A Level H1 Physics Waves Sound Light Quiz

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A-Level Physics H1 Quiz - Waves Sound Light

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 45 minutes
Total Marks: 45

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Show all working clearly. Marks may be awarded for correct working even if the final answer is incorrect.
4. Use $g = 9.81 \text{ m s}^{-2}$, $c = 3.00 \times 10^8 \text{ m s}^{-1}$, $h = 6.63 \times 10^{-34} \text{ J s}$, and $e = 1.60 \times 10^{-19} \text{ C}$ where appropriate.

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Section A: Wave Properties and Superposition (Questions 1–5)

1. Define the term coherence as applied to two wave sources.
[1]

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2. A progressive wave has a frequency of 250 Hz and a wavelength of 1.2 m. Calculate the speed of the wave.
[2]

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3. Explain why sound waves cannot be polarized, whereas light waves can.
[2]

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4. Two coherent sources of sound waves interfere. At a certain point, the path difference between the waves from the two sources is 1.5 wavelengths. State whether constructive or destructive interference occurs at this point and explain your reasoning.
[2]

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5. A stationary wave is formed on a string fixed at both ends. The string vibrates in its fundamental mode. (a) State the relationship between the length of the string $L$ and the wavelength $\lambda$ of the fundamental mode.
[1]

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(b) If the frequency of the fundamental mode is 120 Hz, calculate the frequency of the third harmonic.
[2]

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Section B: Light and Interference (Questions 6–10)

6. In a Young’s double-slit experiment, monochromatic light of wavelength $\lambda$ produces fringes of separation $x$ on a screen at distance $D$ from the slits. The slit separation is $a$. State the formula relating these quantities.
[1]

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7. Using the setup in Question 6, the slit separation $a$ is doubled while $D$ and $\lambda$ remain constant. Describe and explain the change in the fringe separation $x$.
[2]

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8. Light of wavelength 550 nm is incident normally on a diffraction grating with 500 lines per mm. Calculate the angle $\theta$ for the second-order maximum.
[3]

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9. Explain why a diffraction grating produces sharper and brighter maxima compared to a double-slit arrangement.
[2]

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10. White light is passed through a double-slit apparatus. Describe the appearance of the central fringe and the first-order fringes.
[2]

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Section C: Photoelectric Effect (Questions 11–15)

11. Define the work function of a metal.
[1]

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12. State Einstein’s photoelectric equation, defining all symbols used.
[2]

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13. A metal surface has a work function of 2.3 eV. Light of frequency $8.0 \times 10^{14}$ Hz is incident on the surface. (a) Calculate the energy of a single photon of this light in Joules.
[2]

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(b) Determine the maximum kinetic energy of the emitted photoelectrons in Joules.
[2]

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14. In a photoelectric experiment, the intensity of the incident monochromatic light is increased while keeping the frequency constant. (a) State the effect on the maximum kinetic energy of the photoelectrons.
[1]

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(b) State the effect on the photoelectric current.
[1]

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15. Explain why the existence of a threshold frequency supports the particle nature of light rather than the wave nature.
[2]

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Section D: Sound and Doppler Effect (Questions 16–20)

16. A sound wave travels from air into water. State what happens to: (a) the frequency of the wave.
[1]

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(b) the wavelength of the wave.
[1]

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17. An ambulance emits a siren of frequency 1200 Hz. It moves towards a stationary observer at a speed of 25 m s⁻¹. The speed of sound in air is 340 m s⁻¹. Calculate the frequency heard by the observer.
[3]

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18. Explain why the pitch of the siren appears to drop as the ambulance passes the observer and moves away.
[2]

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19. Ultrasound waves are used in medical imaging. State one property of ultrasound that makes it suitable for this application.
[1]

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20. A stationary source emits sound of frequency $f_s$. An observer moves away from the source with speed $v_o$. The speed of sound is $v$. Derive or state the formula for the observed frequency $f_o$.
[2]

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

1. Coherence means the two sources have a constant phase difference (and the same frequency).
[1]

2. Speed $v = f \lambda$
$v = 250 \times 1.2$
$v = 300 \text{ m s}^{-1}$
[1 for formula, 1 for answer]

3. Sound waves are longitudinal (vibrations parallel to direction of propagation).
[1]
Polarization requires transverse vibrations (perpendicular to direction of propagation), which light possesses.
[1]

4. Destructive interference.
[1]
Path difference is $(n + \frac{1}{2})\lambda$ (where $n=1$), which corresponds to waves arriving in antiphase ($180^\circ$ or $\pi$ rad phase difference).
[1]

5. (a) $L = \frac{\lambda}{2}$ (or $\lambda = 2L$)
[1]
(b) Frequency of $n$-th harmonic $f_n = n f_1$.
$f_3 = 3 \times 120 = 360 \text{ Hz}$
[1 for relationship, 1 for answer]

6. $x = \frac{\lambda D}{a}$
[1]

7. Fringe separation $x$ is halved (decreases by half).
[1]
Because $x$ is inversely proportional to slit separation $a$ ($x \propto \frac{1}{a}$).
[1]

8. Grating spacing $d = \frac{1}{500 \times 10^3} \text{ m} = 2.0 \times 10^{-6} \text{ m}$.
[1]
Formula: $d \sin \theta = n \lambda$
$\sin \theta = \frac{n \lambda}{d} = \frac{2 \times 550 \times 10^{-9}}{2.0 \times 10^{-6}}$
$\sin \theta = 0.55$
$\theta = \sin^{-1}(0.55) = 33.4^\circ$
[1 for substitution, 1 for answer]

9. More slits result in more constructive interference at the maxima positions, making them brighter/sharper.
[1]
Destructive interference is more complete between maxima, making the background darker/narrower peaks.
[1]

10. Central fringe is white.
[1]
First-order fringes are spectra (colored), with violet/blue closest to the center and red furthest (due to $\lambda_{violet} < \lambda_{red}$).
[1]

11. The minimum energy required to remove an electron from the surface of a metal.
[1]

12. $hf = \Phi + K_{max}$ (or $hf = \Phi + \frac{1}{2}mv_{max}^2$)
[1]
Where $h$ is Planck's constant, $f$ is frequency, $\Phi$ is work function, $K_{max}$ is max kinetic energy.
[1]

13. (a) $E = hf = 6.63 \times 10^{-34} \times 8.0 \times 10^{14}$
$E = 5.30 \times 10^{-19} \text{ J}$
[1 for formula/sub, 1 for answer]

(b) Convert $\Phi$ to Joules: $2.3 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 3.68 \times 10^{-19} \text{ J}$.
[1]
$K_{max} = E - \Phi = 5.30 \times 10^{-19} - 3.68 \times 10^{-19}$
$K_{max} = 1.62 \times 10^{-19} \text{ J}$
[1 for subtraction, 1 for answer]

14. (a) No change (remains constant).
[1]
(b) Increases (proportional to intensity).
[1]

15. Wave theory predicts energy depends on intensity (amplitude), so any frequency should eventually emit electrons if intense enough.
[1]
Particle theory predicts energy depends on frequency ($E=hf$); if $hf < \Phi$, no emission occurs regardless of intensity, explaining the threshold.
[1]

16. (a) Frequency remains constant.
[1]
(b) Wavelength increases (since speed increases in water and $v=f\lambda$).
[1]

17. Formula for source moving towards observer: $f_o = f_s \left( \frac{v}{v - v_s} \right)$
[1]
$f_o = 1200 \left( \frac{340}{340 - 25} \right)$
$f_o = 1200 \left( \frac{340}{315} \right)$
$f_o = 1295 \text{ Hz}$ (or 1300 Hz to 2 s.f.)
[1 for substitution, 1 for answer]

18. As the ambulance moves away, the wavelength is stretched (increases).
[1]
Since $v = f\lambda$ and $v$ is constant, an increase in $\lambda$ leads to a decrease in frequency (pitch).
[1]

19. High frequency / Short wavelength allows for high resolution imaging.
[1]
(Alternatively: It is non-ionizing / safe for tissue).

20. $f_o = f_s \left( \frac{v - v_o}{v} \right)$
[2]
(1 mark for correct numerator $v - v_o$ indicating moving away, 1 mark for correct structure).