A Level H2 Physics Waves Sound Light Quiz

A-Level Physics H2 Quiz - Waves Sound Light

Answer Key and Marking Scheme

Total Marks: 50

Section A: Short Answer Questions (10 marks)

1. State the principle of superposition for waves. [2 marks]

Answer: When two or more waves meet at a point, the resultant displacement at that point is equal to the vector sum of the individual displacements of the waves at that point. [1 mark for "resultant displacement equals sum of individual displacements"; 1 mark for "vector sum" or equivalent precise wording]

2. Define the term "coherent sources" as applied to wave interference. [2 marks]

Answer: Coherent sources are sources that emit waves with a constant phase difference [1 mark] and the same frequency/wavelength [1 mark].

3. A sound wave travels from air into water. State what happens to its frequency, wavelength, and speed. [3 marks]

Answer:

• Frequency: remains unchanged [1 mark]
• Wavelength: increases [1 mark]
• Speed: increases [1 mark]

Explanation: Frequency is determined by the source and does not change when the wave enters a different medium. Since v = fλ and v increases in water (sound travels faster in water), λ must increase proportionally.

4. Explain why a diffraction grating produces sharper and brighter maxima compared to a double-slit arrangement when used with monochromatic light. [3 marks]

Answer:

• A diffraction grating has many slits (lines) contributing to the interference pattern, whereas a double-slit has only two [1 mark].
• With many slits, the condition for constructive interference is very precise; slight deviations from the exact angle result in rapid destructive interference, producing very narrow (sharp) maxima [1 mark].
• The intensity at the maxima is proportional to N² (where N is the number of slits illuminated), so the maxima are much brighter than those from a double-slit [1 mark].

5. Distinguish between transverse and longitudinal waves, giving one example of each. [2 marks]

Answer:

• Transverse waves: oscillations are perpendicular to the direction of energy propagation [0.5 marks]; example: light/water waves/string waves [0.5 marks].
• Longitudinal waves: oscillations are parallel to the direction of energy propagation [0.5 marks]; example: sound waves [0.5 marks].

Section B: Structured Questions (20 marks)

6. Progressive wave equation: y = 0.050 sin(4.0πt - 0.80πx) [6 marks]

(a) Amplitude [1 mark]
Answer: A = 0.050 m ✓

(b) Wavelength [2 marks]
Answer: k = 2π/λ = 0.80π [1 mark]
λ = 2π/(0.80π) = 2.5 m [1 mark]

(c) Speed [2 marks]
Answer: ω = 4.0π rad s⁻¹ [0.5 marks]
v = ω/k = 4.0π/(0.80π) = 5.0 m s⁻¹ [1 mark]
Alternative: v = fλ; f = ω/(2π) = 2.0 Hz; v = 2.0 × 2.5 = 5.0 m s⁻¹ [0.5 marks for method]

(d) Direction of propagation [1 mark]
Answer: The wave travels in the positive x-direction (since the sign between the t and x terms is negative). ✓

7. Young's double-slit experiment [7 marks]

(a) Fringe separation [2 marks]
Answer: Δy = λD/d [1 mark]
Δy = (589 × 10⁻⁹ × 2.0)/(0.50 × 10⁻³) = 2.36 × 10⁻³ m = 2.4 mm [1 mark]

(b) Effect of shorter wavelength [2 marks]
Answer: The fringe separation would decrease [1 mark].
Since Δy = λD/d, Δy is directly proportional to λ. A shorter wavelength results in smaller fringe separation [1 mark].

(c) New fringe separation in water [3 marks]
Answer: In water, the wavelength becomes λ' = λ/n = 589/1.33 = 443 nm [1 mark]
New fringe separation: Δy' = λ'D/d [1 mark]
Δy' = (443 × 10⁻⁹ × 2.0)/(0.50 × 10⁻³) = 1.77 × 10⁻³ m = 1.8 mm [1 mark]

8. Standing wave on a string [7 marks]

(a) Sketch [2 marks]
Answer: Third harmonic has 4 nodes (including ends) and 3 antinodes.
Sketch should show: string fixed at both ends, 3 complete loops, nodes at ends and at 1/3 and 2/3 of length, antinodes at maxima of each loop. [1 mark for correct number of nodes/antinodes; 1 mark for clear labelling]

(b) Wavelength [2 marks]
Answer: For third harmonic on a string fixed at both ends: L = 3λ/2 [1 mark]
λ = 2L/3 = 2 × 1.2/3 = 0.80 m [1 mark]

(c) Speed of progressive waves [2 marks]
Answer: v = fλ [1 mark]
v = 150 × 0.80 = 120 m s⁻¹ [1 mark]

(d) Fundamental frequency [1 mark]
Answer: f₁ = f₃/3 = 150/3 = 50 Hz ✓

9. Sound wave interference [6 marks]

(a) Explanation of minima [2 marks]
Answer: The microphone receives sound directly from the speaker and also sound reflected from surrounding surfaces (or the sound waves interfere due to path difference from different parts of the speaker cone) [1 mark]. When the path difference between the two waves reaching the microphone is an odd multiple of half-wavelengths, destructive interference occurs, producing a minimum [1 mark].

(b) Wavelength [2 marks]
Answer: Distance between consecutive minima = λ/2 [1 mark]
λ = 2 × 0.50 = 1.0 m [1 mark]
Check: v = fλ → λ = 340/680 = 0.50 m. The distance between minima should be λ/2 = 0.25 m. However, the question states 0.50 m separation, which suggests the minima are from standing wave pattern where distance between nodes = λ/2. Accept λ = 1.0 m based on given data.

(c) Effect of increased frequency [2 marks]
Answer: The separation between minima would decrease [1 mark].
Since v = fλ, increasing frequency decreases wavelength. The separation between minima is λ/2, so this also decreases [1 mark].

10. Diffraction grating [8 marks]

(a) Number of lines per mm [3 marks]
Answer: d sin θ = nλ [1 mark]
d = nλ/sin θ = 3 × 450 × 10⁻⁹/sin 42.0° [1 mark]
d = 3 × 450 × 10⁻⁹/0.6691 = 2.018 × 10⁻⁶ m
Lines per m = 1/d = 4.955 × 10⁵ m⁻¹
Lines per mm = 4.955 × 10⁵/10³ = 496 lines per mm [1 mark]

(b) Highest order observable [2 marks]
Answer: Maximum order when sin θ ≤ 1: n_max ≤ d/λ [1 mark]
n_max ≤ 2.018 × 10⁻⁶/(450 × 10⁻⁹) = 4.48
Highest order = 4 [1 mark]

(c) Angular separation for two wavelengths [3 marks]
Answer: For λ₁ = 450 nm: sin θ₁ = 2 × 450 × 10⁻⁹/(2.018 × 10⁻⁶) = 0.4460 → θ₁ = 26.5° [1 mark]
For λ₂ = 600 nm: sin θ₂ = 2 × 600 × 10⁻⁹/(2.018 × 10⁻⁶) = 0.5946 → θ₂ = 36.5° [1 mark]
Angular separation = 36.5° - 26.5° = 10.0° [1 mark]

Section C: Data Analysis and Application Questions (20 marks)

11. Photoelectric effect [8 marks]

(a) Maximum kinetic energy [3 marks]
Answer: E_photon = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(350 × 10⁻⁹) [1 mark]
E_photon = 5.683 × 10⁻¹⁹ J [0.5 marks]
Work function Φ = 2.3 eV = 2.3 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J [0.5 marks]
K_max = E_photon - Φ = 5.683 × 10⁻¹⁹ - 3.68 × 10⁻¹⁹ = 2.00 × 10⁻¹⁹ J [1 mark]

(b) Stopping potential [2 marks]
Answer: eV_s = K_max [1 mark]
V_s = K_max/e = 2.00 × 10⁻¹⁹/(1.60 × 10⁻¹⁹) = 1.25 V [1 mark]

(c) Effect of doubling intensity [3 marks]
(i) Maximum kinetic energy: remains unchanged [0.5 marks]. The photon energy depends only on frequency/wavelength, not intensity. Each photon still has the same energy, so maximum KE of emitted electrons is unchanged [1 mark].
(ii) Saturation photocurrent: doubles [0.5 marks]. Doubling intensity doubles the number of photons incident per second, which doubles the number of photoelectrons emitted per second, doubling the current [1 mark].

12. Sound wave in different media [4 marks]

(a) Wavelength in air [1 mark]
Answer: λ_air = v/f = 340/500 = 0.68 m ✓

(b) Wavelength in solid [1 mark]
Answer: λ_solid = v/f = 1500/500 = 3.0 m ✓

(c) Frequency unchanged [2 marks]
Answer: Frequency is determined by the source [1 mark]. When a wave crosses a boundary, the frequency must remain the same to ensure continuity of the wave at the boundary; otherwise, wavefronts would be created or destroyed [1 mark].

13. Microwave interference [4 marks]

(a) Fringe separation [2 marks]
Answer: Δy = λD/d [1 mark]
Δy = (3.0 × 10⁻² × 0.50)/(6.0 × 10⁻²) = 0.25 m = 25 cm [1 mark]

(b) Similarity and difference [2 marks]
Answer: Similarity: both produce alternating regions of high and low intensity (bright and dark fringes) due to constructive and destructive interference [1 mark].
Difference: the fringe separation for microwaves is much larger (centimetres) compared to visible light (millimetres or less) because of the much longer wavelength [1 mark].

14. Diffraction grating calculations [5 marks]

(a) Grating spacing [1 mark]
Answer: d = 1/(600 × 10³) = 1.67 × 10⁻⁶ m ✓

(b) First-order angle [2 marks]
Answer: d sin θ = nλ [0.5 marks]
sin θ = (1 × 500 × 10⁻⁹)/(1.67 × 10⁻⁶) = 0.300 [1 mark]
θ = sin⁻¹(0.300) = 17.5° [0.5 marks]

(c) Maximum number of orders [2 marks]
Answer: n_max ≤ d/λ = 1.67 × 10⁻⁶/(500 × 10⁻⁹) = 3.34 [1 mark]
Maximum orders on one side = 3 [1 mark]

15. Progressive wave on a string [5 marks]

(a) Wave speed [1 mark]
Answer: v = fλ = 50 × 0.80 = 40 m s⁻¹ ✓

(b) Wave equation [2 marks]
Answer: General form: y = A sin(ωt - kx) [0.5 marks]
ω = 2πf = 100π rad s⁻¹; k = 2π/λ = 2π/0.80 = 2.5π rad m⁻¹ [0.5 marks]
y = 0.020 sin(100πt - 2.5πx) [1 mark]

(c) Maximum particle speed [2 marks]
Answer: v_max = ωA [1 mark]
v_max = 100π × 0.020 = 2π = 6.28 m s⁻¹ [1 mark]

16. Young's double-slit analysis [4 marks]

(a) Wavelength [2 marks]
Answer: Δy = λD/d, where Δy for third bright fringe from centre = 3λD/d [0.5 marks]
1.2 × 10⁻² = 3 × λ × 1.5/(0.20 × 10⁻³) [0.5 marks]
λ = (1.2 × 10⁻² × 0.20 × 10⁻³)/(3 × 1.5) = 5.33 × 10⁻⁷ m = 533 nm [1 mark]

(b) Effect of covering one slit [2 marks]
Answer: The interference pattern would disappear, and a single-slit diffraction pattern would be observed [1 mark]. This is because interference requires two coherent sources; with one slit covered, only diffraction from a single slit occurs [1 mark].

17. Standing wave in closed tube [6 marks]

(a) Fundamental frequency [2 marks]
Answer: For a tube closed at one end: L = λ/4 → λ = 4L = 4 × 0.85 = 3.4 m [1 mark]
f₁ = v/λ = 340/3.4 = 100 Hz [1 mark]

(b) Third harmonic frequency [2 marks]
Answer: For a closed tube, only odd harmonics exist: f₃ = 3f₁ [1 mark]
f₃ = 3 × 100 = 300 Hz [1 mark]

(c) Explanation of odd harmonics only [2 marks]
Answer: A tube closed at one end must have a node at the closed end and an antinode at the open end [1 mark]. This condition can only be satisfied when an odd number of quarter-wavelengths fit into the tube length (L = nλ/4, where n = 1, 3, 5...), resulting in only odd harmonics [1 mark].

18. White light and diffraction grating [3 marks]

Answer: At the central maximum (n = 0), all wavelengths have a path difference of zero, so all colours interfere constructively at the same position, producing white light [1 mark]. For higher orders (n ≥ 1), the angle of diffraction depends on wavelength (d sin θ = nλ) [1 mark]. Different colours are diffracted at different angles, spreading out into a continuous spectrum with violet deviated least and red deviated most [1 mark].

19. Ripple tank waves [3 marks]

(a) Wave speed [1 mark]
Answer: v = fλ = 20 × 0.015 = 0.30 m s⁻¹ ✓

(b) Effect of reducing gap width [2 marks]
Answer: When the gap width (3.0 cm) is comparable to the wavelength (1.5 cm), significant diffraction occurs, and waves spread out after passing through the gap [1 mark]. When the gap is reduced to 1.0 cm (less than the wavelength), the diffraction effect becomes more pronounced, and the waves spread out almost semicircularly from the gap [1 mark].

20. Laser and double-slit/grating [5 marks]

(a) Distance to fifth dark fringe [3 marks]
Answer: For dark fringes in double-slit: d sin θ = (m + ½)λ, where m = 0, 1, 2... [0.5 marks]
Fifth dark fringe corresponds to m = 4: d sin θ = (4.5)λ [0.5 marks]
For small angles: sin θ ≈ tan θ = y/D [0.5 marks]
d(y/D) = 4.5λ → y = 4.5λD/d [0.5 marks]
y = 4.5 × 632 × 10⁻⁹ × 3.0/(0.25 × 10⁻³) = 3.41 × 10⁻² m = 3.4 cm [1 mark]

(b) Angle for grating second-order [2 marks]
Answer: d = 1/(300 × 10³) = 3.33 × 10⁻⁶ m [0.5 marks]
d sin θ = nλ → sin θ = 2 × 632 × 10⁻⁹/(3.33 × 10⁻⁶) = 0.379 [1 mark]
θ = sin⁻¹(0.379) = 22.3° [0.5 marks]

END OF ANSWER KEY