C.3 is about what waves do at boundaries and when they meet. You need wavefront and ray diagrams for reflection, refraction and diffraction, Snell's law n₁/n₂ = sin θ₂/sin θ₁ = v₂/v₁, the critical angle and total internal reflection.
Interference questions use the superposition of pulses and waves, the need for coherent sources, path difference conditions for constructive and destructive interference, and Young's double-slit result s = λD/d. Single-slit diffraction patterns and diffraction gratings are Higher Level.
25 questions
158 marks
Paper 1A: 12
Paper 1B: 4
Paper 2: 9
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25 practice questions on C.3 Wave phenomena
1C-1A-18
Snell's law·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A wave passes from medium 1 into medium 2. The angle between the ray and the normal is 50° in medium 1 and 30° in medium 2.
What is (speed of the wave in medium 2)/(speed of the wave in medium 1)?
Diagram drawn to scaleShow mark scheme
Marking point
Mark
Notes
Step 1Snell's law: v2/v1 = sin θ2/sin θ1.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2v2/v1 = sin 30°/sin 50° = 0.500/0.766 = 0.65.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis is the ratio of the angles, 30/50. Snell's law relates the sines of the angles, not the angles themselves.
BCorrect: v2/v1 = sin θ2/sin θ1 = sin 30°/sin 50° = 0.65; the wave slows and bends towards the normal.
CThis uses the angles between the ray and the boundary, 40° and 60°, giving sin 60°/sin 40°. The angles in Snell's law are measured from the normal.
DThis is the inverse ratio, sin 50°/sin 30°, which is v1/v2. The wave bends towards the normal, so it slows down.
Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1Command term: Determine
2C-1A-19
Critical angle & TIR·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
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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Notes
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.
✓ 1
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.
Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection Command term: State
3C-1A-20
Refractive index & speed·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
Light travels at 2.25 × 108 m s−1 in a transparent liquid. In a vacuum it travels at 3.00 × 108 m s−1.
What is the refractive index of the liquid?
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Notes
Step 1n = c/v = 3.00 × 10⁸/2.25 × 10⁸ = 1.33.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis is (c − v)/v, the fractional increase in speed, not the ratio of the speeds.
BThis is v/c. A refractive index is never less than 1, because light travels fastest in a vacuum.
CCorrect: n = c/v = 3.00 × 10⁸/2.25 × 10⁸ = 1.33.
DThis squares the ratio of the speeds, (c/v)². The refractive index is the ratio itself.
Syllabus understandingC.3 — Snell's law, and the refractive index of a medium Command term: Calculate
4C-1A-21
Path difference·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
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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Notes
Step 1Bright fringes have path difference nλ; the central fringe is n = 0, so the second bright fringe beyond it, P, has 2λ.
—
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.
✓ 1
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.
Syllabus understandingC.3 — the condition for constructive interference as given by path difference = nλ, and for destructive interference as given by path difference = (n + ½)λCommand term: Determine
5C-1A-22
Young's double slit·C.3 Wave phenomena
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
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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Notes
Step 1Dark fringes lie midway between bright fringes, at 0.5s, 1.5s, … from the centre of the central bright fringe.
—
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.
—
Step 3s = λD/d gives λ = sd/D = yd/(3D).
✓ 1
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.
Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d, where s is the separation of fringes, d is the separation of the slits, and D is the distance from the slits to the screen Command term: Determine
6C-1A-23
Young's double slit·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
In a double-slit experiment with monochromatic light, the distance between the slits and the screen is doubled and the slit separation is halved.
What is (new fringe separation)/(original fringe separation)?
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Notes
Step 1s = λD/d with λ unchanged, so s ∝ D/d.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2D/d is multiplied by 2/(1/2) = 4, so the fringe separation is multiplied by 4.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis is the inverse ratio, as if s ∝ d/D. Moving the screen further away spreads the fringes out.
BThis treats s as proportional to D × d, so that doubling and halving cancel. The slit separation is in the denominator.
COnly the change in D has been applied; halving the slit separation doubles the fringe separation again.
DCorrect: s ∝ D/d, and D/d becomes 2 ÷ ½ = 4 times its original value.
Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/dCommand term: Determine
7C-1A-24
Diffraction·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
Plane water waves in a ripple tank pass through a gap in a barrier. The width of the gap is then reduced until it is about equal to one wavelength.
What happens to the waves beyond the gap?
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Notes
Step 1Diffraction is greatest when the gap width is comparable to the wavelength, so the waves spread out more; the wavelength itself is unaffected.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
ANothing about the medium has changed, so the speed and the frequency — and therefore the wavelength — are the same.
BA narrower gap produces more spreading, not less.
CA gap comparable to the wavelength produces the largest amount of diffraction.
DCorrect: diffraction becomes more pronounced as the gap approaches the wavelength, and diffraction does not change the wavelength.
Syllabus understandingC.3 — diffraction through a single slit and around objects Command term: State
8C-1A-25
Superposition·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
Two pulses, X and Y, travel towards each other along a string. The diagram shows the string at one instant, drawn on a square grid.
Which diagram shows the string at the instant when the two pulses exactly overlap?
Drawn to scale on a square gridDrawn to the same scaleShow mark scheme
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Notes
Step 1By the principle of superposition the displacements add at every point: (+2 squares) + (−1 square).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The pulses have the same shape and width, so the string forms a single upward pulse of that shape, 1 square high.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the displacements add algebraically, 2 + (−1) = +1 square, giving an upward pulse of the same shape.
BPulses of opposite sign cancel completely only when their displacements are equal in size; here they are 2 squares and 1 square.
CThe sign of the result is wrong: the larger pulse is upward, so the resultant displacement is upward.
DThis adds the sizes of the two displacements and ignores the fact that pulse Y is a negative (downward) displacement.
Syllabus understandingC.3 — superposition of waves and wave pulses Command term: Identify
9C-1B-07
Snell's law·C.3 Wave phenomena
Paper 1BEasy9 marks
Data-based question9 steps to full marksDetermine
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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Notes
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
✓ 1
Part (d)
n = sin i/sin r = gradient of the line, found from a large triangle
✓ 1
n = 1.33
✓ 1
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
✓ 1
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)
Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection; Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1Command term: Determine
10C-1B-08
Young's double slit·C.3 Wave phenomena
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
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
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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)
Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d, where s is the separation of fringes, d is the separation of the slits, and D is the distance from the slits to the screen Command term: Determine
11C-2-07
Snell's law·C.3 Wave phenomena
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksExplain
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)
Show mark scheme
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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)
Syllabus understandingC.3 — wavefront-ray diagrams showing refraction; Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; wave behaviour at boundaries in terms of reflection, refraction and transmission (with C.2 — v = fλ) Command term: Explain
12C-2-08
Critical angle & TIR·C.3 Wave phenomena
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine whether
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)
Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection; wave behaviour at boundaries in terms of reflection, refraction and transmission Command term: Determine whether
13C-2-09
Young's double slit·C.3 Wave phenomena
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksExplain
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d; the condition for destructive interference as given by path difference = (n + ½)λ; superposition of waves Command term: Explain
14C-2-10
Superposition·C.3 Wave phenomena
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksExplain
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)
Syllabus understandingC.3 — superposition of waves and wave pulses; that double-source interference requires coherent sources; the condition for constructive interference as given by path difference = nλ; the condition for destructive interference as given by path difference = (n + ½)λ Command term: Explain
15C-1A-26
Coherent sources·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksExplain
Two identical lamps placed side by side illuminate a screen. No interference fringes are seen on the screen.
What is the reason?
Show mark scheme
Marking point
Mark
Notes
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.
✓ 1
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.
Multiple choice · 1 mark2 steps to full marksDetermine
An optical fibre has a core of refractive index 1.50 surrounded by cladding of refractive index 1.45.
What is the critical angle at the boundary between the core and the cladding?
Show mark scheme
Marking point
Mark
Notes
Step 1At the critical angle the refracted ray grazes the boundary (90°): ncore sin c = ncladding sin 90°.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2sin c = 1.45/1.50 = 0.967, so c = 75.2°.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis is 90° − 75.2°, the angle measured from the boundary rather than from the normal.
BThis is the critical angle for a core–air boundary, sin c = 1/1.50. The second medium here is the cladding, not air.
CThis uses sin c = 1/1.45, the critical angle for cladding against air, not for the core–cladding boundary.
DCorrect: sin c = ncladding/ncore = 1.45/1.50, so c = 75.2°.
Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection Command term: Determine
17C-1A-28
Path difference·C.3 Wave phenomena
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
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
Show mark scheme
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Notes
Step 1At X there is no path difference, so the waves arrive with the phase difference of the sources, π: destructive interference, a minimum.
—
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.
✓ 1
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.
Syllabus understandingC.3 — superposition of waves and wave pulses; that double-source interference requires coherent sources Command term: Deduce
18C-1B-15
Refractive index & speed·C.3 Wave phenomena
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
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
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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.
1.468 lies within the range 1.454 to 1.478, so the result is consistent with the manufacturer's value
✓ 1
Allow ECF from (c) and (d). Must refer to the uncertainty.
Part (f)
There is a constant delay (about 150 ns) in the photodiode, the connecting cables or the oscilloscope trigger, which is added to every time
✓ 1
Accept 'the pulse travels through extra fibre/cable not included in L'. Do not accept 'random error'.
Part (g)
sin c = n2/n1 = 1.450/1.468 = 0.9877
✓ 1
Allow ECF from the candidate's n in (c) for n₁ only if it is greater than 1.450.
c = 81.0°
✓ 1
Accept 79° to 83°, or the ECF value (e.g. n = 1.466 gives 81.5°; n = 1.47 gives 80.5°).
Answers: (b) 4.89 × 10−9 s m−1 · (c) n ≈ 1.47 · (d) Δn ≈ ±0.01 · (g) c ≈ 81.0° (the remaining parts are explanations — see the table above)
Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection; Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; the refractive index as the ratio of wave speeds Command term: Determine
19C-1B-16
Young's double slit·C.3 Wave phenomena
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
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)
Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d, where s is the separation of fringes, d is the separation of the slits, and D is the distance from the slits to the screen; Tool 3 — linearizing a power law with a log–log graph Command term: Determine
20C-2-24
Diffraction·C.3 Wave phenomena
Paper 2Medium11 marks
Short answer & extended response11 steps to full marksDiscuss
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)
Syllabus understandingC.3 — wave diffraction around a body and through an aperture; wavefront-ray diagrams showing refraction and diffraction; that waves travelling in two and three dimensions can be described through the concepts of wavefronts and rays Command term: Discuss
21C-2-25
Critical angle & TIR·C.3 Wave phenomena
Paper 2Easy12 marks
Short answer & extended response12 steps to full marksDetermine whether
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)
Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection; Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; wave behaviour at boundaries in terms of reflection, refraction and transmission Command term: Determine whether
22C-2-26
Superposition·C.3 Wave phenomena
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDetermine whether
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)
Syllabus understandingC.3 — superposition of waves and wave pulses; the condition for constructive interference as given by path difference = nλ; the condition for destructive interference as given by path difference = (n + ½)λ; C.2 — the nature of sound waves Command term: Determine whether
23C-2-27
Young's double slit·C.3 Wave phenomena
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksDeduce
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.
(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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d, where s is the separation of fringes, d is the separation of the slits, and D is the distance from the slits to the screen; the condition for constructive interference as given by path difference = nλ; the condition for destructive interference as given by path difference = (n + ½)λ; that double-source interference requires coherent sources Command term: Deduce
24C-2-28
Path difference·C.3 Wave phenomena
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksDetermine
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)
Show mark scheme
Marking point
Mark
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)
Syllabus understandingC.3 — that double-source interference requires coherent sources; the condition for constructive interference as given by path difference = nλ Command term: Determine
25C-1A-45
Refraction & Snell's law·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
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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Marking point
Mark
Notes
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.
✓ 1
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.
Syllabus understandingC.3 — wavefront-ray diagrams showing refraction and diffraction; wave behaviour at boundaries in terms of reflection, refraction and transmission Command term: Deduce
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