A Level H2 Physics Waves Sound Light Quiz

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A-Level Physics H2 Quiz - Waves Sound Light

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 50 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Section A: Short structured questions (1–8). Section B: Calculation and application (9–15). Section C: Data interpretation and extended response (16–20).
• Show all working clearly. Use SI units.
• Marks for each question are shown in brackets.

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Section A: Short Structured Questions (1–8)

1. State what is meant by the term wavelength of a progressive wave. [1]

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2. A sound wave of frequency 512 Hz travels in air at $340\ \text{m s}^{-1}$. Calculate its wavelength. [2]

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3. Explain why polarisation is evidence that light is a transverse wave. [2]

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4. State the principle of superposition of waves. [1]

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5. The intensity of a sound wave from a point source is $I$ at distance $r$. Write the relationship between intensity and distance. [1]

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6. A standing wave is formed on a string. State the meaning of a node. [1]

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7. In double-slit interference, what condition must be satisfied for a bright fringe to be observed at a point? [2]

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8. A diffraction grating has 500 lines per mm. State the grating spacing $d$ in metres. [2]

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Section B: Calculation and Application (9–15)

9. A wave on a string has amplitude $0.040\ \text{m}$ and frequency $25\ \text{Hz}$. The wave speed is $10\ \text{m s}^{-1}$. (a) Calculate the wavelength. [2] (b) Calculate the period. [1]

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10. Two identical loudspeakers are placed $0.80\ \text{m}$ apart and emit sound of frequency $680\ \text{Hz}$ in phase. A listener is $3.0\ \text{m}$ directly in front of one speaker. Speed of sound $= 340\ \text{m s}^{-1}$. (a) Calculate the wavelength of the sound. [1] (b) Determine the path difference to the listener. [2] (c) State whether constructive or destructive interference occurs. [1]

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11. Light of wavelength $600\ \text{nm}$ passes through double slits separated by $0.30\ \text{mm}$. The screen is $2.0\ \text{m}$ away. (a) Calculate the fringe separation $\Delta y$. [3] (b) State what happens to $\Delta y$ if the screen distance is doubled. [1]

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12. A diffraction grating with $1.0 \times 10^5$ lines per metre is used to observe light of wavelength $500\ \text{nm}$. (a) Calculate the grating spacing $d$. [1] (b) Find the angle $\theta$ for the first-order maximum ($n=1$). [3]

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13. Unpolarised light of intensity $I_0$ passes through a polariser then an analyser at $60^\circ$ to the polariser. (a) State the intensity after the polariser. [1] (b) Calculate the intensity after the analyser using Malus' law. [2]

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14. A string fixed at both ends is $1.2\ \text{m}$ long and vibrates in its fundamental mode. (a) State the wavelength of the standing wave. [1] (b) If the wave speed is $48\ \text{m s}^{-1}$, calculate the fundamental frequency. [2]

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15. A sound source emits $0.20\ \text{W}$ uniformly. Calculate the intensity at $4.0\ \text{m}$ from the source. [3]

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Section C: Data Interpretation and Extended Response (16–20)

16. The diagram shows a displacement–distance graph of a transverse wave at time $t=0$.

Image pending generation: graph for Q16.

(a) Determine the amplitude. [1] (b) Determine the wavelength. [1] (c) If the frequency is 5.0 Hz, calculate the wave speed. [2]

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17. The figure shows a standing wave on a string at two instants.

Image pending generation: diagram for Q17.

(a) For the 3-loop mode, state the number of nodes. [1] (b) Calculate the wavelength of the 3-loop mode. [2] (c) Explain how the frequency changes when the string goes from 3-loop to 2-loop mode. [2]

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18. A student places two speakers $1.0\ \text{m}$ apart and plays a $440\ \text{Hz}$ tone. She walks parallel to the line joining speakers at $2.0\ \text{m}$ distance and hears maxima and minima. (a) Explain the formation of maxima using superposition. [3] (b) Calculate the fringe separation if the observation line is $2.0\ \text{m}$ from the midpoint. [3]

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19. Describe an experiment to determine the wavelength of laser light using a diffraction grating. Include procedure, measurements, and how the wavelength is found. [5]

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20. The graph shows intensity against diffraction angle for single-slit diffraction.

Image pending generation: graph for Q20.

(a) State the condition for the first minimum. [1] (b) If slit width is $0.050\ \text{mm}$ and first minimum at $12^\circ$, calculate $\lambda$. [3] (c) Explain how the pattern would change if the slit width is halved. [2]

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A-Level Physics H2 Quiz - Waves Sound Light (Answer Key)

Total Marks: 40
Topic: Waves, Sound & Light (syllabus-first, AI-generated from Stage 4 templates; not claimed as past-year derived)

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Section A

1. [1 mark]
Wavelength is the distance between two consecutive points in phase (e.g. two adjacent crests or compressions).
Teaching note: "In phase" means oscillating together; for a sine wave, crest-to-crest distance = one λ.

2. [2 marks]
$v = f\lambda \Rightarrow \lambda = \frac{v}{f} = \frac{340}{512} = 0.664\ \text{m}$ (allow 0.66 m).
Marks: formula 1, substitution + answer 1.

3. [2 marks]
Polarisation is the restriction of vibration to one plane. Only transverse waves can be polarised because their oscillations are perpendicular to direction of travel; longitudinal waves vibrate along the direction of travel and cannot be plane-polarised.
Marks: transverse nature 1, explanation of why longitudinal cannot 1.

4. [1 mark]
When two or more waves meet, the resultant displacement is the vector sum of the individual displacements.

5. [1 mark]
$I \propto \frac{1}{r^2}$ (intensity inverse-square law for point source).

6. [1 mark]
A node is a point on a standing wave where displacement is always zero (no oscillation).

7. [2 marks]
Path difference $= n\lambda$ (where $n = 0,1,2,\dots$).
Marks: path difference condition 1, integer multiple of λ 1.

8. [2 marks]
500 lines per mm = $500 \times 10^3$ lines per m = $5.0\times10^5\ \text{m}^{-1}$.
$d = \frac{1}{5.0\times10^5} = 2.0\times10^{-6}\ \text{m}$.
Marks: conversion 1, answer 1.

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Section B

9. [3 marks]
(a) $\lambda = \frac{v}{f} = \frac{10}{25} = 0.40\ \text{m}$ [2]
(b) $T = \frac{1}{f} = \frac{1}{25} = 0.040\ \text{s}$ [1]

10. [4 marks]
(a) $\lambda = \frac{340}{680} = 0.50\ \text{m}$ [1]
(b) Path to far speaker $= \sqrt{3.0^2 + 0.80^2} = \sqrt{9.64} = 3.104\ \text{m}$.
Path difference $= 3.104 - 3.0 = 0.104\ \text{m}$ [2]
(c) $0.104 / 0.50 = 0.208 \approx$ not half-integer → constructive (nearer to nλ than (n+½)λ) [1]

11. [4 marks]
(a) $\Delta y = \frac{\lambda D}{a} = \frac{600\times10^{-9} \times 2.0}{0.30\times10^{-3}} = 4.0\times10^{-3}\ \text{m} = 4.0\ \text{mm}$ [3]
(b) $\Delta y$ doubles [1]

12. [4 marks]
(a) $d = \frac{1}{1.0\times10^5} = 1.0\times10^{-5}\ \text{m}$ [1]
(b) $d\sin\theta = n\lambda \Rightarrow \sin\theta = \frac{500\times10^{-9}}{1.0\times10^{-5}} = 0.0500$
$\theta = \sin^{-1}(0.0500) = 2.87^\circ$ [3]

13. [3 marks]
(a) After polariser: $I_0/2$ [1]
(b) $I = (I_0/2)\cos^2 60^\circ = (I_0/2)(0.25) = 0.125 I_0$ [2]

14. [3 marks]
(a) Fundamental: $L = \lambda/2 \Rightarrow \lambda = 2L = 2.4\ \text{m}$ [1]
(b) $f = \frac{v}{\lambda} = \frac{48}{2.4} = 20\ \text{Hz}$ [2]

15. [3 marks]
$I = \frac{P}{4\pi r^2} = \frac{0.20}{4\pi (4.0)^2} = \frac{0.20}{201.1} = 9.95\times10^{-4}\ \text{W m}^{-2}$ [3]

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Section C

16. [4 marks]
(a) Amplitude = $0.05\ \text{m}$ [1]
(b) Wavelength = $0.40\ \text{m}$ [1]
(c) $v = f\lambda = 5.0 \times 0.40 = 2.0\ \text{m s}^{-1}$ [2]

17. [5 marks]
(a) 3-loop mode: nodes at both ends + 2 internal = 4 nodes [1]
(b) 3 loops → $L = 3(\lambda/2) \Rightarrow \lambda = 2L/3 = 1.2/3 = 0.40\ \text{m}$ [2]
(c) Frequency $f = v/\lambda$; λ increases from 0.40 m (3-loop) to 0.60 m (2-loop) so frequency decreases [2]

18. [6 marks]
(a) Maxima occur where waves from two speakers arrive in phase (path difference = nλ); superposition gives larger amplitude [3]
(b) $\lambda = 340/440 = 0.773\ \text{m}$; fringe sep $y \approx \frac{\lambda D}{s} = \frac{0.773 \times 2.0}{1.0} = 1.55\ \text{m}$ [3]

19. [5 marks]
Procedure: shine laser at grating, measure distance D to screen, mark maxima.
Measure x for order n. Use $d\sin\theta = n\lambda$, $\tan\theta = x/D$.
Marks: apparatus 1, measurement 1, formula 1, calculation 1, safety/accuracy 1.

20. [6 marks]
(a) $a\sin\theta = \lambda$ [1]
(b) $\lambda = a\sin\theta = (0.050\times10^{-3})\sin 12^\circ = 1.04\times10^{-5}\ \text{m} = 10.4\ \mu\text{m}$ (accept 10 μm) [3]
(c) Halving a → θ doubles (first minimum farther out); central maximum wider [2]