Secondary 3 Physics Waves Sound Light Quiz

Secondary 3 Physics Quiz - Waves Sound Light (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. Answer: C

Working:
Wave speed $v = f \lambda$
Given: $\lambda = 0.4 \text{ m}$ (distance between adjacent crests = wavelength), $f = 5 \text{ Hz}$
$v = 5 \times 0.4 = 2.0 \text{ m/s}$

Key concept: For any wave, $v = f\lambda$. Distance between adjacent crests is the wavelength.

2. Answer: C

Explanation:

• A is incorrect: Sound waves are longitudinal waves (particles vibrate parallel to wave direction).
• B is incorrect: Sound cannot travel in vacuum; it requires a medium.
• C is correct: Sound waves are mechanical waves that require a medium (solid, liquid, or gas) to propagate.
• D is incorrect: Speed of sound increases with temperature in gases.

3. Answer: B

Working:
Echo travels to cliff and back: total distance = $2 \times 170 = 340 \text{ m}$
Time = 1.0 s
Speed = $\frac{\text{distance}}{\text{time}} = \frac{340}{1.0} = 340 \text{ m/s}$

Common mistake: Forgetting to double the distance (using 170 m instead of 340 m gives 170 m/s, option A).

4. Answer: A

Working:
Snell's Law: $n_1 \sin i = n_2 \sin r$
$n_{\text{air}} = 1$, $n_{\text{glass}} = 1.5$, $i = 30^\circ$
$1 \times \sin 30^\circ = 1.5 \times \sin r$
$0.5 = 1.5 \sin r$
$\sin r = \frac{0.5}{1.5} = \frac{1}{3}$
$r = \sin^{-1}(\frac{1}{3}) \approx 19.5^\circ$

Key concept: Light bends towards the normal when entering a denser medium ($n_2 > n_1$), so $r < i$.

5. Answer: C

Explanation:
Angle of incidence = 40°, so angle of reflection = 40° (Law of Reflection).
Angle between incident ray and reflected ray = $40^\circ + 40^\circ = 80^\circ$.

Visual check: The two rays are on opposite sides of the normal, each at 40° to it.

6. Answer: D

Explanation:
Order of EM spectrum by increasing frequency (decreasing wavelength):
Radio → Microwave → Infrared → Visible → Ultraviolet → X-rays → Gamma rays
Gamma rays have the highest frequency and shortest wavelength.

7. Answer: C

Working:
Frequency $f = \frac{1}{T}$
Period $T = 0.02 \text{ s}$
$f = \frac{1}{0.02} = 50 \text{ Hz}$

Key concept: Period and frequency are reciprocals.

8. Answer: C

Working:
Critical angle $c$: $\sin c = \frac{n_2}{n_1} = \frac{1}{1.33}$ (from water to air)
$\sin c = 0.7519$
$c = \sin^{-1}(0.7519) \approx 48.8^\circ \approx 49^\circ$

Key concept: Total internal reflection occurs when light travels from denser to rarer medium and $i > c$.

9. Answer: B

Explanation:

• Loudness depends on amplitude. Doubling amplitude increases loudness.
• Pitch depends on frequency. Frequency unchanged → pitch unchanged.

Key concept: Amplitude → loudness/energy; Frequency → pitch.

10. Answer: B

Working:
Lens formula: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
$u = 30 \text{ cm}$, $v = 15 \text{ cm}$ (real image, so $v$ positive)
$\frac{1}{f} = \frac{1}{30} + \frac{1}{15} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$
$f = 10 \text{ cm}$

Section B: Structured Questions (18 marks)

11. (a) Amplitude = 4 cm [1]

Explanation: Amplitude is the maximum displacement from equilibrium. From graph, peak displacement = 4 cm.

(b) Wavelength = 1.0 m [1]
Explanation: Two complete waves in 2.0 m → one wavelength = 1.0 m. Or distance between adjacent crests (0.5 m to 1.5 m) = 1.0 m.

(c) Speed = 2.5 m/s [2]
Working:
$v = f \lambda = 2.5 \times 1.0 = 2.5 \text{ m/s}$
Mark breakdown: 1 mark for formula/substitution, 1 mark for correct answer with unit.

12. (a) Refractive index = 1.59 (or 1.6) [2]

Working:
$n = \frac{\sin i}{\sin r} = \frac{\sin 40^\circ}{\sin 25^\circ} = \frac{0.6428}{0.4226} = 1.521 \approx 1.52$
Accept 1.5–1.6 depending on rounding.
Mark breakdown: 1 mark for correct formula/substitution, 1 mark for answer.

(b) Speed in glass = $1.89 \times 10^8 \text{ m/s}$ [2]
Working:
$n = \frac{c_{\text{air}}}{c_{\text{glass}}}$
$c_{\text{glass}} = \frac{c_{\text{air}}}{n} = \frac{3.0 \times 10^8}{1.52} = 1.97 \times 10^8 \text{ m/s}$
(Using $n=1.52$ gives $1.97 \times 10^8$; using $n=1.59$ gives $1.89 \times 10^8$)
Mark breakdown: 1 mark for formula, 1 mark for calculation with unit.

(c) The two surfaces of the rectangular block are parallel. The ray bends towards the normal on entry and away from the normal on exit by the same amount, so the emergent ray is parallel to the incident ray. [1]
Key concept: Parallel-sided block → emergent ray parallel to incident ray but laterally displaced.

13. (a) Total internal reflection occurs. [1]

Explanation: Angle of incidence (50°) > critical angle (~49°), and light travels from denser (water) to rarer (air) medium.

(b) Critical angle = 48.8° (≈ 49°) [2]
Working:
$\sin c = \frac{1}{1.33} = 0.7519$
$c = \sin^{-1}(0.7519) = 48.8^\circ$
Mark breakdown: 1 mark for $\sin c = 1/n$, 1 mark for calculation.

(c) Optical fibres / fibre optics in telecommunications / endoscopes / binoculars / periscopes. [1]
Any one valid application accepted.

14. (a) Image distance = 16.7 cm [2]

Working:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
$\frac{1}{10} = \frac{1}{25} + \frac{1}{v}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{25} = \frac{5 - 2}{50} = \frac{3}{50}$
$v = \frac{50}{3} = 16.7 \text{ cm}$
Mark breakdown: 1 mark for correct substitution, 1 mark for answer with unit.
Note: Positive $v$ → real image on opposite side of lens from object.

(b) Characteristics (any two): [2]

1. Real (formed on opposite side of lens)
2. Inverted
3. Magnified (since $u < 2f$)
4. Formed beyond 2f on the other side

(c) Magnification = 0.67 (or -0.67) [1]
Working:
$m = \frac{v}{u} = \frac{16.7}{25} = 0.67$
Sign convention: negative for inverted real image, but magnitude often accepted.
Key concept: $|m| < 1$ → diminished; $|m| > 1$ → magnified. Here object at 25 cm (between f and 2f) → magnified, but calculation gives 0.67? Wait: $u=25$, $f=10$, $2f=20$. Object at 25 cm is beyond 2f, so image should be diminished (between f and 2f). $v=16.7$ cm is between f and 2f. Magnification = 16.7/25 = 0.67 < 1 → diminished. Correct.

15. (a) Observed frequency = 880 Hz [2]

Working:
Doppler effect (source moving towards stationary observer):
$f' = f \left( \frac{v}{v - v_s} \right)$
$f' = 800 \times \frac{340}{340 - 30} = 800 \times \frac{340}{310} = 800 \times 1.0968 = 877.4 \approx 880 \text{ Hz}$
Mark breakdown: 1 mark for correct formula, 1 mark for calculation.

(b) The source is moving towards the observer, causing wavefronts to compress (wavelength decreases). Since wave speed is constant in the medium, frequency increases ($f = v/\lambda$). [1]
Key concept: Motion of source changes wavelength; observer receives more wavefronts per second.

Section C: Longer Structured Questions (12 marks)

16. (a) Dispersion (of light) [1]

(b) Different colours of light have different wavelengths. In glass, refractive index is higher for shorter wavelengths (violet) and lower for longer wavelengths (red). Since $n = \sin i / \sin r$, different colours refract by different amounts (violet bends most, red least), causing white light to split into a spectrum. [2]
Mark breakdown: 1 mark for different wavelengths/refractive indices, 1 mark for different deviation/bending.

(c) Angle of refraction (red) = 27.7° [2]
Working:
$n_{\text{red}} = 1.51$, $i = 45^\circ$
$\sin r = \frac{\sin 45^\circ}{1.51} = \frac{0.7071}{1.51} = 0.4683$
$r = \sin^{-1}(0.4683) = 27.9^\circ$ (≈ 28°)
Mark breakdown: 1 mark for Snell's law substitution, 1 mark for answer.

(d) Violet has the greater angle of deviation.
Explanation: Violet light has a shorter wavelength and higher refractive index in glass ($n_{\text{violet}} = 1.53 > n_{\text{red}} = 1.51$). It bends more at both surfaces of the prism, resulting in a larger total deviation. [2]
Mark breakdown: 1 mark for identifying violet, 1 mark for explanation linking higher $n$ to greater bending.

17. (a) Bright bands correspond to crests (converging light) and dark bands to troughs (diverging light) of the water waves acting as lenses. [1]

Explanation: Water surface acts like a series of convex (crests) and concave (troughs) lenses focusing/defocusing light onto the screen.

(b) Wavelength = 3.0 cm [2]
Working:
Distance between 5 consecutive bright bands = 4 wavelengths = 12 cm
$\lambda = \frac{12}{4} = 3.0 \text{ cm} = 0.03 \text{ m}$
Mark breakdown: 1 mark for recognizing 5 bands = 4$\lambda$, 1 mark for calculation.

(c) Speed = 0.3 m/s [1]
Working:
$v = f \lambda = 10 \times 0.03 = 0.3 \text{ m/s}$

(d) Wavelength decreases and speed decreases.
Explanation: In shallower water, wave speed decreases ($v \propto \sqrt{d}$ for shallow water waves). Frequency remains constant (determined by source). Since $v = f\lambda$, wavelength must also decrease. [2]
Mark breakdown: 1 mark for both decrease, 1 mark for explanation linking $v$, $f$, $\lambda$.

18. (a) Ray diagram completion [2]

Expected rays:

• Ray 1: From top of object, parallel to principal axis → refracts through F through F through focal point on right side.
• Ray 2: From top of object, through optical centre → continues undeviated.
• Image formed where rays intersect: real, inverted, magnified, beyond 2F on right.
  Mark breakdown: 1 mark for each correct ray, or 2 marks for correct image location with two rays.

(b) Image distance = 60 cm [2]
Working:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
$\frac{1}{15} = \frac{1}{20} + \frac{1}{v}$
$\frac{1}{v} = \frac{1}{15} - \frac{1}{20} = \frac{4 - 3}{60} = \frac{1}{60}$
$v = 60 \text{ cm}$
Mark breakdown: 1 mark for substitution, 1 mark for answer.

(c) Three characteristics: [3]

1. Real (formed on opposite side, $v$ positive)
2. Inverted
3. Magnified (since $v > u$, or $m = v/u = 60/20 = 3$)
   Also acceptable: formed beyond 2F on the other side.

19. (a) $3.0 \times 10^8 \text{ m/s}$ [1]

(b) Visible light (green region) [1]
Explanation: Visible spectrum ≈ 400–700 nm. 500 nm is in the green region.

(c) Use: Sterilisation / disinfection / vitamin D synthesis / fluorescent lamps / security marking
Hazard: Skin cancer / sunburn / premature aging / eye damage (cataracts) / DNA damage
[2] — 1 mark each.

(d) Microwaves can pass through the atmosphere (ionosphere) with little absorption, travel in straight lines (line-of-sight), and can be directed in narrow beams using dish antennas, allowing high-bandwidth communication with satellites. [2]
Key points: Penetrate atmosphere, line-of-sight, directional beams, high frequency → high data capacity.

20. (a) Speed of sound = 340 m/s [2]

Working:
Total distance travelled by sound = $2 \times 6.8 = 13.6 \text{ m}$
Time = 0.04 s
$v = \frac{13.6}{0.04} = 340 \text{ m/s}$
Mark breakdown: 1 mark for doubling distance, 1 mark for calculation with unit.

(b) Any one valid reason: [1]

• Temperature different from standard conditions (speed varies with $\sqrt{T}$)
• Wind affecting sound propagation
• Reaction time / measurement error in data logger
• Humidity effects
• Wall not perfectly reflecting (some absorption)

(c) No, the speed of sound does not change.
Explanation: Speed of sound in a medium depends only on the properties of the medium (temperature, density, elasticity), not on the frequency of the sound wave. [1]

(d) Modifications for water: [2]

• Submerge the loudspeaker and microphone/hydrophone in water
• Use a waterproof sound source (e.g., underwater speaker/pinger) and hydrophone
• Place a large flat reflector (e.g., metal plate) underwater at known distance
• Measure time for echo return and calculate $v = 2d/t$
• Ensure temperature of water is noted/controlled
  Mark breakdown: 1 mark for submerged setup with appropriate transducers, 1 mark for measurement method.

End of Answer Key