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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.
Write your answers in the spaces provided.
Show all working clearly. Marks may be awarded for correct working even if the final answer is incorrect.
Use $g = 9.81 \text{ m s}^{-2}$ where appropriate.
The speed of light in vacuum $c = 3.00 \times 10^8 \text{ m s}^{-1}$ .

Section A: Multiple Choice & Short Concepts (10 Marks)

1. Which of the following statements correctly describes the phase relationship between two points on a stationary wave separated by half a wavelength ( $\lambda/2$ )?
[1]
A. They are in phase.
B. They are in antiphase ( $180^\circ$ out of phase).
C. The phase difference depends on the amplitude.
D. One is at a node and the other at an antinode.

Answer: __________________________

2. A sound wave travels from air into water. Which property of the wave remains unchanged?
[1]
A. Speed
B. Wavelength
C. Frequency
D. Amplitude

Answer: __________________________

3. In a Young’s double-slit experiment, the fringe separation $x$ is given by $x = \frac{\lambda D}{a}$ . If the slit separation $a$ is doubled and the distance to the screen $D$ is halved, what is the new fringe separation in terms of the original $x$ ?
[1]
A. $x/4$
B. $x/2$
C. $x$
D. $2x$

Answer: __________________________

4. Define the term coherence as applied to light sources.
[2]

5. State the condition required for constructive interference to occur between two waves from coherent sources.
[2]

Section B: Wave Properties & Stationary Waves (10 Marks)

6. A pipe of length $L$ is closed at one end and open at the other. Write down the expression for the wavelength $\lambda$ of the fundamental frequency (first harmonic) in terms of $L$ .
[1]

$\lambda =$ __________________________

7. Explain why sound waves cannot be polarized.
[2]

8. A progressive wave is described by the equation:
$y = 0.05 \sin(40\pi t - 2\pi x)$
where $y$ and $x$ are in meters and $t$ is in seconds.

(a) Determine the amplitude of the wave.
[1]
Amplitude = __________________________ m

(b) Calculate the frequency of the wave.
[2]

9. A stationary wave is formed on a string fixed at both ends. The length of the string is $1.2$ m. The speed of the wave on the string is $24$ m s $^{-1}$ .

(a) Calculate the fundamental frequency of the string.
[2]

(b) Calculate the frequency of the third harmonic.
[1]
Frequency = __________________________ Hz

10. Explain how the energy of the wave is distributed in a stationary wave compared to a progressive wave.
[2]

Section C: Interference & Diffraction (15 Marks)

11. In a double-slit interference experiment, monochromatic light of wavelength $550$ nm is used. The slits are separated by $0.40$ mm, and the screen is placed $2.0$ m away.

(a) Calculate the separation between adjacent bright fringes on the screen.
[3]

(b) The monochromatic source is replaced by a white light source. Describe and explain the appearance of the central fringe and the first few fringes on either side.
[3]

12. A diffraction grating has $500$ lines per mm. Monochromatic light is incident normally on the grating. The second-order maximum is observed at an angle of $30^\circ$ to the normal.

(a) Calculate the wavelength of the light.
[3]

(b) Determine the highest order of maximum that can be observed with this light.
[3]

13. Explain why a diffraction grating produces sharper and brighter maxima than a double-slit arrangement.
[2]

14. One of the slits in a double-slit experiment is now covered. Describe the change in the pattern observed on the screen.
[2]

Section D: Sound & Doppler Effect (15 Marks)

15. Ultrasound is used in medical imaging. State the typical frequency range of ultrasound used in medical diagnostics.
[1]

16. Explain the principle of pulse-echo technique used to determine the depth of a tissue boundary.
[3]

17. Why is a coupling gel used between the ultrasound transducer and the patient’s skin? Refer to acoustic impedance in your answer.
[3]

18. A police car emitting a siren of frequency $800$ Hz is moving towards a stationary observer at a speed of $30$ m s $^{-1}$ . The speed of sound in air is $340$ m s $^{-1}$ . Calculate the frequency heard by the observer as the car approaches.
[3]

19. The car passes the observer and continues moving away at the same speed. Calculate the frequency heard by the observer now.
[2]

20. Sketch a graph showing how the observed frequency changes with time as the car passes the observer. Assume the car moves at constant speed.
[2]

Answers

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

1. B
Explanation: Points separated by $\lambda/2$ on a stationary wave are in adjacent loops (separated by a node). Adjacent loops vibrate in opposite directions, so they are in antiphase ( $180^\circ$ out of phase).

2. C
Explanation: Frequency is determined by the source and does not change when a wave crosses a boundary. Speed and wavelength change.

3. A
Explanation: $x = \frac{\lambda D}{a}$ . New $x' = \frac{\lambda (D/2)}{2a} = \frac{1}{4} \frac{\lambda D}{a} = \frac{x}{4}$ .

4.
Answer: Coherence means the waves have a constant phase difference [1] and the same frequency (or wavelength) [1].

5.
Answer: The path difference between the two waves must be an integer multiple of the wavelength [1].
Mathematically: Path Difference $= n\lambda$ , where $n = 0, 1, 2, ...$ [1].

6.
Answer: $\lambda = 4L$
Explanation: For a pipe closed at one end, the fundamental mode has a node at the closed end and an antinode at the open end. This corresponds to $L = \lambda/4$ .

7.
Answer: Sound waves are longitudinal waves [1]. The oscillations of particles are parallel to the direction of energy propagation. Polarization requires transverse oscillations (perpendicular to propagation) to filter specific planes [1].

8.
(a) Amplitude $= 0.05$ m [1]
(b) $\omega = 40\pi$ . $f = \frac{\omega}{2\pi} = \frac{40\pi}{2\pi} = 20$ Hz [2]

9.
(a) Fundamental: $L = \lambda/2 \Rightarrow \lambda = 2L = 2.4$ m.
$f_1 = \frac{v}{\lambda} = \frac{24}{2.4} = 10$ Hz [2]
(b) $f_3 = 3 f_1 = 3 \times 10 = 30$ Hz [1]

10.
Answer: In a stationary wave, energy is stored (trapped) between nodes and does not propagate along the string [1]. In a progressive wave, energy is transmitted in the direction of wave propagation [1].

11.
(a) $x = \frac{\lambda D}{a}$
$\lambda = 550 \times 10^{-9}$ m, $D = 2.0$ m, $a = 0.40 \times 10^{-3}$ m
$x = \frac{550 \times 10^{-9} \times 2.0}{0.40 \times 10^{-3}} = \frac{1100 \times 10^{-9}}{4 \times 10^{-4}} = 275 \times 10^{-5} = 2.75 \times 10^{-3}$ m [1]
$x = 2.75$ mm [1]
Correct units and substitution [1].

(b) Central fringe: White [1].
Side fringes: Spectra / Coloured [1].
Explanation: Different wavelengths (colours) have different fringe separations ( $x \propto \lambda$ ). Red (longer $\lambda$ ) diffracts more than violet (shorter $\lambda$ ), causing the fringes to spread out into spectra [1].

12.
(a) Grating spacing $d = \frac{1}{500 \text{ lines/mm}} = \frac{1}{500 \times 10^3 \text{ lines/m}} = 2.0 \times 10^{-6}$ m [1]
Formula: $d \sin \theta = n \lambda$
$2.0 \times 10^{-6} \times \sin(30^\circ) = 2 \times \lambda$
$2.0 \times 10^{-6} \times 0.5 = 2 \lambda$
$1.0 \times 10^{-6} = 2 \lambda \Rightarrow \lambda = 5.0 \times 10^{-7}$ m [1]
$\lambda = 500$ nm [1]

(b) Max order when $\sin \theta \le 1$ .
$n = \frac{d \sin \theta}{\lambda}$ . Max $n = \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}}{5.0 \times 10^{-7}} = 4$ [1]
Since $\sin \theta = 1$ gives exactly $n=4$ , the 4th order is visible at $90^\circ$ .
Highest order $= 4$ [2] (1 for calculation, 1 for correct integer conclusion).

13.
Answer: Grating has many slits (thousands) [1]. This causes more destructive interference in non-maximum directions, resulting in sharper/narrower peaks, and more constructive interference at maxima, making them brighter [1].

14.
Answer: The interference pattern disappears [1]. It is replaced by a single-slit diffraction pattern (a broad central maximum with weaker side maxima) [1].

15.
Answer: $> 20$ kHz (Typically $1$ MHz to $15$ MHz) [1]

16.
Answer: A short pulse of ultrasound is emitted [1]. It reflects off the tissue boundary [1]. The time delay $t$ between emission and reception of the echo is measured. Depth $d = \frac{vt}{2}$ [1].

17.
Answer: There is a large difference in acoustic impedance between air and skin/tissue [1]. This causes significant reflection at the air-skin boundary, preventing ultrasound from entering the body [1]. The gel has an impedance similar to tissue, minimizing reflection and maximizing transmission [1].

18.
Answer: Source moving towards observer:
$f' = f \left( \frac{v}{v - v_s} \right)$
$f' = 800 \left( \frac{340}{340 - 30} \right) = 800 \left( \frac{340}{310} \right)$
$f' = 800 \times 1.0968 \approx 877$ Hz [3] (1 for formula, 1 for sub, 1 for ans)

19.
Answer: Source moving away:
$f' = f \left( \frac{v}{v + v_s} \right)$
$f' = 800 \left( \frac{340}{340 + 30} \right) = 800 \left( \frac{340}{370} \right)$
$f' = 800 \times 0.9189 \approx 735$ Hz [2]

20.
Answer: Graph:
Y-axis: Frequency, X-axis: Time.
Horizontal line at high freq ( $\approx 877$ Hz) before passing [0.5].
Sharp drop (not vertical, but steep) as it passes [0.5].
Horizontal line at lower freq ( $\approx 735$ Hz) after passing [0.5].
Labels correct [0.5].