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A Level H2 Physics Waves Sound Light Quiz
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Questions
A-Level Physics H2 Quiz - Waves Sound Light
Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 50
Duration: 45 minutes
Total Marks: 50
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- Take g = 9.81 m s⁻² unless otherwise stated.
- Speed of light c = 3.00 × 10⁸ m s⁻¹.
- Planck's constant h = 6.63 × 10⁻³⁴ J s.
Section A: Short Answer Questions (10 marks)
Answer all questions in this section.
1. State the principle of superposition for waves.
[2 marks]
2. Define the term "coherent sources" as applied to wave interference.
[2 marks]
3. A sound wave travels from air into water. State what happens to its frequency, wavelength, and speed.
[3 marks]
4. Explain why a diffraction grating produces sharper and brighter maxima compared to a double-slit arrangement when used with monochromatic light.
[3 marks]
5. Distinguish between transverse and longitudinal waves, giving one example of each.
[2 marks]
Section B: Structured Questions (20 marks)
Answer all questions in this section.
6. A progressive wave is described by the equation: [ y = 0.050 \sin(4.0\pi t - 0.80\pi x) ] where y and x are in metres and t is in seconds.
(a) Determine the amplitude of the wave.
[1 mark]
(b) Calculate the wavelength of the wave.
[2 marks]
(c) Calculate the speed of the wave.
[2 marks]
(d) State the direction of propagation of the wave.
[1 mark]
7. A student sets up Young's double-slit experiment using monochromatic light of wavelength 589 nm. The slits are separated by 0.50 mm and the screen is placed 2.0 m from the slits.
(a) Calculate the fringe separation on the screen.
[2 marks]
(b) The student replaces the light source with one of shorter wavelength. State and explain the effect on the fringe separation.
[2 marks]
(c) The student then immerses the entire apparatus in water (refractive index = 1.33). Calculate the new fringe separation.
[3 marks]
8. A stationary wave is set up on a stretched string of length 1.2 m fixed at both ends. The string vibrates in its third harmonic at a frequency of 150 Hz.
(a) Sketch the standing wave pattern, labelling all nodes and antinodes.
[2 marks]
(b) Calculate the wavelength of the standing wave.
[2 marks]
(c) Calculate the speed of the progressive waves on the string.
[2 marks]
(d) Calculate the fundamental frequency of the string.
[1 mark]
9. A loudspeaker emits sound waves of frequency 680 Hz. A microphone connected to an oscilloscope is moved along a line away from the speaker. The oscilloscope displays two points where the signal amplitude is a minimum, separated by 0.50 m. The speed of sound in air is 340 m s⁻¹.
(a) Explain why minima are observed as the microphone is moved.
[2 marks]
(b) Calculate the wavelength of the sound waves.
[2 marks]
(c) Explain how the separation between minima would change if the frequency of the sound were increased.
[2 marks]
10. A beam of monochromatic light of wavelength 450 nm is incident normally on a diffraction grating. The third-order maximum is observed at an angle of 42.0° to the normal.
(a) Calculate the number of lines per millimetre on the grating.
[3 marks]
(b) Determine the highest order of maximum that can be observed with this grating and light source.
[2 marks]
(c) The light source is replaced with one that emits two wavelengths: 450 nm and 600 nm. Calculate the angular separation between the second-order maxima for these two wavelengths.
[3 marks]
Section C: Data Analysis and Application Questions (20 marks)
Answer all questions in this section.
11. A student investigates the photoelectric effect using a photocell. Monochromatic light of wavelength 350 nm is incident on a metal cathode with work function 2.3 eV.
(a) Calculate the maximum kinetic energy of the emitted photoelectrons in joules.
[3 marks]
(b) Calculate the stopping potential required to reduce the photocurrent to zero.
[2 marks]
(c) The intensity of the incident light is doubled while keeping the wavelength constant. State and explain the effect on: (i) the maximum kinetic energy of the photoelectrons, (ii) the saturation photocurrent.
[3 marks]
12. A sound wave of frequency 500 Hz travels through air at 340 m s⁻¹ and then enters a solid medium where its speed becomes 1500 m s⁻¹.
(a) Calculate the wavelength of the sound wave in air.
[1 mark]
(b) Calculate the wavelength of the sound wave in the solid medium.
[1 mark]
(c) Explain why the frequency remains unchanged when the wave enters the solid medium.
[2 marks]
13. A microwave transmitter emits waves of wavelength 3.0 cm. The waves are directed towards two parallel slits separated by 6.0 cm. A detector is moved along a line 50 cm behind the slits.
(a) Calculate the fringe separation observed by the detector.
[2 marks]
(b) State one similarity and one difference between this microwave interference pattern and that obtained with visible light in Young's double-slit experiment.
[2 marks]
14. A diffraction grating has 600 lines per mm. Light of wavelength 500 nm is incident normally on the grating.
(a) Calculate the grating spacing d.
[1 mark]
(b) Determine the angle at which the first-order maximum is observed.
[2 marks]
(c) Calculate the maximum number of orders visible on one side of the central maximum.
[2 marks]
15. A progressive wave on a string has a frequency of 50 Hz and a wavelength of 0.80 m. The amplitude of the wave is 2.0 cm.
(a) Calculate the speed of the wave.
[1 mark]
(b) Write an equation for the wave, assuming it travels in the positive x-direction and that at t = 0, x = 0, the displacement y = 0.
[2 marks]
(c) Calculate the maximum speed of a particle on the string.
[2 marks]
16. In a Young's double-slit experiment, the distance between the slits is 0.20 mm and the screen is 1.5 m away. The third bright fringe is 1.2 cm from the central fringe.
(a) Calculate the wavelength of the light used.
[2 marks]
(b) State and explain what would happen to the fringe pattern if one of the slits is covered.
[2 marks]
17. A standing wave is set up in a tube closed at one end. The tube length is 0.85 m and the speed of sound in air is 340 m s⁻¹.
(a) Determine the fundamental frequency of the tube.
[2 marks]
(b) Calculate the frequency of the third harmonic.
[2 marks]
(c) Explain why only odd harmonics are present in a tube closed at one end.
[2 marks]
18. White light is incident normally on a diffraction grating. Explain why the central maximum is white, while the higher-order maxima show a spectrum of colours.
[3 marks]
19. A student uses a ripple tank to study water waves. The waves have a frequency of 20 Hz and a wavelength of 1.5 cm.
(a) Calculate the speed of the water waves.
[1 mark]
(b) The waves pass through a gap of width 3.0 cm. State and explain the effect on the wave pattern if the gap width is reduced to 1.0 cm.
[2 marks]
20. A laser emits light of wavelength 632 nm. The light is incident on a double-slit with separation 0.25 mm. The screen is 3.0 m from the slits.
(a) Calculate the distance between the central bright fringe and the fifth dark fringe.
[3 marks]
(b) The double-slit is replaced with a diffraction grating of 300 lines per mm. Calculate the angle of the second-order maximum for the same laser light.
[2 marks]
END OF QUIZ
Answers
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