The questions in this review are original Marlbridge practice questions, not past exam questions, and each combines two or more topics. The rubrics are a suggested Marlbridge rubric, not official scoring. Data: c = 3.00 × 10⁸ m/s; on a string v = √(F_T/μ); filters are ideal; light meets films at normal incidence; visible light runs from about 400 nm to 700 nm. All data are invented.
Big ideas of the unit
- A wave moves energy, not matter. Its speed is set by the medium: √(F_T/μ) on a string, the material and its temperature for sound, c for light in a vacuum (Topic 14.1).
- The source sets f; λ follows. v = fλ and T = 1/f link every calculation in the unit (Topic 14.2).
- At a boundary f stays, v and λ change. A reflection into a slower medium is inverted; only transverse waves can be polarized (Topic 14.3).
- Light is a transverse EM wave. Its E and B fields are perpendicular to each other and to the travel; the spectrum is ordered by wavelength (Topic 14.4).
- Relative motion shifts the received frequency. A shrinking separation raises f; a faster change gives a bigger shift (Topic 14.5).
- Superposition explains beats and standing waves. Boundary conditions choose which wavelengths fit (Topic 14.6).
- Path difference decides bright or dark. One slit: dark at a sin θ = mλ (Topic 14.7). Two slits or a grating: bright at d sin θ = mλ (Topic 14.8).
- Thin films combine both ideas. Count the reflection phase changes, then compare 2t with λ/n (Topic 14.9).
Key relationships and methods
| Idea | Relationship or method | Watch out for |
|---|---|---|
| String speed (14.1) | v = √(F_T/μ) | v ∝ √F_T, not F_T |
| Periodic wave (14.2) | v = fλ; T = 1/f | Reading λ from a time axis |
| Boundary (14.3) | f fixed; λ₂/λ₁ = v₂/v₁; slower → inverted reflection | Inverting the transmitted pulse |
| Polarization (14.3) | Unpolarized → I₀/2; crossed filters → 0 | Using B instead of E for the direction |
| EM waves (14.4) | c = fλ; radio → gamma, λ falls | Giving waves different speeds in a vacuum |
| Doppler (14.5) | Approach: f > f₀; recede: f < f₀ | Qualitative only; no formula needed |
| Beats (14.6) | f_beat = |f₁ − f₂| | Using the average or the sum |
| Standing waves (14.6) | Same ends: fₙ = nv/(2L); node–antinode: fₙ = nv/(4L), n odd | Nodes are λ/2 apart |
| Single slit (14.7) | Dark: a sin θ = mλ; y_min = mλL/a | m = 0 is the bright centre |
| Double slit, grating (14.8) | Bright: d sin θ = mλ; y_max = mλL/d | No small-angle form for gratings |
| Thin film (14.9) | Count 180° changes; compare 2t with λ/n | Using λ in air for the film |
Question 1 (multiple choice · mixed)
Topics 14.5, 14.6. Two identical loudspeakers emit the same steady note. A listener stands still in front of both. Speaker P is fixed. Speaker Q is carried steadily toward the listener, and the listener hears beats. Which statement is correct?
- (A) No beats are possible, because both speakers vibrate at the same frequency.
- (B) Q’s sound arrives above the rest frequency; carrying Q faster makes the beats slower.
- (C) Q’s sound arrives above the rest frequency; carrying Q faster makes the beats faster.
- (D) The beats happen because Q’s sound travels faster through the air than P’s.
Answer and explanation
Answer: (C). Q approaches, so the listener receives it above f₀, while P arrives at f₀. Two slightly different received frequencies give beats at f_beat = |f_Q − f_P|. A faster approach gives a bigger shift, so a bigger difference and faster beats.
- (A) uses the rest frequencies. Beats depend on the frequencies the listener receives.
- (B) reverses the size rule.
- (D) The speed of sound is set by the air, not by the motion of the source.
Question 2 (multiple choice · mixed)
Topics 14.4, 14.7, 14.8. A double slit has slits 0.30 mm apart, each 0.060 mm wide. It is lit first with green light, then with red light. The screen does not move. Compared with green, which is correct for red?
- (A) Wider fringe spacing; same envelope; fewer bright fringes inside the central band
- (B) Wider fringe spacing; wider envelope; the same number of bright fringes inside the envelope’s central band
- (C) Narrower fringe spacing; narrower envelope; the same number inside the central band
- (D) Wider fringe spacing; wider envelope; more bright fringes inside the central band
Answer and explanation
Answer: (B). Red has the longer wavelength. The fringe spacing λL/d and the envelope width 2λL/a both grow by the same factor. The count inside the central band depends only on d/a = 5: the fifth-order fringes are missing, leaving 9 bright fringes for either colour.
- (A) forgets that the envelope also depends on λ.
- (C) reverses the spectrum order: red is longer than green.
- (D) changes the count. Both widths scale together, so the ratio is fixed.
Question 3 (constructed response · mixed)
Topics 14.1, 14.2, 14.3, 14.6. A light string (μ = 0.0040 kg/m) is joined to a heavy string (μ = 0.016 kg/m). Both are under a tension of 40 N.
(a) Calculate the wave speed on each string. (b) A vibrator on the light string sends a steady 125 Hz wave towards the join. Calculate the period and the wavelength on each side of the join. (c) The vibrator is replaced by a single upright pulse. A student predicts: “The reflected pulse will be upright, and the pulse that enters the heavy string will be twice as wide.” Evaluate both parts of the prediction. (d) The heavy string is removed. The light string is clamped at two points 1.20 m apart, at the same tension, and driven at 125 Hz. Show that a standing wave forms. State its harmonic number, the number of nodes and the node spacing.
Worked solution and suggested Marlbridge rubric
(a) v₁ = √(40 ÷ 0.0040) = 100 m/s; v₂ = √(40 ÷ 0.016) = 50 m/s.
(b) f does not change at the join, so T = 1/125 Hz = 8.0 ms on both strings. λ₁ = 100 ÷ 125 = 0.80 m; λ₂ = 50 ÷ 125 = 0.40 m.
(c) Both parts are wrong. The wave slows down at the join, so the reflected pulse is inverted. The transmitted pulse takes the same time to pass the join as the incident pulse, but moves at half the speed, so it is half as wide, not twice as wide.
(d) Fixed at both ends: f₁ = v/(2L) = 100 ÷ 2.40 = 41.7 Hz. 125 Hz = 3 × 41.7 Hz, a whole number, so a standing wave forms: the third harmonic. Check: λ/2 = 0.40 m, and 1.20 m ÷ 0.40 m = 3 half-wavelengths. Three loops need 4 nodes (including the ends), 0.40 m apart.
| Point | What earns it |
|---|---|
| 1 | Both speeds |
| 1 | T = 8.0 ms on both sides, because f is unchanged |
| 1 | λ₁ = 0.80 m and λ₂ = 0.40 m |
| 1 | Reflected pulse inverted, linked to the slower heavy string |
| 1 | Transmitted pulse half as wide, using equal passing time |
| 1 | f₁ = 41.7 Hz and 125 Hz = 3f₁ (or 3 half-wavelengths fit) |
| 1 | 4 nodes, 0.40 m apart |
Total: 7 points. Carry forward a wrong speed once.
Question 4 (constructed response · mixed)
Topics 14.1, 14.2, 14.5, 14.6. Two pipes, A and B, are each 0.50 m long and closed at one end. The air in A is at room temperature (sound speed 340 m/s). The air in B is warmer (sound speed 346 m/s).
(a) Sketch the fundamental of pipe A with N and A marked. Calculate its wavelength and frequency. (b) Explain why B’s fundamental is higher, and calculate it. (c) Both pipes sound their fundamentals together. Calculate the beat frequency and the time between loud moments. (d) Predict the beat frequency when both sound their third harmonics. (e) With the fundamentals sounding, pipe B is wheeled steadily toward the listener. A student says the beats will stop. Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Node at the closed end, antinode at the open end: the pipe holds a quarter-wavelength. λ = 4L = 2.0 m; f_A = 340 ÷ 2.0 = 170 Hz.
(b) The length, and so λ = 2.0 m, is the same. Sound travels faster in warmer air, and f = v/λ, so f_B = 346 ÷ 2.0 = 173 Hz.
(c) f_beat = 173 − 170 = 3 Hz; loud moments are 1/3 s = 0.33 s apart.
(d) Third harmonics: 3 × 170 = 510 Hz and 3 × 173 = 519 Hz, so the beat frequency is 9 Hz. Every harmonic of B is higher by the same factor, so the difference grows with n.
(e) The claim is wrong. B approaches, so the listener receives it above 173 Hz. A is still received at 170 Hz. The difference grows, so the beats get faster, not slower. (They would slow down only if the higher-pitched pipe moved away.)
| Point | What earns it |
|---|---|
| 1 | Sketch with N at the closed end and A at the open end, a quarter-wavelength |
| 1 | λ = 2.0 m and f_A = 170 Hz |
| 1 | B higher because v is higher with λ fixed by the pipe; 173 Hz |
| 1 | 3 Hz and 0.33 s |
| 1 | 9 Hz for the third harmonics |
| 1 | Approaching B is received above 173 Hz |
| 1 | So the difference and the beat frequency increase; claim rejected |
Total: 7 points. Do not award the last point for a correct verdict without the received-frequency reasoning.
Question 5 (constructed response · mixed)
Topics 14.4, 14.7, 14.8. A red laser shines straight onto a grating with 500 lines/mm. On a screen 1.00 m away, each first-order spot is 0.344 m from the central spot.
(a) Calculate the angle of the first-order spots and the wavelength. (b) Show that the small-angle form would give a wavelength about 6% too large. (c) Find the highest order the grating produces for this light. (d) The same laser shines through a single slit onto a screen 2.00 m away. The central bright band is 10.4 mm wide. Calculate the slit width. (e) A violet laser (410 nm) replaces the red one. Predict the new first-order distance on the grating screen and the new central band width for the slit.
Worked solution and suggested Marlbridge rubric
(a) d = (1/500) mm = 2.00 × 10⁻⁶ m. tan θ = 0.344 ÷ 1.00, so θ = 19.0°. λ = d sin θ = 6.51 × 10⁻⁷ m (651 nm), red as expected.
(b) Small-angle: λ = dy/L = (2.00 × 10⁻⁶)(0.344) ÷ 1.00 = 688 nm, which is 688 ÷ 651 = 1.06 times too large. The angle is far above 10°.
(c) m < d/λ = 3.07, so the highest order is m = 3 (at 77°).
(d) Central band = 2λL/a, so a = 2λL ÷ width = 2(6.51 × 10⁻⁷)(2.00) ÷ (1.04 × 10⁻²) = 2.50 × 10⁻⁴ m (0.25 mm).
(e) Grating: sin θ = 4.10 × 10⁻⁷ ÷ 2.00 × 10⁻⁶ = 0.205, θ = 11.8°, y = L tan θ = 0.209 m. Slit: width ∝ λ, so 10.4 mm × (410 ÷ 651) = 6.6 mm. Violet has the shorter wavelength, so both patterns shrink.
| Point | What earns it |
|---|---|
| 1 | d = 2.00 × 10⁻⁶ m and θ = 19.0° from tan θ = y/L |
| 1 | λ = 651 nm |
| 1 | Small-angle value 688 nm, about 6% high, with the angle above 10° as the reason |
| 1 | Highest order 3 from d/λ |
| 1 | a = 0.25 mm, using the full central width 2λL/a |
| 1 | Violet grating spot at about 0.21 m, using d sin θ = mλ and tan θ |
| 1 | Central band about 6.6 mm, linked to the shorter wavelength |
Total: 7 points. Carry forward a wrong λ from (a) once.
Question 6 (constructed response · mixed)
Topics 14.3, 14.4, 14.9. A thin layer of oil (n = 1.45) floats on water (n = 1.33). Its thickness grows smoothly from zero at its edge. It is lit from above by light of wavelength 550 nm.
(a) Using the rule for pulses on strings, explain which of the two reflections has a 180° phase change. (b) Calculate the frequency and the wavelength of the light inside the oil. (c) The reflected light shows dark and bright bands. Explain why the edge is dark, and find the thicknesses at the first two dark bands after the edge. (d) In sunlight, glare from the oil and water surface is partly polarized horizontally. Explain why sunglasses with a vertical transmission axis reduce the glare, and predict what happens if the wearer tilts her head through 90°.
Worked solution and suggested Marlbridge rubric
(a) Light is slower in a medium with higher n (v = c/n). A reflection is inverted when the wave meets a slower medium. Air → oil: 1.00 to 1.45, slower, so 180°. Oil → water: 1.45 to 1.33, faster, so no change. Exactly one phase change.
(b) f = c/λ = (3.00 × 10⁸) ÷ (5.50 × 10⁻⁷) = 5.45 × 10¹⁴ Hz, the same in the oil. λ_oil = 550 ÷ 1.45 = 379 nm.
(c) With one phase change, dark needs 2t = mλ_oil. At the edge t ≈ 0, so the single half-cycle shift makes the reflections cancel: dark. Next: t = λ_oil/2 = 190 nm (m = 1) and t = λ_oil = 379 nm (m = 2).
(d) Light is a transverse wave, and polarization means its E field oscillates along one direction. A vertical axis blocks the horizontally polarized glare but passes half of the unpolarized light from other objects. Tilted through 90°, the axis is horizontal, so the glare passes and looks brighter.
| Point | What earns it |
|---|---|
| 1 | Higher n means slower light, linked to the string rule |
| 1 | 180° at air–oil, none at oil–water |
| 1 | f = 5.45 × 10¹⁴ Hz, unchanged in the oil |
| 1 | λ_oil = 379 nm |
| 1 | Edge dark: t ≈ 0 with one phase change |
| 1 | Dark at 190 nm and 379 nm |
| 1 | Vertical axis blocks horizontal glare; glare returns when tilted |
Total: 7 points. Accept 2nt = mλ for (c).
How did you do?
Questions 1–2 are worth 1 point each and Questions 3–6 are worth 28 in total. There is no pass mark: use the checklist that matches the points you lost.
- wave speed (Q3(a), 4(b)): Topic 14.1 checklist
- v = fλ and period (Q3(b), 4(a)): Topic 14.2 checklist
- boundaries and polarization (Q3(b)–(c), 6(a), 6(d)): Topic 14.3 checklist
- EM waves and the spectrum (Q2, 6(b)): Topic 14.4 checklist
- Doppler effect (Q1, 4(e)): Topic 14.5 checklist
- beats and standing waves (Q1, 3(d), 4): Topic 14.6 checklist
- single slit (Q2, 5(d)): Topic 14.7 checklist
- double slit and grating (Q2, 5(a)–(c)): Topic 14.8 checklist
- thin films (Q6(a)–(c)): Topic 14.9 checklist
For a quicker topic-by-topic check, take the Unit 14 diagnostic.