°. The sines, not the angles, are proportional.
BCorrect: θ2 = sin−1(sin θ1/1.50) rises ever more slowly and reaches 42° (the critical angle) at grazing incidence.
CThis is light going from glass into air, sin θ2 = 1.50 sin θ1: the refracted angle reaches 90° at θ1 = 42°. Light entering glass bends towards the normal.
DThis measures both angles from the boundary instead of from the normal, so a ray along the normal would appear refracted at 48°. A ray along the normal is not deviated.
The speed of sound is 340 m s−1 in air and 1480 m s−1 in water. Total internal reflection of sound can occur at the flat surface of a calm lake.
Which row gives the direction in which the sound must travel to the surface for this to happen, and the critical angle?
Sound travelsCritical angle
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Step 1Total internal reflection can occur only when the wave travels towards a medium in which it moves faster, so that it bends away from the normal. For sound that medium is the water: the sound must travel from the air.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2n1/n2 = sin θ2/sin θ1 = v2/v1 with θ2 = 90°: sin c = 340/1480 = 0.230, so c = 13°.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: sound is faster in water, so it bends away from the normal on entering the water; sin c = 340/1480 gives c = 13°.
BThis copies the case of light, which is slower in water. Sound is faster in water, so sound going from water into air bends towards the normal and is always partly transmitted.
CThe direction is right, but 77° is the angle between the ray and the water surface. The critical angle is measured from the normal.
DBoth are wrong: sound travelling from water into air slows down and cannot be totally internally reflected, and 77° is measured from the surface, not from the normal.
The critical angle for light travelling in medium X towards a boundary with air is 35°. For light travelling in medium Y towards a boundary with air it is 50°.
What is the critical angle for light travelling in X towards a boundary with Y?
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Step 1nX = 1/sin 35° = 1.743 and nY = 1/sin 50° = 1.305.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2For X → Y: sin c = nY/nX = sin 35°/sin 50° = 0.749, so c = 48.5°.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
ACritical angles cannot be subtracted: it is the refractive indices (the reciprocals of the sines) that combine, as a ratio.
BThis multiplies the two sines, sin 35° × sin 50°. The ratio nY/nX equals sin 35° ÷ sin 50°.
CThis is the angle between the critical ray and the boundary, 90° − 48.5°; the critical angle is measured from the normal.
DCorrect: sin c = nY/nX = sin 35°/sin 50° = 0.749, so c = 48.5°.
58IBP2S-022-02-06Path differencePaper 1AHard1 mark
An engineer testing an audio system accidentally reverses the wires to one of two identical loudspeakers, so that they emit in antiphase at the same frequency. Point X is the same distance from both loudspeakers. At point Y, the path difference is half a wavelength. What is heard at X and at Y?
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The sources start with a phase difference of π. At X the path difference is zero, so no extra phase difference is added and the waves arrive in antiphase: this is a minimum. At Y the path difference of λ/2 adds a further π, giving a total of 2π. The waves arrive in phase, so Y is a maximum. Option B is the pattern for in-phase sources, which forgets the reversed wiring.
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Answer A
Answer: A
Exam tip: For antiphase sources, swap the usual conditions. A whole number of wavelengths gives a minimum, and a whole number + ½ gives a maximum.
59IBP2S-022-02-10Path differencePaper 1AHard1 mark
Two small loudspeakers A and B emit sound in phase at the same frequency. They are 3.0 wavelengths apart. A microphone is moved slowly along the straight line from A to B. How many minima does it detect between A and B?
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Let the microphone be a distance x from A. The path difference is x − (3λ − x) = 2x − 3λ, which varies continuously from −3λ to +3λ. Minima occur where the path difference equals ±0.5λ, ±1.5λ or ±2.5λ. That is six positions, spaced λ/2 apart. The answer 5 counts the maxima between the speakers (0, ±λ, ±2λ). The answer 3 counts only one sign of path difference. The answer 7 wrongly includes an extra half-integer value.
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Answer A
Answer: A
Exam tip: Write the path difference as a function of position, find its full range, and then count every half-integer multiple of λ within that range, both positive and negative.
An LED strip along the edge of an aquarium wall sends light inside the glass (refractive index 1.50). A ray in the glass strikes the glass–water boundary (refractive index of water 1.33) at an angle of incidence of 60°. What happens at this boundary?
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The glass–water critical angle is sin−1(1.33/1.50) = 62.5°. Since 60° is smaller than this, total internal reflection does not occur. Snell's law gives sin r = 1.50 sin 60°/1.33 = 0.977, so r = 77.6° ≈ 78°, bending away from the normal. Partial reflection always accompanies refraction. Option A uses the glass–air critical angle. Option C bends the ray towards the normal. Option D ignores the change in wave speed.
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Answer B
Answer: B
Common error: The critical angle depends on both media. Recalculate it for each pair of materials instead of reusing the value for air.
Light travelling in water (refractive index 1.33) passes into a horizontal glass plate (refractive index 1.50) floating on the water. Above the plate is air. As the angle of incidence in the water is gradually increased, light eventually stops emerging into the air. At what angle of incidence in the water does this first happen?
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The boundaries are parallel, so n sin θ is the same at each one: 1.33 sin θwater = 1.50 sin θglass = 1.00 sin θair. Light just fails to emerge when θair = 90°, so sin θwater = 1/1.33 and θwater = 48.8°. The glass has no effect on this result. Option A is the glass–air critical angle, but that is an angle in the glass, not in the water. Option B is the glass–water critical angle, which is irrelevant here because light moves into the glass. Option D overlooks the glass–air boundary.
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Answer C
Answer: C
Exam tip: For parallel layers, n sin θ stays constant through every layer. You can link the first and last media directly and skip the middle layers.
A laser beam in air is directed at the flat surface of water in a transparent tank. The student marks the path of the beam in the water on a card held vertically in the tank and measures the angle of incidence i and the angle of refraction r with a protractor. Each angle has an uncertainty of ±1°.
i / ° ± 1°
r / ° ± 1°
sin i
sin r
15
11
0.259
0.191
25
18
0.423
0.309
35
25
0.574
0.423
45
32
55
38
0.819
0.616
65
43
0.906
0.682
75
47
0.966
0.731
Graph drawn to scale
(a)
Show that the absolute uncertainty in sin r for r = 32° is about 0.015.
(1)
(b)
Complete the table for i = 45°.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the refractive index of water using the graph.
(2)
(e)
The accepted refractive index of water for this light is 1.333. Comment on the student's value.
(1)
(f)
Calculate the critical angle for light travelling from water into air, using your answer to (d).
(2)
(g)
Suggest one change to the procedure that would reduce the uncertainty in r.
(1)
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Part (a)
sin 33° − sin 32° = 0.5446 − 0.5299 = 0.0147
✓ 1
Accept (sin 33° − sin 31°)/2 = 0.0148. Must see the calculation AND an answer to at least 2 s.f.
Part (b)
sin 45° = 0.707 AND sin 32° = 0.530
✓ 1
Both needed. Accept 2 or 4 decimal places.
Part (c)
Single straight line through (or very close to) the origin and within all error bars
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Part (d)
n = sin i/sin r = gradient of the line, found from a large triangle
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n = 1.33
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Accept 1.30 to 1.37. No unit.
Part (e)
The value agrees with 1.333 to within about 1 % (well within the uncertainty suggested by the error bars), so it is consistent with the accepted value
✓ 1
Allow ECF from (d). Must quote a comparison.
Part (f)
sin c = 1/n
✓ 1
Accept n₁/n₂ = sin θ₂/sin θ₁ with θ₂ = 90°.
c = sin−1(1/1.33) = 48.8°
✓ 1
Accept 46.9° to 50.3°. Allow ECF from (d).
Part (g)
Mark two points far apart along the refracted ray on the card and join them to draw a long line before measuring the angle, OR use a larger protractor/semicircular scale with finer divisions
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Do not accept 'repeat the readings' without saying how this reduces the uncertainty.
Answers: (a) 0.0147 · (b) 0.707; 0.530 · (d) n ≈ 1.33 · (f) c ≈ 48.8° (the remaining parts are explanations — see the table above)
A student has five lasers of known wavelength λ and one double-slit slide. The slide is fixed at a distance D = (2.500 ± 0.005) m from a screen. For each laser the student measures the distance across five fringe separations, 5s, with a ruler. Each measurement of 5s has an uncertainty of ±0.5 mm.
The student's hypothesis is that the fringe separation s is proportional to λ.
λ / nm
5s / mm ± 0.5 mm
s / mm
405
17.1
3.42
450
19.0
3.80
520
22.1
4.42
594
25.3
5.06
638
27.1
5.42
Graph drawn to scale
(a)
State the absolute uncertainty in s.
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Outline how the graph supports the hypothesis.
(1)
(d)
Determine the gradient of the line of best fit.
(2)
(e)
Determine the separation d of the slits.
(2)
(f)
The slide is labelled d = 0.30 mm ± 2 %. The uncertainty in the student's value of d is ±3 %. Comment on the student's value.
(1)
(g)
A sixth laser, of unknown wavelength, gives 5s = (22.5 ± 0.5) mm with the same apparatus. Determine, using the graph, its wavelength and the absolute uncertainty in the wavelength.
(2)
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Part (a)
±0.1 mm «0.5 mm/5»
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Accept ±0.10 mm.
Part (b)
Single straight line through all the error bars, which, extended, passes through «or very close to» the origin
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Part (c)
A straight line through the origin «within the error bars» shows s ∝ λ
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OWTTE. Both 'straight' and 'through the origin' needed.
Part (d)
Uses a triangle covering more than half the line
✓ 1
gradient = 8.49 × 10−3 mm nm−1 «= 8.49 × 103»
✓ 1
Accept 8.23 × 10−3 to 8.74 × 10−3 mm nm−1. Unit or conversion to a pure number needed.
Part (e)
gradient = D/d, so d = D/gradient
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d = 2.500/(8.49 × 103) = 2.95 × 10−4 m
✓ 1
Allow ECF from (d). Award [2] for CNA. Accept 2.86 × 10−4 to 3.04 × 10−4 m.
Part (f)
Student: 0.295 mm ± 3 % = 0.286–0.303 mm; label: 0.294–0.306 mm. The ranges overlap, so the value is consistent with the label
✓ 1
Allow ECF from (e). Must refer to the uncertainties.
Part (g)
s = 4.50 mm, so λ = 4.50/8.49 × 10−3 = 530 nm
✓ 1
Allow reading from the graph and ECF from (d). Accept 520–545 nm.
Δs = 0.1 mm gives Δλ = 0.1/8.49 × 10−3 ≈ ±12 nm, so λ = (530 ± 12) nm
✓ 1
Accept ±10 to ±15 nm. Accept the range read from the graph at s = 4.40 mm and 4.60 mm.
Answers: (a) ±0.1 mm · (d) 8.49 × 10−3 mm nm−1 · (e) d = 2.95 × 10−4 m · (g) λ = (530 ± 12) nm (the remaining parts are explanations — see the table above)
64C-1B-08Young's double slitPaper 1BMedium11 marks
Light from a red laser passes through a double slit. The slit separation, measured with a travelling microscope, is d = (0.250 ± 0.005) mm. Fringes are seen on a screen at a distance D from the slits. For each value of D the student measures the distance across 10 fringe separations with a ruler whose uncertainty is ±1 mm, and calculates the fringe separation s.
D / m ± 0.01 m
10s / mm ± 1 mm
s / mm
1.00
26
2.6
1.40
37
3.7
1.80
47
4.7
2.20
58
5.8
2.60
68
6.8
3.00
78
7.8
Graph drawn to scale
(a)
State the absolute uncertainty in s and outline why the student measures across 10 fringe separations.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line of best fit.
(2)
(d)
Determine the wavelength λ of the light.
(2)
(e)
The percentage uncertainty in the gradient is 4 %. Determine the absolute uncertainty in λ.
(2)
(f)
The laser is labelled "650 nm ± 10 nm". Comment on the student's result.
(1)
(g)
Suggest, with a reason, one change that would reduce the uncertainty in λ.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
±0.1 mm
✓ 1
A single fringe separation is only a few mm, so measuring across 10 separations reduces the percentage (fractional) uncertainty by a factor of 10
✓ 1
OWTTE. Do not accept 'more accurate' alone.
Part (b)
Single straight line through all error bars (passing close to the origin)
✓ 1
Part (c)
Uses a triangle covering more than half the line
✓ 1
gradient = 2.60 mm m−1 = 2.60 × 10−3
✓ 1
Accept 2.55 to 2.67 mm m−1. No unit needed if converted (ratio).
Part (d)
s = (λ/d)D, so λ = gradient × d
✓ 1
λ = 2.60 × 10−3 × 0.250 × 10−3 = 6.50 × 10−7 m
✓ 1
Accept 6.4 × 10−7 to 6.7 × 10−7 m. Allow ECF from (c). Award [2] for CNA.
Part (e)
Δλ/λ = 4 % + 0.005/0.250 (2 %) = 6 %
✓ 1
Must add the uncertainty in d.
Δλ = 0.06 × 6.50 × 10−7 ≈ ±4 × 10−8 m, so λ = (6.5 ± 0.4) × 10−7 m
✓ 1
Allow ECF from (d). Accept ±35 nm to ±43 nm. MP2 includes matching the precision of value and uncertainty.
Part (f)
The ranges 611–689 nm and 640–660 nm overlap, so the result is consistent with the label
✓ 1
Allow ECF from (e).
Part (g)
The gradient (4 %) is the larger contribution, so measure across more fringe separations (e.g. 20) OR use larger values of D, OR measure d with a more precise instrument to reduce its 2 %
✓ 1
Change must be linked to the uncertainty it reduces.
Answers: (a) ±0.1 mm · (c) 2.60 mm m−1 · (d) λ ≈ 6.50 × 10−7 m · (e) Δλ ≈ ±4 × 10−8 m (the remaining parts are explanations — see the table above)
65C-1B-15Refractive index & speedPaper 1BMedium11 marks
Short pulses of infrared light from a laser are sent into reels of optical fibre of different lengths L. A fast photodiode at the far end of each fibre is connected to an oscilloscope, which is triggered by the laser. The time t between the trigger and the arrival of the pulse is read from the oscilloscope screen. The uncertainty in each time is ±10 ns, which is too small to show on the graph. The fibre lengths are those printed on the reels.
L / m
t / ns ± 10 ns
100
642
200
1126
300
1612
400
2102
500
2593
600
3084
Graph drawn to scale
(a)
Draw the line of best fit on the graph.
(1)
(b)
Determine the gradient of the line of best fit, in s m−1.
(2)
(c)
Determine the refractive index n of the core of the fibre.
(2)
(d)
The lines of maximum and minimum gradient consistent with the uncertainties have gradients 4.924 ns m−1 and 4.845 ns m−1. Determine the absolute uncertainty in n.
(2)
(e)
The manufacturer states that n = 1.468 for the core. Comment on the student's result.
(1)
(f)
Suggest why the line of best fit does not pass through the origin.
(1)
(g)
The core is surrounded by cladding of refractive index 1.450. Calculate the critical angle at the core–cladding boundary.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Single straight line through all the points, not forced through the origin
✓ 1
Part (b)
Uses a triangle covering more than half the line
✓ 1
gradient = 4.89 × 10−9 s m−1
✓ 1
Accept 4.80 × 10−9 to 4.98 × 10−9 s m−1. Must be in s m−1.
Part (c)
Gradient = 1/v, so v = 1/gradient AND n = c/v = c × gradient
✓ 1
n = 3.00 × 108 × 4.89 × 10−9 = 1.47
✓ 1
Accept 1.44 to 1.49. Allow ECF from (b). Award [2] for CNA.
66C-1B-23Critical angle & TIRPaper 1BMedium11 marks
A student builds a simple refractometer to measure the sugar content of drinks. A semicircular dish with thin, transparent walls is filled with a sugar solution. A laser beam is directed through the curved wall towards the centre of the flat wall, and the dish is rotated on a turntable until the beam refracted into the air at the flat wall just disappears. The angle of incidence at this moment is the critical angle c, read from the turntable scale with an uncertainty of ±0.3°. The thin flat wall does not affect the critical angle, so the refractive index of the solution is n = 1/sin c.
The student uses six solutions of known concentration (grams of sugar per 100 g of solution, %). The graph shows n against concentration with error bars.
concentration / %
c / ° ± 0.3°
n
0
48.5
1.3352
8
48.1
1.3435
16
47.4
24
46.9
1.3696
32
46.2
1.3855
40
45.5
1.4020
Graph drawn to scale
(a)
Show that the absolute uncertainty in n for the 0 % solution is about 0.006.
(1)
(b)
Complete the table for the 16 % solution.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line of best fit, including its unit.
(2)
(e)
The accepted refractive index of pure water is 1.333. Comment on the intercept of your line.
(1)
(f)
For a sports drink, the student measures c = 48.2° ± 0.3°. Determine, using your line, the sugar concentration of the drink and its absolute uncertainty.
(3)
(g)
The label of the drink states 6.0 g of sugar per 100 g. Comment on whether the refractometer can test this claim, and suggest one improvement to the method.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1/sin 48.2° − 1/sin 48.5° = 0.0062 OR half the range = 0.0062 «≈ 0.006»
✓ 1
Must see the calculation AND an answer to at least 2 s.f.
Part (b)
n = 1/sin 47.4° = 1.3585
✓ 1
Accept 3 to 5 decimal places.
Part (c)
Single straight line passing through all the error bars
✓ 1
Do not force through a particular intercept.
Part (d)
Uses a triangle covering more than half the line
✓ 1
gradient = 0.00168 %−1 «≈ 1.68 × 10−3 per %»
✓ 1
Accept 0.00160 to 0.00177 %−1. The unit (%−1 or per %) is required.
Part (e)
Intercept ≈ 1.332, which agrees with 1.333 within the uncertainty in n (±0.006), so the method gives the accepted value for water «no evidence of a systematic error»
✓ 1
Allow ECF from (c). Must quote the intercept and compare.
Part (f)
n = 1/sin 48.2° = 1.3414
✓ 1
concentration = (1.3414 − 1.332)/0.00168 = 5.6 %
✓ 1
Allow reading from the graph and ECF from (d) and (e). Accept 4.1 to 7.1 %.
Δn ≈ ±0.006 gives an uncertainty of ±0.006/0.00168 ≈ ±4 %, so concentration = (6 ± 4) %
✓ 1
Accept ±3 % to ±5 %. MP3 is for matching the precision of value and uncertainty.
Part (g)
6.0 % lies within the range, so the label is consistent with the result, but the uncertainty «±4 %» is almost as large as the value, so the refractometer cannot test the claim precisely
✓ 1
Allow ECF from (f).
Reduce the uncertainty in c: e.g. repeat the reading of c several times and take the mean, use a scale with finer divisions / a vernier, or detect the disappearance of the refracted beam with a light sensor
✓ 1
The improvement must reduce the uncertainty in c (or n). Do not accept 'use more solutions' alone.
Answers: (a) 0.0062 · (b) 1.3585 · (d) 0.00168 %−1 · (e) 1.332 · (f) (6 ± 4) % (the remaining parts are explanations — see the table above)
67C-1B-33Refraction & Snell's lawPaper 1BMedium10 marks
A student tests Snell's law for water waves. A glass plate lies on the floor of a ripple tank so that the water is 12 mm deep on one side of its straight edge and 3 mm deep above it. A straight dipper vibrating at 5.0 Hz sends plane waves across the edge (Figure 1).
For six orientations of the dipper the student photographs the waves with a ruler in the photograph and measures, with a protractor, the angle θ1 between the ray and the normal to the edge in the deep water and the angle θ2 in the shallow water. The uncertainty in each angle is ±2°.
On one photograph the student also measures the distance across four wavelengths: (233 ± 2) mm in the deep water and (145 ± 2) mm in the shallow water.
θ1 / ° ± 2°
θ2 / ° ± 2°
sin θ1
sin θ2
20
12
0.342
0.208
30
18
0.500
0.309
40
23
0.643
0.391
50
28
60
33
0.866
0.545
70
36
0.940
0.588
Figure 1 — plan view, not to scaleFigure 2 — graph drawn to scale; error bars show the uncertainty in sin θ₂
(a)
Outline why the angle between a wavefront and the edge of the plate is equal to the angle between the ray and the normal to the edge.
(1)
(b)
Complete the table for θ1 = 50°.
(1)
(c)
Show that the absolute uncertainty in sin θ2 for θ1 = 40° is about 0.03.
(1)
(d)
Plot the point for θ1 = 50° and draw the line of best fit on Figure 2.
(1)
(e)
Determine the gradient of the line of best fit.
(2)
(f)
Determine the ratio (wavelength in shallow water)/(wavelength in deep water) and its absolute uncertainty.
(2)
(g)
Deduce whether the student's results support Snell's law for these water waves.
(1)
(h)
Suggest one improvement to the measurement of the wavelengths.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The ray is perpendicular to the wavefront and the normal is perpendicular to the edge, so both lines are turned through 90°: the angle between them is unchanged
✓ 1
OWTTE. Accept a labelled sketch.
Part (b)
sin 50° = 0.766 AND sin 28° = 0.469
✓ 1
Both needed. Accept 2 or 4 d.p.
Part (c)
sin 25° − sin 23° = 0.4226 − 0.3907 = 0.0319 «≈ 0.03»
✓ 1
Accept half the range, (sin 25° − sin 21°)/2 = 0.0321. Must see full substitution OR answer to at least 3 s.f.
Part (d)
Point plotted at (0.766, 0.469) AND a single straight line through all the error bars, passing close to the origin
✓ 1
Allow ECF from (b) for the plotted point.
Part (e)
Uses a triangle covering more than half the line
✓ 1
gradient = 0.62
✓ 1
Accept 0.59–0.65. No unit.
Part (f)
λ2/λ1 = 145/233 = 0.622
✓ 1
fractional uncertainty = 2/233 + 2/145 = 0.022, so ratio = 0.62 ± 0.01
✓ 1
Accept ± 0.01 to ± 0.02. Award [2] for CNA. MP2 includes matching the precision of value and uncertainty.
Part (g)
Snell's law predicts sin θ2/sin θ1 = v2/v1 = λ2/λ1 «the frequency is the same in both regions»; the gradient «≈ 0.62» agrees with λ₂/λ₁ = 0.62 ± 0.01 within the uncertainties «lines through the error bars allow gradients from about 0.53 to 0.74», so the results support Snell's law
✓ 1
Allow ECF from (e) and (f). The link between the gradient and the wavelength ratio must be seen.
Part (h)
Measure across more wavelengths «e.g. eight» / use a photograph of larger scale, so that the percentage uncertainty in each wavelength is smaller
✓ 1
The improvement must reduce the uncertainty in the wavelengths. Do not accept 'repeat the experiment' alone.
Answers: (b) 0.766, 0.469 · (e) gradient ≈ 0.62 · (f) 0.62 ± 0.01 (the remaining parts are explanations — see the table above)
68C-1B-35Reflection & transmissionPaper 1BMedium10 marks
A narrow beam of light enters a semicircular glass block through its curved surface, along a radius, so that it meets the centre of the flat surface from inside the glass at an angle of incidence θ. The block stands on a turntable whose scale is read to ±0.5°.
A light sensor measures the power of the beam reflected back into the glass at the flat surface. The student calculates the reflected fraction R = (reflected power)/(power arriving at the flat surface), with an uncertainty of ±0.02.
θ / ° ± 0.5°
10
20
30
34
37
39
40
40.5
41.5
42.5
45
50
R ± 0.02
0.04
0.05
0.07
0.09
0.13
0.23
0.32
0.43
1.00
1.00
0.99
1.00
Graph drawn to scale
(a)
Assuming that no light is absorbed, state the fraction of the light that is transmitted into the air at θ = 30°.
(1)
(b)
Draw the curve of best fit on the graph.
(1)
(c)
Outline how the graph shows that total internal reflection occurs above a certain angle.
(1)
(d)
Estimate, using the graph, the critical angle c and its absolute uncertainty.
(2)
(e)
Determine the refractive index of the glass and its absolute uncertainty.
(3)
(f)
The manufacturer states that the refractive index of the glass is 1.52. Comment on the student's result.
(1)
(g)
Suggest one change to the procedure that would reduce the uncertainty in c.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1 − 0.07 = 0.93
✓ 1
Accept 0.91–0.95 or a percentage.
Part (b)
Smooth curve through all the error bars, almost flat below about 34°, rising ever more steeply and reaching R = 1 between 40.5° and 41.5°, then horizontal at R = 1
✓ 1
Do not credit a straight line joining the points or a curve that overshoots 1.
Part (c)
R = 1 «all the light is reflected, none is transmitted» for every angle above about 41°
✓ 1
OWTTE.
Part (d)
c = 41.0° «where the curve first reaches R = 1»
✓ 1
Accept 40.6°–41.4°. Allow ECF from the candidate's curve.
± 0.5° «the curve reaches 1 between the readings at 40.5° and 41.5° / scale uncertainty»
✓ 1
Accept ±0.4° to ±1°.
Part (e)
n = 1/sin c
✓ 1
n = 1/sin 41.0° = 1.524
✓ 1
Allow ECF from (d).
1/sin 40.5° = 1.540 and 1/sin 41.5° = 1.509, so n = 1.52 ± 0.02
✓ 1
Accept the largest deviation from the best value. MP3 is for matching the precision of value and uncertainty.
Part (f)
1.52 lies within the range 1.51–1.54, so the result agrees with the manufacturer's value
✓ 1
Allow ECF from (e). Must refer to the uncertainty.
Part (g)
Take readings at smaller intervals of θ «e.g. every 0.2°» between about 40° and 42°, where R changes most rapidly / use a turntable with a finer scale «vernier»
✓ 1
Do not accept 'repeat the readings' alone.
Answers: (a) 0.93 · (d) c = (41.0 ± 0.5)° · (e) n = 1.52 ± 0.02 (the remaining parts are explanations — see the table above)
69C-1B-37Critical angle & TIRPaper 1BMedium11 marks
Acrylic rods are used as light guides. A student shines light along a straight rod of diameter D and bends it round cylinders of decreasing radius R (Figure 1). R is the radius of the inner surface of the bend. The student records the largest radius at which light is first seen escaping from the outer surface of the bend. D is measured with callipers (±0.02 mm); the cylinders are available in steps of 0.5 mm and the uncertainty in each recorded R is ±1.0 mm.
A simple model, considering the ray that grazes the inner surface, predicts that light escapes when R < D/(n − 1), where n is the refractive index of the acrylic.
D / mm ± 0.02 mm
R / mm ± 1.0 mm
3.00
5.5
4.00
7.0
5.00
9.5
6.00
11.5
8.00
15.5
10.00
19.5
Figure 1 — not to scaleFigure 2 — graph drawn to scale
(a)
Explain, with reference to the critical angle, why light escapes from the rod when it is bent too tightly.
(2)
(b)
Draw the line of best fit on Figure 2.
(1)
(c)
Determine the gradient of the line of best fit.
(2)
(d)
Determine, using your gradient, the refractive index of the acrylic.
(2)
(e)
The lines of maximum and minimum gradient through the error bars have gradients 2.29 and 1.75. Determine the absolute uncertainty in the refractive index.
(2)
(f)
The accepted refractive index of acrylic is 1.49. Comment on the student's result.
(1)
(g)
The line of best fit does not pass through the origin. Suggest one reason for this.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The tighter the bend «smaller R», the smaller the angle of incidence at which light meets the outer surface
✓ 1
OWTTE.
When the angle of incidence falls below the critical angle, total internal reflection no longer occurs and light is refracted out of the rod
✓ 1
Do not accept 'light hits the surface at the critical angle' without the comparison.
Part (b)
Single straight line through all the error bars, not forced through the origin
✓ 1
Part (c)
Uses a triangle covering more than half the line
✓ 1
gradient = 2.03
✓ 1
Accept 1.95–2.11. No unit.
Part (d)
gradient = 1/(n − 1), so n = 1 + 1/gradient
✓ 1
n = 1 + 1/2.03 = 1.49
✓ 1
Allow ECF from (c). Award [2] for CNA.
Part (e)
n = 1 + 1/2.29 = 1.438 AND n = 1 + 1/1.75 = 1.571
✓ 1
Δn = (1.571 − 1.438)/2 ≈ ±0.07
✓ 1
Accept ±0.05 to ±0.09.
Part (f)
1.49 lies within 1.43–1.56, so the result is consistent with the accepted value
✓ 1
Allow ECF from (d) and (e). Must refer to the uncertainty.
Part (g)
Systematic error: only a little light escapes just below the critical radius, so the student notices the escaping light only at a radius smaller than D/(n − 1) «every R is recorded too small by about the same amount»
✓ 1
Accept a constant error in R, e.g. R measured to the wrong surface of the cylinder. Do not accept 'random error'.
Answers: (c) 2.03 · (d) n = 1.49 · (e) ±0.07 (the remaining parts are explanations — see the table above)
A student uses a green laser and six double-slit slides with different slit separations d. The screen is fixed at a distance D = 2.400 m from the slits. For each slide the student measures the distance across 10 fringe separations, 10s, and calculates log d and log s.
The student's hypothesis is that the fringe separation s is inversely proportional to d.
d / mm
10s / mm ± 1 mm
log( d / mm )
log( s / mm )
0.100
128
−1.000
1.107
0.150
85
−0.824
0.929
0.200
64
−0.699
0.806
0.300
43
0.400
32
−0.398
0.505
0.500
26
−0.301
0.415
Graph drawn to scale
(a)
Complete the table for d = 0.300 mm.
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line of best fit.
(2)
(d)
Explain how the graph supports the hypothesis.
(2)
(e)
Determine the wavelength of the laser light using the graph.
(3)
(f)
Suggest, with a reason, one change that would improve the data for the slides with the largest d.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
log( d / mm ) = log 0.300 = −0.523
✓ 1
Accept 2 or 4 decimal places.
s = 4.3 mm, so log( s / mm ) = 0.633
✓ 1
Must divide 10s by 10 before taking the log.
Part (b)
Single straight line with points evenly distributed on both sides, extended to log( d / mm ) = 0
✓ 1
Part (c)
Uses a triangle covering more than half the line
✓ 1
gradient = −0.99
✓ 1
Accept −0.96 to −1.04.
Part (d)
A straight line on a log–log graph shows a power law, s = Kdn, with n equal to the gradient
✓ 1
OWTTE.
The gradient is −1 (within the precision of the graph), so s ∝ d−1, i.e. s is inversely proportional to d
✓ 1
Allow ECF from (c).
Part (e)
Intercept at log( d / mm ) = 0 read as 0.114
✓ 1
Accept a point on the line used in log s = −log d + log(λD).
λD = 100.114 = 1.299 mm2
✓ 1
Recognises that s × d = λD.
λ = 1.299/2400 mm = 5.41 × 10−7 m
✓ 1
Accept 527 nm to 557 nm. Allow ECF. Award [3] for CNA.
Part (f)
Increase D: for large d the fringes are closest together, so 10s is small and its percentage uncertainty is largest; a larger D spreads the fringes out
✓ 1
Accept 'measure across more fringes (e.g. 20)' with the same reason.
Answers: (a) −0.523; 0.633 · (c) −0.99 · (e) λ ≈ 5.41 × 10−7 m (the remaining parts are explanations — see the table above)
Two small loudspeakers, 0.80 m apart, are connected to the same signal generator so that they emit sound of the same frequency f in phase. The experiment is done on a playing field to avoid reflections. A microphone, about 2.5 m in front of the loudspeakers, is moved sideways from the position of maximum loudness on the centre line until it detects the first minimum. The distances r1 and r2 from the microphone to the two loudspeakers are then measured with a tape measure, each with an uncertainty of ±0.5 cm, and the path difference Δ = r2 − r1 is calculated.
The air temperature is 20 °C. The graph shows Δ against 1/f with error bars.
f / Hz
(1/f) / ms
r1 / cm ± 0.5 cm
r2 / cm ± 0.5 cm
Δ / cm
500
2.000
240.5
274.8
34.3
600
1.667
243.7
271.8
28.1
750
1.333
248.3
270.9
22.6
900
1.111
252.2
270.8
18.6
1200
0.833
252.9
267.3
14.4
1500
0.667
261.2
272.5
11.3
Graph drawn to scale
(a)
Explain why the microphone detects a minimum where the path difference is half a wavelength.
(2)
(b)
State the absolute uncertainty in Δ.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the speed of sound using the gradient of your line.
(3)
(e)
Draw the lines of maximum and minimum gradient that pass through all the error bars, and hence determine the absolute uncertainty in the speed of sound.
(2)
(f)
The accepted speed of sound in air at 20 °C is 343 m s−1. Comment on the student's result.
(1)
(g)
Suggest, with a reason, how the student could reduce the percentage uncertainty in Δ at the higher frequencies without changing the apparatus.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The loudspeakers are in phase, so a path difference of λ/2 means the two waves arrive at the microphone half a cycle out of phase «in antiphase / phase difference π»
✓ 1
Accept path difference = (n + ½)λ with n = 0.
The waves superpose destructively: their displacements «of similar amplitude» cancel, giving a minimum
✓ 1
OWTTE.
Part (b)
±1.0 cm «the uncertainties of r1 and r2 add»
✓ 1
Accept ±1 cm.
Part (c)
Single straight line through all the error bars, passing close to the origin
✓ 1
Part (d)
gradient = 17.1 cm ms−1 «= 171 m s−1»
✓ 1
Accept 16.7 to 17.5 cm ms−1. Must use a large triangle.
Δ = λ/2 = v/(2f), so gradient = v/2
✓ 1
v = 2 × 171 = 341 m s−1
✓ 1
Accept 334 to 350 m s−1. Allow ECF. Award [3] for CNA with gradient seen.
Part (e)
Lines through all error bars with gradients of about 18.7 and 15.7 cm ms−1, giving v = 375 m s−1 and 315 m s−1
✓ 1
Accept ±0.4 cm ms−1 on each gradient.
Δv = (375 − 315)/2 ≈ ±30 m s−1
✓ 1
Accept ±24 to ±36 m s−1. Accept the larger difference from the best value.
Part (f)
343 m s−1 lies within the range 311 to 371 m s−1, so the result is consistent with the accepted value
✓ 1
Allow ECF from (d) and (e). Must refer to the uncertainty.
Part (g)
Move the microphone to the second minimum, where Δ = 3λ/2: Δ is three times larger but its absolute uncertainty is the same, so the percentage uncertainty is a third as large
✓ 1
Accept any later minimum, (n + ½)λ, with the same reasoning. Do not accept 'repeat' alone.
Answers: (b) ±1.0 cm · (d) v ≈ 341 m s−1 · (e) Δv ≈ ±30 m s−1(the remaining parts are explanations — see the table above)
A student investigates road mirages: on a hot day the road far ahead seems to be covered with water. The student takes values of the refractive index n of dry air at atmospheric pressure, for yellow light, from an online physical-properties database. For a gas, (n − 1) is proportional to the density of the gas.
The student's hypothesis is that (n − 1) is inversely proportional to the absolute temperature T, as the ideal gas model predicts. The database values are given to ±0.1 in the second column.
T / K
(n − 1) × 106
log(T / K)
log((n − 1) × 106)
240
332.3
2.380
2.522
260
306.8
2.415
2.487
280
284.9
2.447
2.455
300
265.9
320
249.2
2.505
2.397
340
234.6
2.531
2.370
360
221.6
2.556
2.346
Graph drawn to scale; the uncertainties are too small to show
(a)
Complete the table for T = 300 K.
(1)
(b)
Plot the point for T = 300 K and draw the line of best fit.
(1)
(c)
Determine the gradient of the line of best fit.
(2)
(d)
Explain how the result of (c) supports the student's hypothesis.
(2)
(e)
In a simple model of a mirage, air at 300 K lies above a thin layer of air at 330 K next to the road, with a sharp boundary between them. Show that the critical angle for light travelling in the cooler air towards the boundary is about 89.6°.
(2)
(f)
The eyes of a driver are 1.20 m above the road. Light from the sky that is totally internally reflected at the boundary at the critical angle reaches the driver's eyes. Calculate the distance from the driver to the point where the mirage is seen.
(1)
(g)
The temperature of the hot layer is known only to ±5 K. Determine the absolute uncertainty in the distance in (f).
(2)
(h)
Suggest one limitation of the sharp-boundary model of the mirage.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
log 300 = 2.477 AND log 265.9 = 2.425
✓ 1
Both needed. Accept 3 or 4 d.p.
Part (b)
Point plotted correctly AND a single straight line through all the points
✓ 1
Allow ECF from (a).
Part (c)
Uses a triangle covering more than half the line
✓ 1
gradient = −1.00
✓ 1
Accept −0.97 to −1.03. Negative sign required.
Part (d)
A straight line on a log–log graph shows (n − 1) = kTp, where p is the gradient
✓ 1
OWTTE.
p = −1 «within the precision of the data», so (n − 1) ∝ 1/T; this matches the ideal gas model, in which the density at constant pressure is ρ = PM/(RT) ∝ 1/T
✓ 1
Allow ECF from (c).
Part (e)
n = 1 + 265.9 × 10−6 at 300 K and 1 + 241.7 × 10−6 at 330 K «from the graph or from (n − 1)T = constant»
✓ 1
Accept values read from the graph within ±1 × 10−6.
sin c = 1.0002417/1.0002659 = 0.9999758, so c = 89.60°
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (f)
angle to the road = 90° − 89.60° = 0.40°, so distance = 1.20/tan 0.40° = 173 m
Accept ±10 to ±16 m. Accept the larger difference from the value in (f).
Part (h)
The temperature, and so n, changes gradually with height, so the light bends along a curved path instead of reflecting at one boundary / the temperature of the layer is not uniform along the road
✓ 1
OWTTE.
Answers: (a) 2.477, 2.425 · (c) −1.00 · (e) c = 89.60° · (f) 173 m · (g) ±13 m (the remaining parts are explanations — see the table above)
A small lamp L is fixed to the floor of a swimming pool, 1.80 m below the flat surface of the water. The refractive index of the water is 1.33 and that of air is 1.00. The figure, drawn to scale, shows two rays, A and B, from L reaching the surface, and a point S on the surface 2.5 m horizontally from the point directly above L.
Diagram drawn to scale
(a)
(i)
Calculate the speed of light in the water.
(1)
(ii)
Show that the critical angle for the boundary between the water and air is about 49°.
(1)
(b)
Determine whether light travelling directly from L to S can leave the water at S.
(3)
(c)
Light from L leaves the water only through a circular area of the surface. Calculate the radius of this circle.
(2)
(d)
Ray A meets the surface at an angle of incidence of 30°. Calculate the angle of refraction of ray A in the air.
(2)
(e)
On the figure, draw the paths of rays A and B after they reach the surface.
(2)
(f)
Salt is dissolved in the water, which increases its refractive index. State the effect on the radius of the circle in (c).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = c/n = 3.00 × 108/1.33 = 2.26 × 108 m s−1
✓ 1
Accept 2.3 × 108 m s−1.
Part (a)(ii)
sin c = 1/1.33, so c = 48.8°
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)
Angle of incidence at S: tan θ = 2.5/1.80, so θ = 54°
✓ 1
54° is greater than the critical angle «48.8°»
✓ 1
Allow ECF from (a)(ii).
so the light is totally internally reflected at S and does not leave the water
✓ 1
MP3 only scores if an angle of incidence has been calculated.
Part (c)
r = 1.80 × tan c = 1.80 × tan 48.8°
✓ 1
Allow ECF from (a)(ii).
r = 2.1 m
✓ 1
Accept 2.0–2.1 m. Award [2] for CNA.
Part (d)
1.33 sin 30° = 1.00 sin θ
✓ 1
θ = 42°
✓ 1
Accept 41.7°. Award [2] for CNA.
Part (e)
A: refracted into the air, bent away from the normal «at about 42° to it»
✓ 1
Accept a weak reflected ray drawn as well. Allow ECF from (d).
B: totally internally reflected back into the water, with the angle of reflection equal to 60°
✓ 1
Do not credit a refracted ray for B.
Part (f)
The radius decreases «the critical angle is smaller»
✓ 1
Answers: (a)(i) 2.26 × 108 m s−1 · (a)(ii) 48.8° · (b) θ = 54° > c: no light leaves at S · (c) 2.1 m · (d) 42° (the remaining parts are explanations — see the table above)
74C-2-42Refraction & Snell's lawPaper 2Easy11 marks
Ocean swell of period 10.0 s approaches a coral reef. Over the deeper water the waves travel at 15.0 m s−1; over the shallow reef they travel at 9.0 m s−1. The edge of the reef is straight. Figure 1 is a plan view, drawn to scale, of the wavefronts in the deeper water, which meet the edge of the reef at an angle of incidence of 40°.
Figure 1 — Plan view drawn to scale
(a)
Outline the relationship between a ray and the wavefronts of a wave.
(1)
(b)
Calculate the wavelength of the swell in the deeper water.
(1)
(c)
(i)
Calculate the wavelength of the swell over the reef.
(1)
(ii)
Calculate the angle between a ray and the normal to the edge of the reef, over the reef.
(2)
(d)
On Figure 1, draw three wavefronts over the reef.
(3)
(e)
Explain, with reference to a wavefront, why the direction of travel of the swell changes at the edge of the reef.
(2)
(f)
Near a beach the water becomes gradually shallower. Suggest why waves that approach a beach at an angle usually arrive with their wavefronts almost parallel to the shoreline.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A ray is perpendicular to the wavefronts «and shows the direction in which the wave travels / energy is transferred»
✓ 1
Part (b)
λ = vT = 15.0 × 10.0 = 150 m
✓ 1
Part (c)(i)
λ = 9.0 × 10.0 = 90 m «the period is unchanged»
✓ 1
Part (c)(ii)
sin θ2 = (v2/v1) sin θ1 = (9.0/15.0) × sin 40°
✓ 1
θ2 = 22.7°
✓ 1
Award [2] for CNA. Accept 22–23°.
Part (d)
Three straight, parallel wavefronts over the reef that join the deep-water wavefronts at the edge of the reef
✓ 1
Do not award MP1 if the wavefronts are curved or not continuous at the edge.
Spacing between wavefronts smaller than in the deeper water «90 m: just under the length of the 100 m scale bar»
✓ 1
Accept spacing 80–100 m.
Wavefronts turned towards the edge of the reef: angle between each wavefront and the edge ≈ 23° «less than 40°»
✓ 1
Allow ECF from (c)(ii). Accept 19–27°.
Part (e)
One end of a wavefront reaches the shallow water first and slows down
✓ 1
while the rest of the wavefront still travels faster in the deeper water, so the wavefront turns «and the ray bends towards the normal»
✓ 1
OWTTE.
Part (f)
The waves slow down progressively as the depth decreases, so they keep refracting towards the normal to the shoreline
✓ 1
OWTTE.
Answers: (b) 150 m · (c)(i) 90 m · (c)(ii) 22.7° (the remaining parts are explanations — see the table above)
When a guitar string is plucked, the sound contains several harmonics. A microphone records the note from one string. In a simple model, the sound at the microphone consists of only two components: the first harmonic and the second harmonic.
Figure 1 shows how the displacement of the air at the microphone that each component would produce on its own varies with time t. The speed of sound in air is 340 m s−1.
Figure 1 — drawn to scale. Solid: first harmonic; dashed: second harmonic
(a)
Determine the frequency of each component.
(2)
(b)
On Figure 1, draw the resultant displacement of the air at the microphone from t = 0 to t = 8.0 ms.
(3)
(c)
Estimate, using your graph, the maximum displacement of the air at the microphone.
(1)
(d)
The vibrating length of the string is 0.640 m.
(i)
State the wavelength of the second harmonic on the string.
(1)
(ii)
Calculate the wavelength in air of the second-harmonic component of the sound.
(1)
(iii)
Explain why the wavelengths in (d)(i) and (d)(ii) are different.
(1)
(e)
The guitarist plucks the string at a different point, so that the second harmonic has the same amplitude as the first. Outline why the shape of the resultant changes but the pitch (frequency) of the note does not.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 8.0 ms and 4.0 ms «read from the graph»
✓ 1
f = 1/T: 125 Hz and 250 Hz
✓ 1
Award [2] for both frequencies correct.
Part (b)
Resultant found by adding the two displacements at each time «principle of superposition»: 0 at t = 0, 4.0 ms and 8.0 ms; +4.0 μm at 2.0 ms and −4.0 μm at 6.0 ms
✓ 1
Accept ±0.3 μm.
About +4.8 μm at 1.0 ms and +0.8 μm at 3.0 ms; about −0.8 μm at 5.0 ms and −4.8 μm at 7.0 ms
✓ 1
Accept ±0.3 μm.
Smooth curve, repeating once in 8.0 ms, with its peak in the first 2 ms and its trough in the last 2 ms
✓ 1
Do not credit a sinusoidal curve of period 8.0 ms and amplitude 6 μm.
Part (c)
5.2 μm «at t ≈ 1.3 ms»
✓ 1
Accept 4.9–5.4 μm. Allow ECF from the candidate's curve.
Part (d)(i)
λ = L = 0.640 m «two half-wavelengths fit the string»
✓ 1
Part (d)(ii)
λ = 340/250 = 1.36 m
✓ 1
Allow ECF from (a).
Part (d)(iii)
The frequency is the same, but the wave speed on the string is different from the speed of sound in air, so λ = v/f is different
✓ 1
OWTTE.
Part (e)
The displacements being added are different, so the resultant has a different shape
✓ 1
The resultant still repeats every 8.0 ms «the period of the first harmonic», so the frequency of the note is unchanged
✓ 1
OWTTE.
Answers: (a) 125 Hz and 250 Hz · (c) ≈ 5.2 μm · (d)(i) 0.640 m · (d)(ii) 1.36 m (the remaining parts are explanations — see the table above)
76C-2-64Reflection & transmissionPaper 2Easy11 marks
Glaciologists measure the thickness of a glacier with radar carried by a helicopter (Figure 1). The antenna sends short pulses of radio waves of frequency 60 MHz vertically downwards and detects the echoes. The refractive index of ice for these waves is 1.78.
For one pulse, the echo from the surface of the ice arrives 0.400 μs after the pulse is sent, and the echo from the rock bed arrives 3.10 μs after the pulse is sent.
Figure 1 — not to scale
(a)
Outline what happens to a radio pulse when it reaches the boundary between the air and the ice.
(2)
(b)
Calculate the height H of the antenna above the ice.
(1)
(c)
(i)
Calculate the speed of the radio waves in the ice.
(1)
(ii)
Determine the thickness h of the ice.
(2)
(d)
Calculate the wavelength of the radio waves in the ice.
(1)
(e)
Each pulse lasts 50 ns. Two echoes can be told apart only if one has finished arriving before the next begins.
(i)
Determine the smallest thickness of a layer within the ice whose top and bottom surfaces give echoes that can be told apart.
(2)
(ii)
Outline what the antenna detects when two echoes overlap in time.
(1)
(f)
Suggest one reason why the echo from the rock bed is much weaker than the echo from the ice surface.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Part of the pulse is reflected «back up, giving the first echo» and part is transmitted into the ice
✓ 1
The transmitted pulse travels more slowly in the ice, with the same frequency and a shorter wavelength
✓ 1
OWTTE.
Part (b)
H = 3.00 × 108 × 0.400 × 10−6/2 = 60.0 m
✓ 1
Do not accept 120 m.
Part (c)(i)
v = c/n = 3.00 × 108/1.78 = 1.69 × 108 m s−1
✓ 1
Part (c)(ii)
time in the ice «there and back» = 3.10 − 0.40 = 2.70 μs
✓ 1
h = 1.69 × 108 × 1.35 × 10−6 = 228 m
✓ 1
Accept 227–228 m. Allow ECF from (c)(i). Award [1] max for 455 m «time not halved».
Part (d)
λ = 1.69 × 108/60 × 106 = 2.81 m
✓ 1
Allow ECF from (c)(i).
Part (e)(i)
the extra path is 2 × thickness, so 2d = v × 50 ns
✓ 1
d = 1.69 × 108 × 25 × 10−9 = 4.2 m
✓ 1
Award [2] for CNA. Award [1] max for 8.4 m.
Part (e)(ii)
The two echoes superpose: their displacements add to give a single «longer, distorted» pulse, so they cannot be told apart
✓ 1
OWTTE.
Part (f)
Only part of the pulse is transmitted at the ice surface and only part is reflected at the bed / energy is absorbed in the ice / the pulse spreads out over a larger area on the longer path
✓ 1
Any one.
Answers: (b) 60.0 m · (c)(i) 1.69 × 108 m s−1 · (c)(ii) h = 228 m · (d) 2.81 m · (e)(i) 4.2 m (the remaining parts are explanations — see the table above)
77C-2-66Refraction & Snell's lawPaper 2Easy10 marks
A thermal camera forms images of people from the infrared radiation emitted by their skin, whose temperature is 307 K. The camera has a window of germanium, which transmits infrared radiation. The refractive index of germanium for this radiation is 4.00.
Figure 1 shows plane wavefronts of the infrared radiation in air meeting the flat front surface of the window. The rays make an angle of 40° with the normal.
Figure 1 — wavefront spacing drawn to scale
(a)
Show that the wavelength at which the skin emits the most power is about 9.4 μm.
(1)
(b)
(i)
Calculate the speed of the infrared radiation in the germanium.
(1)
(ii)
Calculate the wavelength in the germanium of the radiation in (a).
(1)
(c)
Calculate the angle between the refracted rays and the normal.
(2)
(d)
On Figure 1, draw three wavefronts of the radiation in the germanium.
(2)
(e)
(i)
Calculate the critical angle for radiation travelling from germanium into air.
(1)
(ii)
The back surface of the window is parallel to the front surface. Explain why radiation that enters through the front surface can never be totally internally reflected at the back surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λmax = 2.9 × 10−3/307 = 9.45 × 10−6 m
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(i)
v = 3.00 × 108/4.00 = 7.50 × 107 m s−1
✓ 1
Part (b)(ii)
λ = 9.45/4.00 = 2.36 μm «frequency unchanged»
✓ 1
Allow ECF from (a). Accept 2.35 μm.
Part (c)
sin r = sin 40°/4.00 = 0.161
✓ 1
r = 9.2°
✓ 1
Accept 9.2°–9.3°. Award [2] for CNA.
Part (d)
Three straight, parallel wavefronts in the germanium, each joining an incident wavefront at the surface, turned so that they make a small angle «9°» with the surface
✓ 1
Allow ECF from (c).
Spacing about one quarter of the spacing of the wavefronts in air
✓ 1
Accept one fifth to one third.
Part (e)(i)
sin c = 1/4.00, c = 14.5°
✓ 1
Part (e)(ii)
The largest possible angle of refraction at the front surface «for grazing incidence» is sin−1(1/4.00) = 14.5°, which equals the critical angle
✓ 1
With parallel faces the angle of incidence at the back equals the angle of refraction at the front, so it can never exceed the critical angle
✓ 1
OWTTE.
Answers: (a) 9.45 × 10−6 m · (b)(i) 7.50 × 107 m s−1 · (b)(ii) 2.36 μm · (c) 9.2° · (e)(i) 14.5° (the remaining parts are explanations — see the table above)
In a public aquarium, a visitor shines a narrow torch beam from the air onto the flat acrylic wall of a shark tank. The angle of incidence is 40°. The refractive index of acrylic is 1.49. The speed of light in air is 3.00 × 108 m s−1.
Diagram not to scale
(a)
Calculate the angle of refraction in the acrylic.
(2)
(b)
State what happens to the wavelength of the light as it enters the acrylic.
(1)
(c)
Outline why the frequency of the light does not change as it enters the acrylic.
(1)
(d)
Calculate the speed of light in the acrylic.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses Snell's law: sin r = sin 40° / 1.49 = 0.431
✓ 1
r = 25.6° (accept 26°)
✓ 1
Part (b)
The wavelength decreases
✓ 1
(A statement that it decreases by a factor of 1.49 is also correct, but the factor is not required for the mark.)
—
Part (c)
The frequency is set by the source, so the number of waves arriving at the boundary each second equals the number leaving it each second
✓ 1
Part (d)
v = c/n = 3.00 × 108 / 1.49 = 2.01 × 108 m s−1
✓ 1
Common error: Make sure your calculator is in degree mode. In radian mode, sin 40 gives a meaningless value.
79IBP2S-022-00-12Critical angle & TIRPaper 2Easy8 marks
An automatic rain sensor on a car windscreen sends an infrared beam into the glass, which has refractive index 1.50. The beam strikes the outer surface of the glass at an angle of incidence of 45°. If the beam is totally internally reflected, it returns to a detector. The refractive index of air is 1.00 and that of rainwater is 1.33.
Diagram not to scale
(a)
State what is meant by the critical angle.
(1)
(b)
Calculate the critical angle for the glass–air boundary.
(2)
(c)
Explain why, when the windscreen is dry, the beam reaches the detector.
(2)
(d)
A layer of rainwater now covers the outer surface. Determine whether the beam is still totally internally reflected, and state the effect on the detector signal.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The angle of incidence, in the denser medium, for which the angle of refraction is 90° .
✓ 1
Part (b)
sin c = 1.00/1.50 .
✓ 1
c = 41.8° .
✓ 1
Part (c)
The angle of incidence, 45°, is greater than the critical angle, 41.8° .
✓ 1
So total internal reflection occurs and the beam is reflected back through the glass to the detector .
✓ 1
ECF from (b).
—
Part (d)
sin c = 1.33/1.50 = 0.887 .
✓ 1
c = 62.5° .
✓ 1
45° is less than 62.5°, so there is no TIR: light refracts into the water and the detector signal falls .
✓ 1
Common error: When the outer medium changes from air to water, recalculate the critical angle using n₂/n₁. Do not reuse 1/n.
80IBP2S-022-00-13Refraction & Snell's lawPaper 2Easy5 marks
In a ripple tank used to model sea waves approaching a harbour, a straight vibrator produces plane waves. These pass from a deep region into a shallow region. A student records these data:
Region | frequency / Hz | wavelength / cm Deep | 8.0 | 2.4 Shallow | 8.0 | 1.5
In the deep region, the wavefronts make an angle of 40° with the straight boundary between the regions.
(a)
Calculate the wave speed in each region.
(2)
(b)
Outline the relationship between a ray and the wavefronts it represents.
(1)
(c)
Calculate the angle that the wavefronts make with the boundary in the shallow region.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Deep: v = fλ = 8.0 × 0.024 = 0.19 m s−1 .
✓ 1
Shallow: v = 8.0 × 0.015 = 0.12 m s−1 .
✓ 1
Part (b)
A ray is perpendicular to the wavefronts and shows the direction in which energy travels .
✓ 1
Part (c)
The angle between wavefront and boundary equals the angle of incidence, and sin θ2 = sin 40° × (v2/v1) = 0.643 × 0.625 .
✓ 1
θ2 = 23.7°, accept 24° .
✓ 1
ECF from (a).
—
Exam tip: The angle between a wavefront and the boundary equals the angle between the ray and the normal. You can use either in Snell's law.
81IBP2S-022-01-11Refractive index & speedPaper 2Easy7 marks
A medical endoscope carries light to the inside of a patient's body along an optical fibre. The fibre core has a refractive index of 1.52. The light used has a frequency of 4.74 × 1014 Hz. The speed of light in a vacuum is 3.00 × 108 m s−1.
(a)
State what is meant by the refractive index of a material.
(1)
(b)
Calculate the speed of light in the fibre core.
(2)
(c)
Calculate the wavelength of the light in the fibre core.
(2)
(d)
The fibre is 1.80 m long. Calculate the minimum time for light to travel from one end of the fibre to the other, assuming the light travels along the axis of the fibre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The ratio of the speed of light in a vacuum to the speed of light in the material, n = c/v.
✓ 1
Part (b)
v = c/n = 3.00 × 108/1.52
✓ 1
v = 1.97 × 108 m s−1
✓ 1
Part (c)
λ = v/f = 1.97 × 108/4.74 × 1014 (ECF from (b))
✓ 1
λ = 4.16 × 10−7 m (416 nm)
✓ 1
Part (d)
t = L/v = 1.80/1.97 × 108 (ECF from (b))
✓ 1
t = 9.12 × 10−9 s (accept 9.12–9.14 × 10−9 s, depending on whether the rounded value of v is used)
✓ 1
Exam tip: The minimum time corresponds to light travelling along the axis. Light that zig-zags by total internal reflection covers a longer path and takes longer.
82IBP2S-022-01-12Young's double slitPaper 2Easy6 marks
A lighting technician checks the wavelength of a laser used in a stage show. The laser beam is directed at a double slit with slit separation 0.25 mm. Fringes form on a wall 3.20 m from the slits. The distance across 10 fringe spacings is measured as 72 mm.
Diagram not to scale
(a)
State why laser light produces a clear, stable interference pattern.
(1)
(b)
Calculate the wavelength of the laser light.
(2)
(c)
Outline why the technician measures across 10 fringe spacings instead of 1.
(1)
(d)
The laser is replaced by a blue laser of wavelength 450 nm. The arrangement is otherwise unchanged. Calculate the new fringe spacing.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The light from the two slits is coherent: it has the same frequency and a constant phase difference.
✓ 1
Part (b)
s = 72/10 = 7.2 mm and λ = sd/D = (7.2 × 10−3 × 0.25 × 10−3)/3.20
✓ 1
λ = 5.6 × 10−7 m (563 nm)
✓ 1
Part (c)
This reduces the percentage uncertainty in s, because the same absolute uncertainty in the length measurement is spread over 10 spacings.
✓ 1
Part (d)
s = λD/d = (450 × 10−9 × 3.20)/(0.25 × 10−3), or scaling s = 7.2 mm × 450/563
✓ 1
s = 5.8 × 10−3 m (5.8 mm)
✓ 1
Common error: Convert the slit separation from mm to m before substituting into the equation. Missing this conversion changes the answer by a factor of 1000.
An underwater LED lamp in a swimming pool sends a ray of light to the flat water surface. The ray meets the surface at 35° to the normal. The refractive index of the water is 1.33 and the speed of light in a vacuum is 3.00×108 m s−1.
Diagram not to scale
(a)
Calculate the speed of light in the water.
(1)
(b)
Calculate the angle of refraction of the ray in the air.
(2)
(c)
State what happens to some of the light at the surface, apart from refraction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = c/n = 3.00×108/1.33 = 2.26×108 m s−1
✓ 1
Part (b)
1.33 sin 35° = 1.00 sin r, so sin r = 0.763
✓ 1
r = 49.7° (accept 50°)
✓ 1
Part (c)
Some of the light is (partially) reflected back into the water
✓ 1
Common error: Put each refractive index next to the angle in its own medium. Here 1.33 goes with the 35° angle in the water and 1.00 goes with the angle in the air.
A periscope uses two right-angled glass prisms with angles of 90°, 45° and 45°. Figure 1 shows one of the prisms. A ray of light enters face AB along the normal and meets face BC at X.
Refractive index of the glass = 1.52.
Figure 1 — Diagram NOT to scale
(a)
Outline what is meant by the critical angle.
(1)
(b)
Show that the critical angle for the glass–air boundary is about 41°.
(1)
(c)
State the angle of incidence at X and explain why all of the light is reflected there.
(2)
(d)
Draw, on Figure 1, the path of the ray after X until it has left the prism.
(2)
(e)
A cheaper prism of the same shape is made from a plastic of refractive index 1.38.
(i)
Determine whether this prism would reflect all of the light at X.
(3)
(ii)
Determine the smallest refractive index that the material of a prism of this shape must have for all of the light to be reflected at X.
(2)
(f)
Suggest one advantage of using a glass prism rather than a plane mirror to reflect the light in a periscope.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the angle of incidence «in the medium of higher refractive index» for which the angle of refraction is 90°
✓ 1
OWTTE
Part (b)
sin c = 1/1.52 «c = 41.1°»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
45°
✓ 1
45° is greater than the critical angle «41°», so total internal reflection occurs «no light is refracted into the air»
✓ 1
Allow ECF from (b)
Part (d)
reflected at X with the angle of reflection equal to the angle of incidence «the ray turns through 90° and travels parallel to AB towards AC»
✓ 1
leaves through face AC along the normal, without deviation
✓ 1
Do not award MP2 for a ray bent at AC
Part (e)(i)
sin c = 1/1.38, so c = 46.4°
✓ 1
the angle of incidence «45°» is less than this critical angle
✓ 1
Allow ECF from MP1
so total internal reflection does not occur: some light is refracted out through BC and the prism would not reflect all the light «the image is dimmer»
✓ 1
MP3 only scores if a comparison has been made
Part (e)(ii)
total internal reflection at X needs the critical angle to be no more than 45°: sin c = 1/n ≤ sin 45°
✓ 1
n ≥ 1/sin 45° = 1.41 «√2»
✓ 1
Accept 1.414. Award [2] for CNA
Part (f)
total internal reflection reflects «almost» 100 % of the light / a prism needs no reflective coating that can tarnish or be scratched
✓ 1
Accept any one sensible advantage
Answers: (b) c = 41.1° · (e)(i) c = 46.4° > 45°: not all of the light is reflected · (e)(ii) n = 1.41 (the remaining parts are explanations — see the table above)
This question is about the superposition of waves.
Figure 1
(a)
Figure 1(a) shows two pulses, X and Y, travelling towards each other along a long rope at time t = 0. Each pulse travels at 0.10 m s−1.
(i)
On Figure 1(b), draw the shape of the rope at t = 0.40 s.
(2)
(ii)
Calculate the time at which the whole rope is momentarily straight.
(1)
(iii)
Explain why, although the rope is straight at the time in (a)(ii), the two pulses are seen again a short time later.
(2)
(b)
Two dippers, S1 and S2, in a ripple tank vibrate in phase with a frequency of 12 Hz. The waves travel at 0.18 m s−1.
(i)
Calculate the wavelength of the waves.
(1)
(ii)
Point P is 9.0 cm from S1 and 12.0 cm from S2. Determine whether P is a point of maximum or minimum disturbance.
(2)
(iii)
The dippers are now made to vibrate in antiphase. State and explain what is now observed at P.
(3)
(c)
Suggest why the water at a point of minimum disturbance is not completely still.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
each pulse has moved 4.0 cm: rope flat except between 6 cm and 12 cm, with a maximum of +2 squares at 8 cm and a minimum of −2 squares at 10 cm
✓ 1
Measure positions from the left edge of the grid. Accept ±0.5 square
straight sections: rising from 6 cm to 8 cm, falling steeply through zero at 9 cm to 10 cm, rising back to zero at 12 cm
✓ 1
MP2 only scores if MP1 scores. Do not award MP2 for two separate unchanged pulses
Part (a)(ii)
the centres «4 cm and 14 cm» meet after each has moved 5.0 cm: t = 0.050/0.10 = 0.50 s
✓ 1
Award [0] for 0.40 s
Part (a)(iii)
at that instant the displacements cancel but the rope is moving «the section from 7 cm to 9 cm moves down and the section from 9 cm to 11 cm moves up»: the energy of the pulses is kinetic energy of the rope
✓ 1
OWTTE
each pulse travels on as if the other were not there «superposition only adds the displacements», so X emerges moving right and Y moving left, unchanged
✓ 1
OWTTE
Part (b)(i)
λ = 0.18/12 = 0.015 m «1.5 cm»
✓ 1
Part (b)(ii)
path difference = 12.0 − 9.0 = 3.0 cm = 2.0λ
✓ 1
Allow ECF from (b)(i)
a whole number of wavelengths, so the waves arrive in phase: maximum disturbance
✓ 1
MP2 only scores if MP1 scores
Part (b)(iii)
P is now a point of minimum disturbance
✓ 1
the sources now have a phase difference of π «half a cycle»
✓ 1
the path difference of 2λ adds no further phase difference, so the waves arrive in antiphase and cancel «destructive interference»
✓ 1
OWTTE
Part (c)
the two waves arriving at the point have different amplitudes «they have travelled different distances and spread out / lost energy», so they do not cancel completely
✓ 1
OWTTE
Answers: (a)(ii) t = 0.50 s · (b)(i) λ = 0.015 m · (b)(ii) path difference = 2λ: maximum (the remaining parts are explanations — see the table above)
Sea waves of wavelength 40 m and period 5.0 s approach a harbour. The harbour wall has an entrance 120 m wide. The figure is a plan view, drawn to scale, of the wavefronts approaching the entrance. The depth of the water is the same inside and outside the harbour.
Plan view drawn to scale
(a)
Calculate the speed of the waves.
(1)
(b)
On the figure, draw three wavefronts that have passed through the entrance.
(2)
(c)
State and explain why the wavelength of the waves inside the harbour is the same as outside.
(2)
(d)
During a storm, waves of wavelength 100 m reach the harbour. Describe and explain how the pattern of waves inside the harbour differs from the pattern you drew in (b).
(2)
(e)
The harbour master proposes narrowing the entrance to 40 m to protect the boats moored behind the harbour wall, on either side of the entrance. Discuss this proposal for the waves of wavelength 40 m.
(3)
(f)
Suggest why the waves inside the harbour have a smaller amplitude than the waves outside.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = λ/T = 40/5.0 = 8.0 m s−1
✓ 1
Part (b)
Three wavefronts with the same spacing as outside «40 m, accept 30–50 m on the scale of the figure»
✓ 1
Straight in the middle «opposite the entrance» and curving round at the ends into the regions behind the wall on both sides
✓ 1
Do not credit wavefronts that are straight across their whole length or fully semicircular. OWTTE.
Part (c)
The frequency «period» of the waves is set by the source and does not change
✓ 1
The speed does not change because the depth is the same, so λ = v/f is unchanged
✓ 1
OWTTE.
Part (d)
The wavelength «100 m» is now comparable to the width of the entrance «120 m»
✓ 1
so the waves diffract much more: the wavefronts are almost semicircular and spread into the regions behind the wall
✓ 1
OWTTE.
Part (e)
A narrower entrance lets less of each wavefront «less wave energy» into the harbour
✓ 1
but the entrance would then equal the wavelength, so the waves that enter diffract strongly and spread round behind the wall, where boats were sheltered before
✓ 1
So the waves become smaller overall but may become larger at the sheltered moorings: the proposal may not protect these boats
✓ 1
MP3 needs a judgement consistent with MP1 or MP2. OWTTE.
Part (f)
Only part of the wave energy passes through the entrance, and this energy spreads out over a wider area as the waves diffract
✓ 1
Accept 'part of the energy is reflected by the wall'. OWTTE.
Answers: (a) 8.0 m s−1(the remaining parts are explanations — see the table above)
Noise-cancelling headphones contain a microphone that detects the noise arriving at each ear cup. A small loudspeaker inside the ear cup then emits a sound that superposes with the noise at the ear. The graph shows how the displacement of the air at the ear, caused by a noise of frequency 200 Hz, varies with time. The speed of sound in air is 340 m s−1.
Graph drawn to scale
(a)
State the principle of superposition.
(1)
(b)
On the graph, draw the displacement of the air at the ear caused by the loudspeaker when the noise is cancelled completely.
(2)
(c)
Calculate the wavelength of the noise in air.
(1)
(d)
In a cheap design, the sound from the loudspeaker is the correct sound for (b) but it reaches the ear 0.50 ms late.
(i)
State this delay as a fraction of the period of the 200 Hz noise.
(1)
(ii)
Determine whether these headphones reduce or increase the loudness of noise of frequency 1000 Hz.
(3)
(e)
Explain why a single loudspeaker in a room cannot cancel a noise at every point in the room.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
When two or more waves meet, the resultant displacement at any point is the «vector» sum of the individual displacements
✓ 1
OWTTE.
Part (b)
Sinusoidal curve with the same amplitude «3 units» and the same period «5.0 ms» as the noise
✓ 1
In antiphase with the noise: a minimum at 1.25 ms where the noise has a maximum «the curve is the noise curve reflected in the time axis»
✓ 1
Do not credit MP2 for a curve shifted by less than half a period.
Part (c)
λ = 340/200 = 1.7 m
✓ 1
Part (d)(i)
T = 5.0 ms, so the delay is 0.10 T «one tenth of a period»
✓ 1
Part (d)(ii)
For 1000 Hz, T = 1.0 ms, so the delay is half a period
✓ 1
The inverted wave «antiphase» delayed by half a period is in phase with the noise
✓ 1
The waves interfere constructively, so the amplitude increases «doubles» and the noise is louder
✓ 1
MP3 only scores if MP1 or MP2 scores.
Part (e)
The path difference between the waves from the noise source and from the loudspeaker is different at different points
✓ 1
so the phase difference changes from point to point: the waves are in antiphase «destructive» at some points but in phase «constructive» at others
✓ 1
OWTTE.
Answers: (c) 1.7 m · (d)(i) 0.10 T · (d)(ii) louder — constructive interference (the remaining parts are explanations — see the table above)
The rain sensor on a car windscreen sends infrared light from an LED into the glass through a coupling block, as shown in Figure 1. The light meets the surfaces of the glass at an angle of incidence of 45°. When the glass is dry, all of the light is reflected between the surfaces until it leaves through a second coupling block to a photodiode. When rain falls, the wipers start automatically.
Refractive index of the glass = 1.50; refractive index of water = 1.33; refractive index of air = 1.00.
Figure 1 — Diagram NOT to scale
(a)
Show that the critical angle for the boundary between the glass and air is about 42°.
(1)
(b)
Determine whether all of the light is reflected at Y, where there is a drop of water on the glass.
(3)
(c)
Calculate the angle of refraction of the light that enters the drop at Y.
(2)
(d)
On Figure 1, draw the paths of the light that leaves Y.
(2)
(e)
Explain how the signal from the photodiode shows whether it is raining heavily or lightly.
(3)
(f)
Explain why the sensor would not work if the light met the surfaces of the glass at an angle of incidence of 65°.
(2)
(g)
Suggest one situation, other than rain, in which the wipers might start.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
sin c = 1.00/1.50 «= 0.667», c = 41.8°
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)
For glass–water: sin c = 1.33/1.50 «= 0.887»
✓ 1
c = 62.5°
✓ 1
Award [2] for CNA.
45° is less than this critical angle, so the light is NOT all reflected: some is refracted into the drop
✓ 1
MP3 only if MP2 is scored.
Part (c)
1.50 sin 45° = 1.33 sin r
✓ 1
r = 52.9°
✓ 1
Award [2] for CNA. Accept 52.8–53.0°.
Part (d)
Refracted ray into the drop, bent away from the normal «at about 53° to the normal»
✓ 1
Allow ECF from (c). Must be on the far side of the normal from the incident ray.
Weaker reflected ray back into the glass at 45° to the normal
✓ 1
Accept a thinner or dashed line for 'weaker'.
Part (e)
Where a drop touches the glass the angle of incidence is less than the critical angle, so light is refracted out of the glass
✓ 1
so less light is reflected on to the photodiode «the signal falls»
✓ 1
The more drops «heavier rain», the more light escapes and the smaller the signal «so the wipers can be set to move faster»
✓ 1
OWTTE.
Part (f)
65° is greater than the critical angle for glass–water «62.5°» as well as for glass–air
✓ 1
so the light would be totally internally reflected whether or not there is water on the glass «the signal would not change»
✓ 1
Allow ECF from (b).
Part (g)
Any material touching the outer surface with a refractive index greater than about 1.06 «1.50 sin 45°», e.g. dirt, an insect, a leaf, ice or frost
✓ 1
Accept any one valid example.
Answers: (a) 41.8° · (b) critical angle 62.5° > 45°: not all reflected · (c) 52.9° (the remaining parts are explanations — see the table above)
During a sound check, a loudspeaker emits a steady note of frequency 425 Hz. The loudspeaker is far behind a straight wall, on the line through the midpoint between two open doorways, A and B, in the wall. The doorways are each 0.90 m wide and their centres are 3.00 m apart. Sound passes through both doorways into a large hall. Point P in the hall is 6.00 m from the centre of A and 7.20 m from the centre of B, as shown in Figure 1. The speed of sound is 340 m s−1.
Figure 1 — Plan view drawn to scale
(a)
Calculate the wavelength of the sound.
(1)
(b)
On Figure 1, draw three wavefronts of the sound that has passed through doorway A.
(2)
(c)
Explain why the sound waves that emerge from the two doorways are coherent.
(2)
(d)
Determine whether a listener at P hears a loud sound or a quiet sound.
(3)
(e)
Suggest two reasons why the listener at P can still hear the note.
(2)
(f)
The frequency of the note is changed. Determine the lowest frequency for which P is a point of maximum loudness.
(2)
(g)
The loudspeaker, still emitting 425 Hz, is moved so that it is nearer to doorway A than to doorway B. Explain why P is then not necessarily a point of minimum loudness.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
λ = 340/425 = 0.800 m
✓ 1
Part (b)
Curved «circular» wavefronts centred on doorway A, spreading into the regions beside the doorway «beyond the edges of the gap»
✓ 1
Do not award MP1 for straight wavefronts of the doorway's width.
Spacing between wavefronts unchanged «0.80 m on the scale»
✓ 1
Accept 0.7–0.9 m.
Part (c)
Both come from the same source, so they have the same frequency
✓ 1
The doorways are the same distance from the loudspeaker, so the waves leave them with a constant «zero» phase difference
✓ 1
OWTTE.
Part (d)
Path difference = 7.20 − 6.00 = 1.20 m
✓ 1
= 1.5 λ
✓ 1
Allow ECF from (a).
Half-integer number of wavelengths, so the waves meet in antiphase «destructive interference»: P is a quiet point
✓ 1
MP3 only if MP2 is scored.
Part (e)
The waves from the two doorways have different amplitudes at P «P is further from B», so they do not cancel completely
✓ 1
Award [1] per valid reason, [2 max].
Sound reflected from the walls / floor / ceiling of the hall reaches P OR sound passes through the wall itself
✓ 1
OWTTE.
Part (f)
For a maximum, 1.20 m = nλ; the lowest frequency has the longest wavelength, n = 1: λ = 1.20 m
✓ 1
f = 340/1.20 = 283 Hz
✓ 1
Award [2] for CNA.
Part (g)
The sound now reaches doorway A before doorway B, so the waves leave the doorways with a phase difference «they are still coherent»
✓ 1
The type of interference at P depends on the total path difference from the loudspeaker, which is no longer 1.20 m «1.5λ»
✓ 1
OWTTE.
Answers: (a) 0.800 m · (d) path difference 1.20 m = 1.5λ: quiet · (f) 283 Hz (the remaining parts are explanations — see the table above)
A straight breakwater AB, 200 m long, is built in the sea parallel to a beach and 300 m from it, to shelter the beach. Swell of wavelength 54 m and period 6.0 s approaches with its wavefronts parallel to the breakwater. The depth of the water is the same everywhere in the region shown.
Figure 1 is a plan view, drawn to scale. Point P on the beach is 332 m from end A and 305 m from end B. Q is on the beach directly behind the centre of the breakwater.
Figure 1 — plan view drawn to scale
(a)
Calculate the speed of the swell.
(1)
(b)
On Figure 1, draw three wavefronts of the swell that has passed end A of the breakwater.
(2)
(c)
On Figure 1, draw a ray to show the direction in which the waves from end A travel when they reach P.
(1)
(d)
Explain why waves of significant amplitude reach Q, even though Q is directly behind the breakwater.
(2)
(e)
Determine whether the water at P oscillates with a large or a small amplitude.
(3)
(f)
The breakwater is to be lengthened to 400 m. Suggest how this would affect the waves reaching Q.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = λ/T = 54/6.0 = 9.0 m s−1
✓ 1
Part (b)
Curved wavefronts centred on «near» end A, spreading round the end into the region behind the breakwater
✓ 1
Do not credit straight wavefronts that stop at the line of A.
Spacing unchanged «54 m on the scale of the figure, the same as the incoming wavefronts»
✓ 1
Accept 45–65 m.
Part (c)
Straight line from A to P «along a radius of the curved wavefronts», with an arrow pointing towards P
✓ 1
Must be perpendicular to the candidate's wavefronts where they cross it.
Part (d)
The swell diffracts round both ends of the breakwater «which is only about four wavelengths long» and spreads into the region behind it
✓ 1
Q is the same distance from A and from B, and the waves leave A and B in phase «the same wavefronts reach both ends together», so they arrive at Q in phase and interfere constructively
✓ 1
OWTTE.
Part (e)
path difference = 332 − 305 = 27 m
✓ 1
= 0.5 λ
✓ 1
Allow ECF from an incorrect subtraction.
Path difference = (n + ½)λ with n = 0: the waves arrive in antiphase and interfere destructively, so P has a small amplitude
✓ 1
MP3 only scores if MP2 scores.
Part (f)
Q is further from the ends, and the breakwater is now much larger compared with the wavelength, so less of the diffracted wave reaches the region behind its centre
✓ 1
The waves at Q would have a smaller amplitude «Q would be better sheltered»
✓ 1
MP2 needs a reason.
Answers: (a) v = 9.0 m s−1 · (e) path difference = 27 m = 0.5λ: small amplitude (the remaining parts are explanations — see the table above)
A ground station sends signals to a geostationary communications satellite. A feed horn at the focus F of a parabolic dish of diameter 1.20 m emits microwaves of frequency 12.0 GHz towards the dish. The dish reflects them into a narrow beam travelling parallel to its axis (Figure 1).
Figure 1 — cross-section drawn to scale
(a)
Calculate the wavelength of the microwaves.
(1)
(b)
On Figure 1, draw two wavefronts of the microwaves between F and the dish and two wavefronts of the reflected beam.
(2)
(c)
At X, the angle between the ray FX and the normal to the surface of the dish is 20°.
(i)
State the angle between the reflected ray and the normal at X.
(1)
(ii)
The reflected ray is parallel to the axis. Deduce the angle between FX and the axis.
(1)
(d)
Explain why the reflected beam spreads out slightly instead of remaining exactly parallel.
(2)
(e)
Suggest the effect on the signal received by the satellite if the dish were replaced by one of half the diameter, with the same feed horn and frequency.
(1)
(f)
A geostationary satellite has an orbital period of 24.0 hours. The mass of the Earth is 5.97 × 1024 kg.
(i)
Show that the radius of the satellite's orbit is about 4.2 × 107 m.
(2)
(ii)
The ground station is 3.80 × 107 m from the satellite. Calculate the time taken for the signal to reach the satellite.
Between F and the dish: arcs of circles «spheres» centred on F
✓ 1
Two wavefronts needed for each mark.
Reflected beam: straight lines perpendicular to the axis «parallel to each other, plane wavefronts», spanning the width of the dish
✓ 1
Do not credit curved reflected wavefronts.
Part (c)(i)
20° «angle of reflection = angle of incidence»
✓ 1
Part (c)(ii)
The incident and reflected rays are each at 20° to the normal, so the angle between them is 40°; the reflected ray is parallel to the axis, so FX makes 40° with the axis
✓ 1
Accept a correct geometric argument from the figure.
Part (d)
The waves diffract «spread out» as they leave the dish, which acts as an aperture «the edges of the beam are limited by the edge of the dish»
✓ 1
The wavelength «2.5 cm» is small compared with the diameter but not negligible, so the spreading is small but noticeable
✓ 1
OWTTE.
Part (e)
The beam spreads more «diffraction increases as the ratio of wavelength to aperture increases», so less of the power reaches the satellite: the received signal is weaker
✓ 1
Reason required.
Part (f)(i)
GMm/r2 = m4π2r/T2, so r3 = GMT2/4π2
✓ 1
r = (6.67 × 10−11 × 5.97 × 1024 × (86400)2/4π2)1/3 = 4.22 × 107 m
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (f)(ii)
t = 3.80 × 107/3.00 × 108 = 0.127 s
✓ 1
Answers: (a) λ = 2.50 × 10−2 m · (c)(i) 20° · (c)(ii) 40° · (f)(i) 4.22 × 107 m · (f)(ii) 0.127 s (the remaining parts are explanations — see the table above)
Ecologists locate calling birds in a forest with pairs of microphones. Two microphones, M1 and M2, are placed 20.0 m apart on level ground. A bird calls from a point far from both microphones. The call reaches M2 35.0 ms after it reaches M1. The speed of sound in the air is 343 m s−1.
Figure 1 shows wavefronts of the call arriving at the microphones.
Figure 1 — plan view, not to scale
(a)
Describe the shape of the wavefronts of the call close to the bird and at the microphones.
(2)
(b)
On Figure 1, draw a ray that shows the direction of travel of the sound arriving at M2.
(1)
(c)
Show that the difference between the distances travelled by the sound to M2 and to M1 is about 12 m.
(1)
(d)
Determine the angle between the direction of the bird and the normal to the line M1M2.
(2)
(e)
Outline why a single pair of microphones cannot tell whether the bird is on one side of the line M1M2 or the other.
(1)
(f)
The call is a pure note of frequency 3.43 kHz. Explain why the time difference cannot be found by comparing only the phases of the continuous note received at M1 and M2.
(2)
(g)
Suggest why the method gives an inaccurate direction for a bird only a few metres from the microphones.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Close to the bird: spheres «circles in plan view» centred on the bird
✓ 1
At the microphones, far away, the radius is so large that over 20 m the wavefronts are almost plane «straight»
✓ 1
OWTTE.
Part (b)
Straight line through M2, perpendicular to the wavefronts, with an arrow towards M2
✓ 1
Part (c)
path difference = vΔt = 343 × 0.0350 = 12.0 m
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (d)
path difference = M₁M₂ × sin θ «plane wavefronts», so sin θ = 12.0/20.0 = 0.600
✓ 1
Accept a correct triangle drawn on the figure.
θ = 37°
✓ 1
Award [2] for CNA. Accept 53° only if it is clearly the angle to the line M₁M₂.
Part (e)
A bird at the mirror-image position on the other side of the line gives the same path difference «the same time difference»
✓ 1
OWTTE.
Part (f)
λ = 343/3430 = 0.100 m, so the path difference is about 120 wavelengths
✓ 1
Path differences that differ by a whole number of wavelengths give the same phase difference, so the phase alone does not fix the time difference
✓ 1
OWTTE.
Part (g)
The wavefronts are then noticeably curved across the 20 m between the microphones, so the rays to M₁ and M₂ are not parallel and the path difference is not M₁M₂ × sin θ
✓ 1
OWTTE.
Answers: (c) 12.0 m · (d) θ = 37° · (f) λ = 0.100 m (the remaining parts are explanations — see the table above)
A manufacturer tests a new transparent polymer for phone screens. A ray of light is directed from air into a semicircular block of the polymer. The angle of incidence i and the angle of refraction r are recorded.
i / °: 20.0, 35.0, 50.0, 65.0 r / °: 13.0, 22.0, 30.5, 36.5
(a)
Outline how a graph of these data could be used to determine the refractive index of the polymer.
(2)
(b)
Using the data for i = 50.0°, calculate the refractive index of the polymer.
(2)
(c)
Calculate the critical angle for light travelling from the polymer into air. Use your answer to (b).
(2)
(d)
The polymer is bonded to the display using an adhesive of refractive index 1.40. Determine the critical angle at the polymer–adhesive boundary.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Plot sin i on the y-axis against sin r on the x-axis
✓ 1
The graph should be a straight line through the origin, and its gradient equals the refractive index n
✓ 1
Part (b)
sin 50.0° = 0.766 and sin 30.5° = 0.508
✓ 1
n = 0.766/0.508 = 1.51
✓ 1
Part (c)
sin c = 1/n = 1/1.51 = 0.662
✓ 1
c = 41.5° (accept 41.4°–41.5°)
✓ 1
Allow error carried forward from (b)
—
Part (d)
sin c = 1.40/1.51 = 0.927
✓ 1
c = 68° (accept 68.0°–68.1°)
✓ 1
Allow error carried forward from (b)
—
Exam tip: When two non-air media meet, the critical angle comes from the ratio of the two refractive indices: sin c = n₂/n₁, where n₁ is the denser medium. The closer the two indices are, the larger the critical angle.
94IBP2S-022-02-13Path differencePaper 2Medium9 marks
An acoustic engineer tests a sound system in an anechoic test room. Two loudspeakers S1 and S2 are connected to the same signal generator and emit in phase at 680 Hz. The speed of sound is 340 m s−1. A microphone at point P is 4.25 m from S1 and 5.00 m from S2.
Diagram not to scale
(a)
Outline why S1 and S2 are coherent sources.
(2)
(b)
Calculate the wavelength of the sound.
(1)
(c)
Determine whether P is a point of maximum or minimum intensity.
(2)
(d)
Explain why the intensity detected at P is not zero.
(2)
(e)
The frequency is gradually increased from 680 Hz. Determine the lowest frequency above 680 Hz at which P is a maximum.
(2)
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Notes
Part (a)
Both are driven by the same signal generator, so they have the same frequency
✓ 1
There is a constant phase difference between them (here zero)
✓ 1
Part (b)
λ = v/f = 340/680 = 0.500 m
✓ 1
Part (c)
Path difference = 5.00 − 4.25 = 0.75 m, which equals 1.5λ
✓ 1
The waves arrive in antiphase, so destructive interference occurs and P is a minimum
✓ 1
Allow ECF from (b)
—
Part (d)
P is at different distances from the two speakers, so the amplitudes of the two waves at P are not equal
✓ 1
The waves do not cancel completely: the resultant amplitude is the difference of the two amplitudes, which is not zero
✓ 1
Part (e)
For a maximum, 0.75 = nλ with n a whole number; at 680 Hz n = 1.5, so the next value is n = 2, giving λ = 0.375 m
✓ 1
f = 340/0.375 = 907 Hz (accept 906.7 Hz)
✓ 1
Exam tip: When the frequency changes, the path difference at a fixed point stays the same, but the number of wavelengths it contains changes. Find the next whole number of wavelengths (for a maximum) or the next half-integer number (for a minimum).
In medical ultrasound imaging, a beam of frequency 3.5 MHz travels through a layer of fat and meets a boundary with muscle. The speed of ultrasound is 1450 m s−1 in fat and 1580 m s−1 in muscle. Snell's law applies to sound in the form sin θ1 / sin θ2 = v1 / v2.
(a)
Calculate the wavelength of the ultrasound in fat.
(1)
(b)
The beam meets the boundary at an angle of incidence of 25° in the fat. Calculate the angle of refraction in the muscle.
(2)
(c)
Determine the critical angle for ultrasound at the fat–muscle boundary.
(2)
(d)
Explain why ultrasound travelling from muscle into fat cannot undergo total internal reflection at this boundary.
(2)
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Notes
Part (a)
λ = v/f = 1450 / 3.5 × 106 = 4.1 × 10−4 m
✓ 1
Part (b)
sin θ2 = sin 25° × 1580/1450 = 0.461
✓ 1
θ2 = 27° (27.4°)
✓ 1
Part (c)
At the critical angle θ2 = 90°, so sin c = 1450/1580 = 0.918
✓ 1
c = 66.6°
✓ 1
Part (d)
Going from muscle to fat, the wave slows down, so it bends towards the normal
✓ 1
The angle of refraction is always less than the angle of incidence, so it can never reach 90° and there is no critical angle
✓ 1
Exam tip: Total internal reflection needs the wave to be moving into a medium where it travels faster. For sound, that is not the same as moving into a 'less dense' material.
96IBP2S-022-00-14Young's double slitPaper 2Medium7 marks
A hair-care laboratory measures the thickness of a single human hair. The hair is placed in a laser beam of wavelength 650 nm. The hair produces the same diffraction pattern as a single slit of equal width. On a screen 2.40 m from the hair, the central bright maximum is 3.1 cm wide.
Diagram not to scale
(a)
Describe how the intensity of the pattern varies across the screen. You may describe a sketch in words.
(2)
(b)
Determine the width of the hair.
(3)
(c)
A thicker hair is tested with the same laser. State and explain the change in the width of the central maximum.
(2)
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Notes
Part (a)
There is a bright central maximum, twice as wide as each of the other maxima .
✓ 1
The subsidiary maxima are much less intense and become fainter further from the centre. They are separated by minima of zero intensity .
✓ 1
Part (b)
The half-width of the central maximum is 1.55 cm, so θ ≈ tan θ = 0.0155/2.40 = 6.46×10−3 rad .
✓ 1
Uses θ = λ/b for the first minimum (small angle) .
✓ 1
b = 650×10−9/6.46×10−3 = 1.0×10−4 m .
✓ 1
Part (c)
The central maximum becomes narrower .
✓ 1
θ = λ/b with λ unchanged, so a larger b gives a smaller angle to the first minimum .
✓ 1
Common error: The angle to the first minimum uses half the width of the central maximum. Using the full width gives a hair width that is half the true value.
97IBP2S-022-00-16Young's double slitPaper 2Medium7 marks
An engineer tests a double-slit arrangement for an underwater optical sensor. A green laser of wavelength 532 nm (in air) illuminates two slits separated by 0.25 mm. Fringes form on a screen 1.80 m away. The refractive index of water is 1.33.
Diagram not to scale
(a)
Outline why the light emerging from the two slits is coherent.
(1)
(b)
Calculate the fringe spacing on the screen when the apparatus is in air.
(2)
(c)
Explain, in terms of path difference, how a dark fringe is formed.
(2)
(d)
The whole region between the slits and the screen is filled with water. Determine the new fringe spacing.
(2)
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Notes
Part (a)
Both slits are illuminated by the same laser wavefront, so the waves from them have the same frequency and a constant phase difference .
✓ 1
Part (b)
s = λD/d = 532×10−9 × 1.80/0.25×10−3 .
✓ 1
s = 3.8×10−3 m .
✓ 1
Part (c)
The path difference from the two slits is a half-integer number of wavelengths, (n + ½)λ .
✓ 1
So the waves arrive in antiphase and superpose destructively .
✓ 1
Part (d)
λwater = 532/1.33 = 400 nm .
✓ 1
s = 2.9×10−3 m, accept 2.88×10−3 m .
✓ 1
ECF from (b).
—
Common error: In a medium, the wavelength becomes λ/n while the frequency is unchanged, so the fringes move closer together. Do not multiply by n.
98IBP2S-022-01-13Path differencePaper 2Medium7 marks
An acoustic engineer tests two loudspeakers in an open field. The speakers are driven in phase by the same signal generator at a frequency of 680 Hz. The speed of sound is 340 m s−1. A microphone at point P is 6.00 m from speaker A and 6.75 m from speaker B.
Diagram not to scale
(a)
Calculate the wavelength of the sound.
(1)
(b)
Determine whether P is a point of maximum or minimum loudness.
(2)
(c)
Explain why the sound detected at P is not completely silent.
(2)
(d)
The frequency is gradually increased from 680 Hz. Determine the lowest frequency above 680 Hz at which P becomes a point of maximum loudness.
(2)
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Notes
Part (a)
λ = v/f = 340/680 = 0.50 m
✓ 1
Part (b)
The path difference is 6.75 − 6.00 = 0.75 m, which equals 1.5λ (ECF from (a))
✓ 1
This is a half-integer number of wavelengths, so the waves arrive in antiphase and interfere destructively: P is a minimum
✓ 1
Part (c)
Sound spreads out, so its amplitude falls with distance. The wave from B travels further, so it has a smaller amplitude at P than the wave from A
✓ 1
The two waves arrive in antiphase but with unequal amplitudes, so the resultant amplitude is not zero
✓ 1
Accept for the first mark: sound reflected from the ground or other objects also reaches P, so the two direct waves do not cancel completely.
—
Part (d)
A maximum requires a path difference of nλ, so 0.75 = n(340/f), giving f = 453n Hz
✓ 1
n = 1 gives 453 Hz, which is below 680 Hz, so n = 2 and f = 907 Hz
✓ 1
Accept the equivalent method: the path difference must change from 1.5λ to 2λ, so f = 680 × 2/1.5 = 907 Hz.
—
Exam tip: When a question asks for the next frequency that gives a maximum, first write the general condition (path difference = nλ). Then test successive values of n against the condition given in the question.
99IBP2S-022-01-14Critical angle & TIRPaper 2Medium7 marks
In a medical ultrasound scan, an ultrasound beam of frequency 3.5 MHz passes from a layer of fat into muscle. The speed of ultrasound is 1450 m s−1 in fat and 1580 m s−1 in muscle. The beam strikes the fat–muscle boundary at an angle of incidence of 20.0°.
(a)
State, with a reason, whether the frequency of the ultrasound changes as it enters the muscle.
(1)
(b)
Calculate the angle of refraction in the muscle.
(2)
(c)
Calculate the wavelength of the ultrasound in the muscle.
(1)
(d)
Total internal reflection can occur at this boundary. Determine the critical angle, and state in which tissue the ultrasound must be travelling for total internal reflection to occur.
(3)
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Notes
Part (a)
The frequency is unchanged, because it is determined by the source (the transducer), not by the medium.
✓ 1
Part (b)
Use sin i / sin r = v1/v2, giving sin r = sin 20.0° × 1580/1450 = 0.373
✓ 1
r = 21.9°
✓ 1
Part (c)
λ = v/f = 1580 / (3.5 × 106) = 4.5 × 10−4 m
✓ 1
Part (d)
sin c = 1450/1580 = 0.918
✓ 1
c = 66.6°
✓ 1
The ultrasound must be travelling in the fat (the medium with the lower wave speed) towards the muscle.
✓ 1
Common error: Snell's law applies to all waves, not only light. In terms of speeds, sin i / sin r = v₁/v₂. Take care not to put the speed ratio the wrong way round.
An engineer is testing a new polymer for smartphone camera lenses. She shines a narrow laser beam from air into a semicircular polymer block and measures the angle of incidence i and the angle of refraction r.
i / °: 20.0, 30.0, 40.0, 50.0, 60.0 r / °: 12.6, 18.5, 23.9, 29.1, 33.2
(a)
Outline why a graph of sin i against sin r is a suitable way to analyse these data.
(2)
(b)
Determine the refractive index of the polymer using the data for i = 50.0°.
(2)
(c)
Calculate the critical angle for light inside the polymer at a polymer–air boundary.
(2)
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Notes
Part (a)
From Snell's law, sin i = n sin r, so sin i is proportional to sin r
✓ 1
The graph is a straight line through the origin whose gradient is n, which averages the readings and shows up anomalies
✓ 1
Part (b)
n = sin 50.0°/sin 29.1° = 0.766/0.486
✓ 1
n = 1.58
✓ 1
Part (c)
sin θc = 1/n = 1/1.58
✓ 1
θc = 39.3° (accept 39.4° from unrounded n); ECF from (b)
✓ 1
Exam tip: When asked to 'outline' a graphical method, state which quantity the gradient represents and why the line should pass through the origin.
A 3.5 MHz ultrasound beam from a medical scanner passes through a layer of fat and then into muscle. The speed of ultrasound is 1450 m s−1 in fat and 1590 m s−1 in muscle. The beam meets the fat–muscle boundary at an angle of incidence of 25°.
(a)
Calculate the wavelength of the ultrasound in the fat.
(1)
(b)
Calculate the angle of refraction in the muscle.
(2)
(c)
Explain how the wavelength of the ultrasound changes as it passes from the fat into the muscle.
(2)
(d)
Determine the critical angle for ultrasound travelling from fat towards muscle.
(2)
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Notes
Part (a)
λ = v/f = 1450/3.5×106 = 4.1×10−4 m
✓ 1
Part (b)
sin r = (vmuscle/vfat) sin 25° = (1590/1450) × 0.423 = 0.463
✓ 1
r = 27.6°
✓ 1
Part (c)
The frequency is unchanged, because it is fixed by the source
✓ 1
The speed increases, so λ = v/f increases
✓ 1
Part (d)
sin θc = vfat/vmuscle = 1450/1590 = 0.912
✓ 1
θc = 65.8°
✓ 1
Common error: For sound, Snell's law uses speeds: sin θ₁/sin θ₂ = v₁/v₂. Using the speeds the wrong way round makes the beam bend in the wrong direction.
102IBP2S-022-03-14Critical angle & TIRPaper 2Medium7 marks
A simple model of a bicycle reflector is a right-angled isosceles prism (angles 45°, 45° and 90°). Light from a car headlamp enters one of the shorter faces along the normal. It then strikes the longest face (the hypotenuse) at an angle of incidence of 45°, where it is meant to be totally internally reflected.
Diagram not to scale
(a)
Show that, with air outside the prism, total internal reflection at the hypotenuse requires the refractive index of the prism to be at least 1.41.
(2)
(b)
The prism is made of acrylic of refractive index 1.49. In rain, a film of water (refractive index 1.33) covers the hypotenuse. Explain whether total internal reflection still occurs at the hypotenuse.
(3)
(c)
Calculate the angle of refraction of the light that enters the water film.
(2)
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Notes
Part (a)
Total internal reflection needs the angle of incidence (45°) to be at least the critical angle, so sin θc = 1/n ≤ sin 45°
✓ 1
n ≥ 1/sin 45° = 1/0.7071 = 1.414, which is at least 1.41
✓ 1
Part (b)
sin θc = 1.33/1.49 = 0.893
✓ 1
θc = 63.2°
✓ 1
45° is less than 63.2°, so there is no total internal reflection at the acrylic–water boundary, and light refracts into the water
✓ 1
Part (c)
1.49 sin 45° = 1.33 sin r, so sin r = 0.792
✓ 1
r = 52.4°
✓ 1
Exam tip: In a 'show that' part, write every step and give the result to more significant figures than the value quoted in the question.
103IBP2S-022-03-15Refraction & Snell's lawPaper 2Medium6 marks
Surfers ride waves that break over a shallow reef. Ocean swell of period 8.0 s travels at 12 m s−1 in deep water. Over the reef its speed falls to 9.0 m s−1. The edge of the reef is a straight boundary between the deep and shallow regions.
Diagram not to scale. Rays are shown.
(a)
Calculate the wavelength of the swell in deep water.
(1)
(b)
The waves in deep water travel at 50° to the normal to the reef edge. Calculate the angle to the normal at which they travel over the reef.
(2)
(c)
Closer to the beach the water becomes steadily shallower. Explain why the wave crests become almost parallel to the shoreline.
(2)
(d)
Determine the wavelength of the swell over the reef.
(1)
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Notes
Part (a)
λ = vT = 12 × 8.0 = 96 m
✓ 1
Part (b)
sin r = (9.0/12) sin 50° = 0.575
✓ 1
r = 35.1°
✓ 1
Part (c)
The wave speed keeps decreasing as the depth decreases, so the waves keep refracting towards the normal
✓ 1
The direction of travel becomes almost perpendicular to the shore, so the crests line up almost parallel to it
✓ 1
Part (d)
The period is unchanged, so λ = 9.0 × 8.0 = 72 m
✓ 1
Common error: Angles in Snell's law are measured between the direction of travel and the normal. The angle between a wave crest and the boundary is the same as this, but do not mix crest angles and ray angles in one equation.
A diver wears a mask with a flat glass faceplate. Light from a fish travels through the water, through the glass and into the air inside the mask. A ray from the fish meets the outer surface of the faceplate at an angle of incidence of 30.0°. Figure 1 shows a section through the faceplate.
Refractive index of water = 1.33; refractive index of glass = 1.52.
Figure 1 — Diagram NOT to scale
(a)
(i)
Calculate the angle of refraction of the ray in the glass.
(2)
(ii)
Calculate the speed of light in the water.
(1)
(b)
Determine the angle between the ray in the air inside the mask and the normal to the faceplate.
(2)
(c)
Explain why the answer to (b) does not depend on the refractive index or the thickness of the glass.
(2)
(d)
Draw, on Figure 1, the path of the ray through the glass and into the air inside the mask.
(2)
(e)
The fish is red. Red light from the fish has a wavelength of 650 nm in air.
(i)
Calculate the wavelength of this light in the water.
(1)
(ii)
A diver without a mask, whose eyes are in contact with the water, also sees the fish as red. Explain why.
(2)
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Notes
Part (a)(i)
1.33 × sin 30.0° = 1.52 × sin θg
✓ 1
θg = 25.9°
✓ 1
Accept 26°. Award [2] for CNA
Part (a)(ii)
v = 3.00 × 108/1.33 = 2.26 × 108 m s−1
✓ 1
Part (b)
angle of incidence at the inner surface = 25.9° «parallel faces», and 1.52 × sin 25.9° = 1.00 × sin θ
✓ 1
Allow ECF from (a)(i)
θ = 41.7°
✓ 1
Accept 41.6–41.8°. Award [2] for CNA
Part (c)
the faces of the glass are parallel, so the angle of refraction at the first face equals the angle of incidence at the second face
✓ 1
so nw sin 30.0° = ng sin θg = nair sin θ: ng cancels and θ depends only on nw, nair and 30.0°
✓ 1
OWTTE. Accept a numerical demonstration with a second value of ng for MP2
Part (d)
in the glass: ray bent towards the normal «about 26° to it» and reaching the inner surface
✓ 1
in the air: ray bent away from the normal, at a larger angle to the normal than in the water «about 42°»
✓ 1
Allow ECF from (a)(i) and (b)
Part (e)(i)
λ = 650/1.33 = 489 nm
✓ 1
Part (e)(ii)
the frequency of the light does not change when it passes from one medium into another «only the speed and the wavelength change»
✓ 1
the colour seen depends on the frequency «photon energy», which is the same as in air, so the fish still looks red
✓ 1
Do not accept 'the wavelength is the same'
Answers: (a)(i) 25.9° · (a)(ii) 2.26 × 108 m s−1 · (b) 41.7° · (e)(i) 489 nm (the remaining parts are explanations — see the table above)
Light of wavelength 650 nm from a laser is incident normally on a double slit. Interference fringes are observed on a screen 2.40 m from the slits.
Figure 1 shows the central part of the fringe pattern next to a millimetre scale. A and B are the centres of two bright fringes, with seven bright fringes between them.
Figure 1Figure 2 — for part (d)
(a)
Outline why the light from the two slits is coherent.
(1)
(b)
Determine the separation of the two slits.
(3)
(c)
Calculate the path difference between the light from the two slits at the centre of the third dark fringe from the centre of the central maximum.
(2)
(d)
Sketch, on the axes in Figure 2, a graph to show how the intensity of the light varies with distance y from the centre of the central maximum. Assume that all the bright fringes in this region have the same intensity.
(2)
(e)
A filter is placed over one slit so that the amplitude of the light from that slit is halved. Explain the changes, if any, to the fringe pattern.
(3)
(f)
Suggest one change to the arrangement that would increase the fringe spacing, and state why it has this effect.
(1)
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Notes
Part (a)
both slits are illuminated by the same wavefront from one laser, so the waves have the same frequency and a constant phase difference
✓ 1
OWTTE
Part (b)
s = (46.0 − 4.0)/8 = 5.25 mm
✓ 1
Accept 5.2–5.3 mm
d = λD/s = 650 × 10−9 × 2.40/5.25 × 10−3
✓ 1
d = 2.97 × 10−4 m
✓ 1
Accept 2.9–3.0 × 10−4 m. Allow ECF for s. Award [2] max if s = 42.0/9 is used
Part (c)
path difference = (n + ½)λ with n = 2 «2.5λ»
✓ 1
= 2.5 × 650 nm = 1.63 × 10−6 m
✓ 1
Accept 1.6 × 10−6 m or 1625 nm. Award [1] max for 1.95 × 10−6 m
Part (d)
maxima at y = 0, ±5.25 mm and ±10.5 mm
✓ 1
Allow ECF from (b). Accept ±5 mm, ±10 mm
smooth curve with zero-intensity minima midway between maxima of equal height
✓ 1
Part (e)
the fringe positions «and separation» are unchanged, because λ, d and D are unchanged
✓ 1
at the bright fringes the resultant amplitude is smaller «1.5A instead of 2A, where A is the amplitude from one unfiltered slit», so they are less bright
✓ 1
at the dark fringes the two waves no longer cancel completely, so they are not dark / the contrast between bright and dark fringes is reduced
✓ 1
OWTTE
Part (f)
move the screen further from the slits / use slits that are closer together / use light of a longer wavelength, because s = λD/d
✓ 1
The change and the reason are both needed
Answers: (b) d = 2.97 × 10−4 m · (c) 1.63 × 10−6 m (the remaining parts are explanations — see the table above)
In a double-slit experiment, two narrow slits 0.40 mm apart are illuminated with laser light, and fringes are observed on a screen 2.00 m from the slits. Two lasers are available: a red laser of wavelength 640 nm and a blue laser of wavelength 480 nm.
Diagram not to scale
(a)
Calculate the fringe separation for each laser.
(2)
(b)
The beams from both lasers are now combined, so that the slits are illuminated with red and blue light at the same time.
(i)
Determine the smallest distance from the central bright fringe at which a red bright fringe and a blue bright fringe coincide.
(3)
(ii)
Point X on the screen is 4.8 mm from the centre of the central bright fringe. Deduce the colour of the light seen at X.
(3)
(iii)
Explain why the red light and the blue light do not interfere with each other.
(2)
(c)
The fringe separation is measured with a ruler. Suggest two changes to the apparatus that would reduce the percentage uncertainty in the measured fringe separation.
(2)
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Notes
Part (a)
Red: s = λD/d = 640 × 10−9 × 2.00/0.40 × 10−3 = 3.2 × 10−3 m
✓ 1
Blue: s = 480 × 10−9 × 2.00/0.40 × 10−3 = 2.4 × 10−3 m
✓ 1
Accept s = 3.2 mm × 480/640.
Part (b)(i)
A coincidence needs nrsr = nbsb «equivalently nr × 640 nm = nb × 480 nm»
✓ 1
Smallest whole numbers: the third red bright fringe and the fourth blue bright fringe «3 × 640 nm = 4 × 480 nm = 1920 nm»
✓ 1
Distance = 3 × 3.2 mm = 9.6 mm
✓ 1
Allow ECF from (a). Award [3] for CNA.
Part (b)(ii)
For red, 4.8/3.2 = 1.5: the path difference is 1.5λ, so the red waves interfere destructively at X
✓ 1
Accept 'X is at a red dark fringe'.
For blue, 4.8/2.4 = 2.0: the path difference is 2λ, so the blue waves interfere constructively at X
✓ 1
X is blue
✓ 1
MP3 only scores if MP1 or MP2 scores.
Part (b)(iii)
The two colours have different frequencies, so the phase difference between them is not constant
✓ 1
so they are not coherent: the screen shows the red pattern and the blue pattern simply added together
✓ 1
OWTTE.
Part (c)
Increase the distance D between the slits and the screen
✓ 1
Use slits with a smaller separation d
✓ 1
Accept 'use light of longer wavelength' as one change. Do not accept 'repeat the measurement'.
Answers: (a) 3.2 × 10−3 m (red), 2.4 × 10−3 m (blue) · (b)(i) 9.6 mm · (b)(ii) blue (the remaining parts are explanations — see the table above)
Two radio transmitters, A and B, stand 6.0 km apart on a straight coastline. They are driven by the same oscillator, so they emit radio waves of frequency 1.5 MHz exactly in phase. A ship sails along a straight course parallel to the coast and 8.0 km from it, and its receiver records the strength of the combined signal. Point O on the course is the same distance from A and from B, and point Q is 4.0 km further along the course. The figure is a map drawn to scale.
Map drawn to scale
(a)
(i)
Calculate the wavelength of the radio waves.
(1)
(ii)
Explain why the signal received at O is strong.
(2)
(iii)
Outline why A and B must be driven by the same oscillator.
(1)
(b)
On the map, draw the line along which the path difference from A and B is zero.
(1)
(c)
Determine the number of positions of maximum signal strength that the ship passes as it sails from O to Q, not counting O itself.
(3)
(d)
The frequency of both transmitters is doubled. Determine the number of maxima that the ship now passes between O and Q.
(2)
(e)
Suggest one limitation of using these maxima to locate a ship.
(1)
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Notes
Part (a)(i)
λ = 3.00 × 108/1.5 × 106 = 200 m
✓ 1
Part (a)(ii)
O is the same distance from A and B, so the path difference is zero
✓ 1
The waves, emitted in phase, arrive in phase and interfere constructively «their amplitudes add»
✓ 1
Part (a)(iii)
To keep the phase difference between the sources constant «coherent», so that the maxima and minima stay in fixed positions
✓ 1
OWTTE.
Part (b)
A straight line through O perpendicular to the coast «the perpendicular bisector of AB»
✓ 1
Must pass through O and the midpoint of AB.
Part (c)
Path difference at Q = √(7.0² + 8.0²) − √(1.0² + 8.0²) = 10.63 − 8.06 = 2.57 km
✓ 1
The path difference increases by one wavelength from one maximum to the next: 2568/200 = 12.8
✓ 1
Allow ECF from (a)(i).
12 maxima «the path difference reaches 12λ but not 13λ»
✓ 1
Do not accept 13. Award [3] for CNA.
Part (d)
λ = 100 m, so 2568/100 = 25.7
✓ 1
25 maxima
✓ 1
Do not accept 24 «double the answer to (c)». Allow ECF from (c).
Part (e)
Any one, e.g.: a maximum fixes a line, not a point, so the ship must count maxima continuously from a known position; OR the pattern is the same on both sides of the line through O, so the side is ambiguous; OR reflections «e.g. from cliffs» shift the positions of the maxima
✓ 1
OWTTE.
Answers: (a)(i) 200 m · (c) path difference 2.57 km; 12 maxima · (d) 25 maxima (the remaining parts are explanations — see the table above)
108C-2-59Refraction & Snell's lawPaper 2Hard13 marks
A primary rainbow is formed by sunlight that is refracted into spherical raindrops, reflected once inside them and refracted out again. Figure 1 shows a ray of sunlight meeting a raindrop at A with an angle of incidence of 59.0°. The refractive index of water is 1.331 for red light and 1.343 for violet light.
The light refracted at A meets the back of the drop at B, where some of it is reflected towards a point C on the surface. It leaves the drop at C. It can be shown that the angle between the incoming sunlight and the light leaving at C is (4r − 2i), where i is the angle of incidence at A and r is the angle of refraction at A.
Figure 1 — drawn to scale
(a)
(i)
Calculate the angle of refraction at A for red light.
(2)
(ii)
Explain why the angle of incidence of the red light at B is equal to the angle of refraction at A.
(1)
(b)
Determine whether the red light is totally internally reflected at B.
(2)
(c)
On Figure 1, draw the path of the red light inside the drop and the light that leaves the drop at B and at C.
(3)
(d)
(i)
Show that the angle between the incoming sunlight and the red light leaving at C is about 42°.
(1)
(ii)
Calculate the corresponding angle for violet light.
(2)
(e)
An observer stands with the Sun behind her. Explain why red light forms the outer edge of the rainbow that she sees and violet light forms the inner edge.
(2)
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Notes
Part (a)(i)
sin r = sin 59.0°/1.331 = 0.6440
✓ 1
r = 40.1°
✓ 1
Award [2] for CNA. Accept 40°.
Part (a)(ii)
OA and OB are both radii, so triangle AOB is isosceles: the angle between AB and the normal OB equals the angle between AB and the normal OA
✓ 1
OWTTE. Accept a labelled sketch.
Part (b)
critical angle: sin c = 1/1.331, c = 48.7°
✓ 1
40.1° is less than 48.7°, so the light is NOT totally internally reflected: only part is reflected and the rest is transmitted out of the drop at B
✓ 1
Allow ECF from (a). MP2 only scores if a comparison with a calculated critical angle is made.
Part (c)
Refracted ray at A bent towards the normal, crossing the drop to B on the far side, below the level of A
✓ 1
At B: a reflected ray making equal angles with the normal OB, going to C in the lower part of the drop, AND a transmitted ray leaving at B, bent away from the normal
✓ 1
The transmitted ray may be shown fainter.
At C: the ray leaves bent away from the normal «at 59° to it», travelling back towards the side the sunlight came from and downwards
✓ 1
Allow ECF from (a)(i).
Part (d)(i)
4 × 40.09° − 2 × 59.0° = 42.4° «≈ 42°»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (d)(ii)
sin r = sin 59.0°/1.343 so r = 39.66°
✓ 1
4 × 39.66° − 2 × 59.0° = 40.6°
✓ 1
Award [2] for CNA. Accept 40.5°–40.7°. A value obtained with r rounded to 39.7° gives 40.8°: accept.
Part (e)
Red light leaves the drops at a larger angle «42.4°» to the direction of the sunlight than violet «40.6°»
✓ 1
Allow ECF from (d).
So the red light that reaches her eye comes from drops at a larger angle from the direction of her shadow «higher in the sky», and violet from drops at a smaller angle «lower»: red on the outside
✓ 1
OWTTE.
Answers: (a)(i) r = 40.1° · (b) c = 48.7° · (d)(i) 42.4° · (d)(ii) 40.6° (the remaining parts are explanations — see the table above)
A student uses microwaves of frequency 10.5 GHz to model Young's double-slit experiment. The microwaves pass through two slits, S1 and S2, each 3.0 cm wide, whose centres are d = 8.0 cm apart. A receiver is moved along a line parallel to the slits at a distance D = 0.600 m from them (Figure 1). Position y is measured from O, which is opposite the midpoint of the slits.
The student finds the centre of the first maximum next to the central maximum at y = 23.0 cm.
Figure 1 — not to scaleFigure 2
(a)
Outline why the microwaves leaving the two slits are coherent.
(1)
(b)
(i)
Calculate the wavelength of the microwaves.
(1)
(ii)
Show that the equation s = λD/d predicts the first maximum at about 21 cm from O.
(1)
(c)
(i)
Calculate the distance from each slit to the receiver when it is at y = 23.0 cm.
(2)
(ii)
Deduce that the student's measurement is consistent with constructive interference.
(1)
(d)
Explain why s = λD/d does not give the position of the first maximum accurately in this arrangement.
(2)
(e)
The student suggests moving the receiver line further from the slits so that the equation becomes accurate. Discuss this suggestion.
(2)
(f)
On Figure 2, sketch how the received signal varies with y between −0.30 m and +0.30 m. Assume that the maxima in this region have equal heights.
(2)
(g)
Slit S2 is covered with a metal plate. Describe and explain how the received signal now varies with y.
(2)
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Notes
Part (a)
Both slits are reached by the same wavefronts from one transmitter, so the waves leaving them have the same frequency and a constant phase difference
Must see full substitution OR answer to at least 3 s.f.
Part (c)(i)
S1: √(0.6002 + 0.1902) = 0.6294 m
✓ 1
Award [1] max if both distances use y measured from one slit.
S2: √(0.6002 + 0.2702) = 0.6580 m
✓ 1
Accept the slits labelled the other way round.
Part (c)(ii)
path difference = 0.6580 − 0.6294 = 0.0286 m, equal to λ «0.0286 m» within the precision of the data: the waves arrive in phase
✓ 1
Allow ECF from (b)(i) and (c)(i).
Part (d)
The equation assumes that the angle between the direction to the receiver and the normal to the slits is small «D much greater than y»
✓ 1
Here y = 23 cm is a large fraction of D «angle ≈ 21°», so the approximations used to obtain the equation are not valid and the prediction is several per cent too small
✓ 1
OWTTE.
Part (e)
Moving the line further away increases y in proportion to D, so the angle of the first maximum is unchanged «it depends only on λ/d»
✓ 1
So the suggestion does not work; the angle is made small by using slits further apart or a shorter wavelength «λ/d ≪ 1»
✓ 1
MP2 needs the reason from MP1 or an equivalent argument.
Part (f)
Maxima at y = 0 and at about ±0.23 m, of equal height
✓ 1
Allow ECF from the stem value.
Minima «close to zero» midway, at about ±0.11 m, with a smooth symmetrical curve
✓ 1
Accept minima at ±0.10 to ±0.12 m.
Part (g)
There are no longer any minima: the signal varies smoothly, largest opposite S1 and falling off gradually on both sides
✓ 1
Only one source remains, so there is no interference; the slit «3.0 cm» is about one wavelength wide, so the microwaves diffract through it and spread over a wide angle
✓ 1
OWTTE.
Answers: (b)(i) λ = 2.86 × 10−2 m · (b)(ii) 0.214 m · (c)(i) 0.6294 m and 0.6580 m · (c)(ii) path difference = 0.0286 m ≈ λ(the remaining parts are explanations — see the table above)
110IBP2S-022-02-16Reflection & transmissionPaper 2Hard9 marks
A radio antenna stands on a sea cliff at a height h = 120 m above a calm sea. It receives radio waves of wavelength 1.50 m from a distant source in space as the source rises above the horizon. Waves reach the antenna directly, and also by reflection off the sea surface. Reflection at the sea surface introduces a phase change of π. The source is at an elevation angle θ above the horizontal. The reflected waves behave as if they came from an image of the antenna a distance h below the sea surface.
Diagram not to scale. Distant source at elevation θ.
(a)
Explain why the direct and reflected waves are coherent.
(2)
(b)
Show that the path difference between the reflected and direct waves is approximately 2h sin θ.
(2)
(c)
Explain why the received signal is a minimum when the source is on the horizon (θ = 0).
(2)
(d)
Determine the smallest elevation angle at which the signal is a maximum.
(2)
(e)
Estimate the angular separation between successive maxima as the source rises.
(1)
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Notes
Part (a)
Both waves originate from the same source
✓ 1
They have the same frequency and a constant phase difference at the antenna for a given θ
✓ 1
Part (b)
The antenna and its image are separated by a vertical distance 2h, and the incoming rays are parallel at angle θ to the horizontal
✓ 1
The extra distance travelled by the reflected ray is the component of 2h along the ray direction, which is 2h sin θ
✓ 1
Part (c)
At θ = 0 the path difference is zero
✓ 1
The phase change of π on reflection means the waves arrive in antiphase, giving destructive interference
✓ 1
Part (d)
Because of the π phase change, a maximum needs 2h sin θ = ½λ, so sin θ = 1.50/(4 × 120) = 3.1 × 10−3
✓ 1
θ = 0.18°
✓ 1
Part (e)
The path difference changes by λ between successive maxima, so for small angles Δθ ≈ λ/(2h) = 1.50/240 = 6.25 × 10−3 rad ≈ 0.36°
✓ 1
Common error: Forgetting the phase change of π on reflection swaps the positions of the maxima and minima. Always add any phase change on reflection to the phase difference from the path difference.
111IBP2S-022-00-11Refractive index & speedPaper 2Hard8 marks
A data-centre link uses an optical fibre 2.00 km long. The core has refractive index 1.50. The critical angle at the core–cladding boundary is 80.0°, measured from the normal to the boundary. Two short pulses of light enter the fibre at the same instant. Pulse X travels along the axis. Pulse Y follows a zigzag path, striking the core–cladding boundary at exactly the critical angle. Take c = 3.00×108 m s−1.
Diagram not to scale. Pulse X travels along the axis; pulse Y zig-zags.
(a)
Calculate the refractive index of the cladding.
(2)
(b)
Show that pulse X takes about 1.0×10−5 s to travel the length of the fibre.
(2)
(c)
Determine the difference between the arrival times of pulses X and Y.
(3)
(d)
Outline why this effect limits how closely together successive pulses can be sent.
(1)
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Notes
Part (a)
Uses sin c = nclad/ncore .
✓ 1
nclad = 1.50 × sin 80.0° = 1.48 .
✓ 1
Part (b)
v = c/n = 3.00×108/1.50 = 2.00×108 m s−1 .
✓ 1
t = 2000/(2.00×108) = 1.00×10−5 s .
✓ 1
Part (c)
The angle of incidence (80.0°) is measured from the normal, so ray Y makes 10.0° with the axis. Its path length is 2000/sin 80.0° (or 2000/cos 10.0°) = 2.03×103 m .
✓ 1
tY = 2031/(2.00×108) = 1.015×10−5 s .
✓ 1
Δt = 1.54×10−7 s; accept any answer from 1.5×10−7 s to 1.55×10−7 s .
✓ 1
Allow error carried forward from (b).
—
Part (d)
Each pulse spreads out in time, so pulses sent too close together would overlap and could not be distinguished .
✓ 1
Exam tip: For zigzag rays, the path length is L/sin c, because the ray makes an angle (90° − c) with the fibre axis. Keep intermediate times to 4 s.f. before subtracting.
112IBP2S-022-01-16Young's double slitPaper 2Hard9 marks
A satellite-communications lab demonstrates two-source interference using microwaves of wavelength 2.8 cm. The microwaves pass through two narrow slits 8.0 cm apart in a metal sheet. A detector is moved along a straight line parallel to the sheet, 1.20 m from it.
Diagram not to scale
(a)
State the condition for an intensity maximum in terms of the path difference from the two slits.
(1)
(b)
Calculate the angle between the central maximum and the first-order maximum.
(2)
(c)
Explain why the equation s = λD/d would give an inaccurate distance from the central maximum to the first-order maximum in this experiment.
(2)
(d)
Determine the total number of maxima that the detector could find on the far side of the sheet.
(2)
(e)
Calculate the distance along the detector's line from the central maximum to the first-order maximum.
(2)
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Notes
Part (a)
The path difference must be a whole number of wavelengths, nλ, where n = 0, 1, 2, …
✓ 1
Part (b)
Use d sin θ = nλ with n = 1: sin θ = λ/d = 2.8/8.0 = 0.35
✓ 1
θ = 20.5°
✓ 1
Part (c)
The equation assumes small angles, so that sin θ ≈ tan θ ≈ θ
✓ 1
Here θ is about 20°, so sin θ (0.350) and tan θ (0.374) differ significantly and the approximation is not valid
✓ 1
Part (d)
sin θ < 1 requires n < d/λ = 2.86, so the highest order is n = 2
✓ 1
Orders 0, ±1 and ±2 give 5 maxima
✓ 1
Part (e)
y = D tan θ = 1.20 × tan 20.5° (allow error carried forward from the angle calculated earlier)
✓ 1
y = 0.45 m (0.448 m)
✓ 1
Common error: With microwaves, λ is comparable to d, so the angles are large. Use d sin θ = nλ and y = D tan θ rather than the small-angle fringe equation.
113IBP2S-022-03-16Refractive index & speedPaper 2Hard8 marks
A hospital links its imaging centre to a data archive 2.0 km away with a step-index optical fibre. The core has refractive index 1.48 and the cladding has refractive index 1.46. The flat end face of the fibre is perpendicular to its axis and surrounded by air. The speed of light in a vacuum is 3.00×108 m s−1.
Diagram not to scale
(a)
Calculate the critical angle at the core–cladding boundary.
(2)
(b)
Determine the largest angle of incidence in air at the end face for which light is still guided along the core by total internal reflection.
(3)
(c)
One ray travels straight along the axis of the fibre. A second ray zig-zags, meeting the core–cladding boundary at exactly the critical angle. Determine the difference in their arrival times at the archive.
(3)
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Notes
Part (a)
sin θc = 1.46/1.48 = 0.986
✓ 1
θc = 80.6°
✓ 1
Part (b)
The largest angle of refraction at the end face is 90° − 80.6° = 9.4°
✓ 1
1.00 sin θ = 1.48 sin 9.4° = 0.242
✓ 1
θ = 14.0°; ECF from (a)
✓ 1
Part (c)
Axial ray: t = nL/c = 1.48 × 2000/3.00×108 = 9.87×10−6 s
✓ 1
Zig-zag path length = L/sin θc = 2000/0.986 = 2027 m, so t = 1.000×10−5 s
✓ 1
Δt ≈ 1.35×10−7 s (accept 1.3–1.4×10−7 s); ECF from (a)
✓ 1
Exam tip: In part (c), find the time difference with full-precision values and round only at the end. Subtracting two nearly equal rounded times can lose most of the significant figures.
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