A Level H1 Physics Waves Sound Light Quiz

A-Level Physics H1 Quiz - Waves Sound Light (Answer Key)

Total Marks: 40
Topic: Waves, Sound & Light

Section A: Short Answer

1. [1 mark]
Answer: A longitudinal wave is one in which the particle displacement is parallel to the direction of wave propagation.
Teaching note: Contrast with transverse waves where displacement is perpendicular. Common mistake: saying particles move with the wave (they oscillate about fixed positions).

2. [1 mark]
Answer: $v = f\lambda$
Teaching note: Speed equals frequency times wavelength. This is the fundamental wave relation.

3. [1 mark]
Answer: Approximately $340\ \text{m s}^{-1}$ (or $330–343\ \text{m s}^{-1}$).
Teaching note: At $20^\circ\text{C}$, $343\ \text{m s}^{-1}$ is typical; $340$ is acceptable.

4. [1 mark]
Answer: Refractive index $n = \frac{c}{v}$ where $c$ is speed in vacuum and $v$ is speed in medium; or $n = \frac{\sin i}{\sin r}$.
Teaching note: It measures how much light slows in a medium.

5. [1 mark]
Answer: Light must travel from a denser to a rarer medium (higher $n$ to lower $n$), and angle of incidence > critical angle.
Teaching note: State both conditions for full credit in exams; one condition sufficient for this short answer.

Section B: Structured Response

6. [3 marks]
(a) [2] $\lambda = \frac{v}{f} = \frac{340}{512} = 0.664\ \text{m}$ (allow 0.66 m).
(b) [1] Yes, audible range is $20\ \text{Hz} – 20\ \text{kHz}$; $512\ \text{Hz}$ is within.
Teaching note: Use wave equation; check units.

7. [2 marks]
Answer: Air particles vibrate parallel to direction of travel, compressing and rarefying neighbouring layers, transferring energy without net mass movement.
[1] for particle oscillation parallel; [1] for compression/rarefaction mechanism.

8. [3 marks]
(a) [1] Perpendicular to propagation.
(b) [2] $\lambda = v/f$; if $f$ doubles and $v$ constant, $\lambda$ halves.
Teaching note: Common error: saying wavelength doubles.

9. [3 marks]
(a) [2] $n_1\sin i = n_2\sin r \Rightarrow 1.00\sin30^\circ = 1.50\sin r \Rightarrow \sin r = 0.333 \Rightarrow r = 19.5^\circ$.
(b) [1] Towards the normal (entering denser medium).
Teaching note: Bends toward normal when going air to glass.

10. [3 marks]
(a) [1] Node is a point of zero displacement.
(b) [2] Produced by superposition of two waves of same frequency travelling in opposite directions, causing destructive interference at that point.
Teaching note: Stationary waves need two coherent oppositely travelling waves.

11. [3 marks]
(a) [1] Interference (or two-source interference).
(b) [2] At soft positions, waves arrive out of phase (path difference = $(m+½)\lambda$) causing destructive interference.
Teaching note: Loud = constructive; soft = destructive.

12. [3 marks]
(a) [2] $v = f\lambda = 2.0 \times 0.40 = 0.80\ \text{m s}^{-1}$.
(b) [1] $\lambda = v/f = 0.80/4.0 = 0.20\ \text{m}$.
Teaching note: Speed constant, so wavelength inversely proportional to frequency.

13. [2 marks]
Answer: Sound diffracts noticeably around doors/walls because its wavelength is comparable to obstacles; light wavelength is tiny so diffraction is only seen with small slits.
[1] for sound diffracts easily; [1] for light needs small aperture.

Section C: Data & Diagram Interpretation

14. [2 marks]
(a) [1] Amplitude = $5.0\ \text{cm}$.
(b) [1] Wavelength = $2.0\ \text{m}$.
From placeholder: peak at +5 cm, trough –5 cm, peak-to-peak 2.0 m.

15. [2 marks]
(a) [1] $T = 4.0/10 = 0.40\ \text{s}$.
(b) [1] $f = 1/T = 2.5\ \text{Hz}$.

16. [3 marks]
(a) [1] Enters perpendicular to surface so incidence = 0°, no bending.
(b) [2] At curved surface: $n\sin i = 1\sin r \Rightarrow 1.50\sin45^\circ = \sin r \Rightarrow \sin r = 1.06 > 1$ → total internal reflection occurs, no refraction.
Teaching note: If student computes arcsin(1.06) invalid, state TIR. (Marks: [2] for correct identification of TIR.)

17. [3 marks]
(a) [2] $\Delta x = \frac{\lambda D}{d} = \frac{600\times10^{-9} \times 2.0}{0.50\times10^{-3}} = 2.4\times10^{-3}\ \text{m} = 2.4\ \text{mm}$.
(b) [1] Fringe separation increases (inversely proportional to $d$).

18. [2 marks]
(a) [1] Fringe spacing = $2\ \text{cm}$ (peak-to-peak).
(b) [1] Yes, coherent (stable pattern).

19. [3 marks]
(a) [2] Total distance = $v t = 340 \times 0.020 = 6.8\ \text{m}$; wall distance = $3.4\ \text{m}$.
(b) [1] Ultrasound has shorter wavelength, better resolution for small objects / less absorbed by air? Actually bats use for localization.

20. [3 marks]
(a) [1] $L = \lambda/4$ (first harmonic closed pipe).
(b) [2] $\lambda = 4L = 1.36\ \text{m}$; $f = v/\lambda = 340/1.36 = 250\ \text{Hz}$.
Teaching note: Closed pipe only odd harmonics.