A Level H2 Physics Waves Sound Light Quiz

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].