A Level H2 Physics Waves Sound Light Quiz

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

Total Marks: 40
Topic: Waves, Sound & Light

Section A

1. [1 mark]
Wavelength is the distance between two consecutive points in phase (e.g. two adjacent crests or troughs) of a progressive wave.
Teaching note: Progressive waves transfer energy; wavelength $\lambda$ is measured in metres.

2. [2 marks]
Use $v = f\lambda \Rightarrow \lambda = v/f$.
$\lambda = 340 / 440 = 0.773\ \text{m}$ (3 s.f.).
Marks: 1 for formula, 1 for answer with unit.

3. [2 marks]
Transverse waves have oscillations perpendicular to direction of travel, so a plane of vibration exists and can be filtered by a polariser. Longitudinal waves oscillate parallel to travel, so no unique plane to select.
Marks: 1 for transverse explanation, 1 for longitudinal.

4. [1 mark]
With 3 antinodes on a string fixed at both ends, there are 4 nodes (including the two ends).

5. [2 marks]
Path difference from the two sources = $n\lambda$ (where $n = 0,1,2,\dots$) and the waves arrive in phase.
Marks: 1 path diff, 1 in phase.

6. [3 marks]
$d = 1/(5.0\times10^5) = 2.0\times10^{-6}\ \text{m}$.
$n\lambda = d\sin\theta \Rightarrow \sin\theta = \lambda/d = 600\times10^{-9} / 2.0\times10^{-6} = 0.300$.
$\theta = \sin^{-1}(0.300) = 17.5^\circ$.
Marks: 1 grating spacing, 1 substitution, 1 angle.

7. [2 marks]
$I = P/(4\pi r^2) = 2.0 / (4\pi \times 4.0^2) = 2.0 / 201.1 = 9.95\times10^{-3}\ \text{W m}^{-2}$.
Marks: 1 formula, 1 answer.

8. [1 mark]
$I = I_0 \cos^2\theta$ where $I_0$ is intensity after polariser, $\theta$ angle between polariser and analyser.

9. [2 marks]
Compare $y = A\sin(\omega t)$: $A = 0.02\ \text{m}$, $\omega = 4\pi\ \text{rad s}^{-1}$.
Marks: 1 each.

10. [2 marks]
Diffraction: spreading of waves past an obstacle/aperture. Interference: superposition of two/more coherent waves to form maxima/minima.
Marks: 1 each.

Section B

11. [3 marks]
$\lambda = d\Delta x / D = (0.50\times10^{-3} \times 1.44\times10^{-3}) / 1.20 = 6.0\times10^{-7}\ \text{m} = 600\ \text{nm}$.
Marks: 1 convert units, 1 formula, 1 answer.

12. [2 marks]
(a) Wavelength = 20 cm (two loops over 40 cm → $\lambda/2 = 20$ cm, so $\lambda = 40$ cm? Wait: 2 complete loops = 2 half-wavelengths? Actually 2 loops = 1 full wavelength if loop = half-wave; 2 loops = $\lambda$ = 40 cm). From labels: nodes at 0,20,40 → distance between nodes = 20 cm = $\lambda/2$, so $\lambda = 40$ cm. [1]
(b) Adjacent nodes = 20 cm. [1]

13. [3 marks]
$d = 1/(2.0\times10^5) = 5.0\times10^{-6}\ \text{m}$.
Using $n=1$: $\lambda = d\sin18^\circ = 5.0\times10^{-6}\times0.309 = 1.545\times10^{-6}\ \text{m}$.
Check $n=2$: $2\lambda = d\sin37.5^\circ = 5.0\times10^{-6}\times0.609 = 3.045\times10^{-6}$ → $\lambda = 1.52\times10^{-6}\ \text{m}$. Avg $\approx 1.53\times10^{-6}\ \text{m}$.
Marks: 1 spacing, 1 calc, 1 consistency.

14. [3 marks]
(a) $T = 4\ \text{ms} \Rightarrow f = 1/T = 250\ \text{Hz}$. [1]
(b) $v = f\lambda = 250 \times 0.80 = 200\ \text{m s}^{-1}$. [2]

15. [2 marks]
Intensity decreases with distance according to inverse square law: $I \propto 1/r^2$.
Marks: 1 law, 1 statement.

Section C

16. [5 marks]
(a) Fundamental: $\lambda = 2L = 1.60\ \text{m}$, $f = v/\lambda = 160/1.60 = 100\ \text{Hz}$. [2]
(b) Second harmonic = $2\times100 = 200\ \text{Hz}$. [1]
(c) Incident and reflected waves of same frequency/amplitude superpose, forming nodes (zero displacement) and antinodes. [2]

17. [5 marks]
Apparatus: laser, double slit ($d$), screen, metre rule.
Procedure: shine laser through slits onto screen, measure fringe separation $\Delta x$ over several fringes, measure $D$.
Use $\lambda = d\Delta x/D$. Repeat for accuracy.
Mark descriptors: apparatus (1), procedure (2), formula use (1), reducing error (1).

18. [5 marks]
(a) After polariser: $I_0/2$. [1]
(b) $I = (I_0/2)\cos^2 30^\circ = 0.5 I_0 \times 0.75 = 0.375 I_0$. [2]
(c) Glare is horizontally polarised; polarising sunglasses block that plane, reducing reflected light. [2]

19. [4 marks]
Each slit acts as source; waves superpose. Principal maxima when $d\sin\theta = n\lambda$. Larger $n$ → larger $\sin\theta$ → larger $\theta$.
Marks: superposition (1), condition (1), angle relation (2).

20. [5 marks]
(a) Closed at one end: $\lambda = 4L = 4\times(340/512) = 2.66\ \text{m}$. [2]
(b) $L = \lambda/4 = 0.664\ \text{m}$. [1]
(c) Node at closed end, antinode at open; only quarter-wave odd multiples fit. [2]