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. C — Sound waves in air are longitudinal waves. [1]
Explanation: In longitudinal waves, particles vibrate parallel to the direction of wave travel. Sound waves in air consist of compressions and rarefactions, making them longitudinal. Transverse waves (like light) have particles vibrating perpendicular to the wave direction. Light is a transverse electromagnetic wave and does not require a medium.

2. B — 300 m/s [1]
Working: $v = f \lambda = 250 \times 1.2 = 300 \text{ m/s}$
Key concept: Wave speed = frequency × wavelength ($v = f\lambda$).

3. B — 4 cm [1]
Explanation: Amplitude is the maximum displacement from the equilibrium position. On a displacement-distance graph, it is the vertical distance from the equilibrium line to a peak (or trough). The graph shows peaks at +4 cm and troughs at -4 cm.

4. C — Steel at 20°C [1]
Explanation: Sound travels fastest in solids (steel ~5000 m/s), slower in liquids (water ~1500 m/s), and slowest in gases (air ~340 m/s). Sound cannot travel in a vacuum.

5. B — 340 m/s [1]
Working: Total distance travelled by sound = $2 \times 170 = 340 \text{ m}$. Time = 1.0 s. Speed = $\frac{340}{1.0} = 340 \text{ m/s}$.
Common mistake: Forgetting to double the distance (echo travels to cliff and back).

6. D — Gamma rays [1]
Explanation: Electromagnetic spectrum order (increasing wavelength / decreasing frequency): Gamma rays < X-rays < UV < Visible < Infrared < Microwaves < Radio waves. Gamma rays have the shortest wavelength and highest frequency.

7. C — The wavelength of light decreases. [1]
Explanation: When light enters a denser medium (glass), its speed decreases ($v = c/n$), frequency remains constant, so wavelength decreases ($\lambda = v/f$). The ray bends towards the normal, not away.

8. C — It undergoes total internal reflection. [1]
Explanation: Total internal reflection occurs when light travels from denser to rarer medium and angle of incidence > critical angle. Here, $50° > 42°$, so TIR occurs.

9. B — Real, inverted, diminished [1]
Explanation: For a convex lens: object at $u = 30 \text{ cm}$, $f = 10 \text{ cm}$. Since $2f < u < \infty$ (i.e., $20 < 30 < \infty$), image is real, inverted, diminished, and located between $f$ and $2f$ on the opposite side.

10. C — Thermal imaging cameras [1]
Explanation: Infrared radiation is emitted by warm objects. Thermal imaging cameras detect this radiation to create temperature maps. Sterilisation uses UV, satellite communication uses microwaves, X-ray photography uses X-rays.

Section B: Structured Questions (18 marks)

11. (a) A wavefront is an imaginary line or surface that joins all points in a wave that are in the same phase (e.g., all crests or all troughs). [1]
Teaching note: Wavefronts are perpendicular to the direction of wave propagation (rays).

(b) [2]
Marking points:

• 1 mark: Wavefronts become circular/semicircular after the gap (diffraction).
• 1 mark: Wavelength (spacing between wavefronts) remains unchanged.
  Expected diagram: Circular wavefronts spreading out from the gap, centred on the gap.

(c) The amount of diffraction decreases (less spreading) when the gap is made wider. [1]
Key concept: Diffraction is most significant when gap size ≈ wavelength. Wider gap → less diffraction.

12. (a) Depth = 600 m [2]
Working:
Total distance travelled by ultrasound = $v \times t = 1500 \times 0.80 = 1200 \text{ m}$
This is the round trip (down and up), so depth = $\frac{1200}{2} = 600 \text{ m}$
Mark breakdown: 1 mark for correct use of $v = \frac{d}{t}$, 1 mark for halving the distance.

(b) Ultrasound has a higher frequency (shorter wavelength) than audible sound, so it:

• Produces a narrower, more directional beam for accurate detection.
• Is not audible to humans, avoiding disturbance.
• Less diffraction around obstacles. [1]
  Any one valid reason accepted.

13. (a) [1]
Marking: Angle $i$ labelled between incident ray and normal in air; angle $r$ labelled between refracted ray and normal in glass. Both on the same side of the normal.

(b) $r = 25.4°$ (or 25°) [2]
Working:
$n = \frac{\sin i}{\sin r}$
$1.5 = \frac{\sin 40°}{\sin r}$
$\sin r = \frac{\sin 40°}{1.5} = \frac{0.6428}{1.5} = 0.4285$
$r = \sin^{-1}(0.4285) = 25.4°$
Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(c) The two surfaces of the rectangular block are parallel. The ray bends towards the normal on entry (air → glass) and bends away from the normal by the same amount on exit (glass → air), 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).

14. (a) The ray enters along the radius, which is normal to the curved surface. Angle of incidence = 0°, so no refraction occurs (ray passes straight through without deviation). [1]
Key concept: $n = \frac{\sin i}{\sin r}$, if $i = 0$, then $r = 0$.

(b) $r = 48.6°$ (or 48.6°) [2]
Working: Light goes from glass to air: $n_{\text{glass}} \sin i = n_{\text{air}} \sin r$
$1.5 \times \sin 30° = 1.0 \times \sin r$
$1.5 \times 0.5 = \sin r$
$\sin r = 0.75$
$r = \sin^{-1}(0.75) = 48.6°$
Mark breakdown: 1 mark for correct Snell's law application (glass to air), 1 mark for correct answer.

(c) Critical angle $c = 41.8°$ [2]
Working:
$\sin c = \frac{n_2}{n_1} = \frac{1.0}{1.5} = \frac{2}{3} = 0.6667$
$c = \sin^{-1}(0.6667) = 41.8°$
Mark breakdown: 1 mark for correct formula $\sin c = \frac{1}{n}$, 1 mark for correct calculation.

15. (a) [2]
Marking points:

• 1 mark: Ray from top of object parallel to principal axis, refracts through focal point F on right side.
• 1 mark: Ray from top of object through optical centre C, continues undeviated.
• Image I located where rays intersect (between F and 2F on right side), inverted.
  Expected result: Real, inverted image between F (15 cm) and 2F (30 cm).

(b) Real, inverted, diminished [1]
Reasoning: Object at 25 cm (between 2F and F for f=15 cm? Wait: 2F = 30 cm, F = 15 cm. Object at 25 cm is between F and 2F. So image is beyond 2F, real, inverted, magnified. Let me recalculate.)
Correction: $f = 15 \text{ cm}$, $u = 25 \text{ cm}$. Since $f < u < 2f$ (15 < 25 < 30), image is real, inverted, magnified, located beyond 2F ($v > 30 \text{ cm}$).
Marking: Accept "real, inverted, magnified" based on correct position analysis.

(c) $v = 37.5 \text{ cm}$ [2]
Working:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
$\frac{1}{15} = \frac{1}{25} + \frac{1}{v}$
$\frac{1}{v} = \frac{1}{15} - \frac{1}{25} = \frac{5 - 3}{75} = \frac{2}{75}$
$v = \frac{75}{2} = 37.5 \text{ cm}$
Mark breakdown: 1 mark for correct substitution, 1 mark for correct answer with unit (cm).
Check: $v = 37.5 \text{ cm} > 30 \text{ cm}$ (beyond 2F), consistent with magnified image.

16. (a) Visible light [1]

(b) Similarity: Both are transverse electromagnetic waves that travel at $3.0 \times 10^8 \text{ m/s}$ in vacuum. [1]
Difference: X-rays have much shorter wavelength / higher frequency / higher energy than visible light, so they can penetrate materials that visible light cannot. [1]
Other valid differences: X-rays are ionising, visible light is not; X-rays used for medical imaging, visible light for vision.

(c) Excessive UV exposure can cause skin cancer / sunburn / premature skin ageing / eye damage (cataracts). [1]
Any one valid danger accepted.

Section C: Longer Structured Questions (12 marks)

17. (a) Bright bands correspond to wave crests (constructive interference) where water is deeper, acting like converging lenses focusing light. Dark bands correspond to troughs (destructive interference) where water is shallower, acting like diverging lenses spreading light. [2]
Mark breakdown: 1 mark for linking bright bands to crests/focusing, 1 mark for linking dark bands to troughs/diverging.

(b) Wavelength = 5 cm [2]
Working: Distance between 5 successive bright bands = 4 wavelengths (since 5 bands have 4 gaps between them).
$4\lambda = 20 \text{ cm}$
$\lambda = 5 \text{ cm} = 0.05 \text{ m}$
Mark breakdown: 1 mark for recognising 5 bands = 4λ, 1 mark for correct calculation.

(c) Speed = 0.6 m/s [1]
Working: $v = f\lambda = 12 \times 0.05 = 0.6 \text{ m/s}$
ECF allowed from (b).

(d) Wavelength: Decreases (since $v$ constant, $\lambda = v/f$, $f$ increases → $\lambda$ decreases). [1]
Wave speed: Remains the same (wave speed in water depends only on water depth, not frequency). [1]
Key concept: For water waves in constant depth, speed is constant; frequency and wavelength are inversely proportional.

18. (a) Critical angle $c = 48.8°$ [2]
Working:
$\sin c = \frac{n_{\text{air}}}{n_{\text{water}}} = \frac{1.00}{1.33} = 0.7519$
$c = \sin^{-1}(0.7519) = 48.8°$
Mark breakdown: 1 mark for correct formula $\sin c = \frac{1}{n}$, 1 mark for correct calculation.

(b) [1]
Marking: Ray shows total internal reflection — reflects back into water at angle of reflection = 60° (equal to angle of incidence). No refracted ray in air.

(c) For total internal reflection to occur at the core-cladding boundary, light must travel from a denser medium (higher refractive index) to a rarer medium (lower refractive index). If the core had a lower refractive index than the cladding, light would refract out into the cladding instead of being totally internally reflected, and the light would not be guided along the fibre. [2]
Mark breakdown: 1 mark for stating core must be denser (higher n) than cladding, 1 mark for explaining TIR condition (light from denser to rarer medium).

19. (a) $v = -15 \text{ cm}$ [2]
Working:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
$\frac{1}{10} = \frac{1}{6} + \frac{1}{v}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{6} = \frac{3 - 5}{30} = -\frac{2}{30} = -\frac{1}{15}$
$v = -15 \text{ cm}$
Mark breakdown: 1 mark for correct substitution (including sign convention: $u$ positive for real object), 1 mark for correct negative answer indicating virtual image.
Sign convention: Real is positive for object distance $u$; virtual image has negative $v$.

(b) Magnification $m = 2.5$ [1]
Working: $m = \frac{v}{u} = \frac{-15}{6} = -2.5$ → magnitude = 2.5 (or $m = 2.5$, virtual and upright)
Accept: 2.5 or -2.5 with explanation.

(c) Virtual, upright, magnified [1]
Reasoning: Negative $v$ → virtual; negative $m$ (or positive magnitude with upright) → upright; $|m| > 1$ → magnified.

(d) A magnifying glass produces a magnified virtual image only when the object is placed within the focal length ($u < f$). In this position, the rays leaving the lens diverge as if coming from a larger virtual image on the same side as the object. If the object is at or beyond the focal point, a real image is formed (which cannot be used as a simple magnifier for direct viewing). [2]
Mark breakdown: 1 mark for stating condition $u < f$, 1 mark for explaining diverging rays form virtual image on same side as object.

20. (a) Wavelength = 0.8 m [1]
Given directly on diagram / from $v = f\lambda$: $\lambda = \frac{v}{f} = \frac{4.0}{5} = 0.8 \text{ m}$.

(b) At $t = 0.05 \text{ s}$ (which is $T/4$ since $T = 1/f = 0.2 \text{ s}$), particle P has moved to its maximum positive displacement (amplitude = 0.05 m). [2]
Explanation: Period $T = 0.2 \text{ s}$. At $t = 0$, P is at equilibrium moving up. After $T/4 = 0.05 \text{ s}$, it reaches maximum positive displacement (crest).
Mark breakdown: 1 mark for determining period / time fraction, 1 mark for correct description of position.

(c) Phase difference = $\frac{\pi}{2}$ rad (or 90°) [2]
Working:
Distance between P and Q = 0.2 m. Wavelength = 0.8 m.
Fraction of wavelength = $\frac{0.2}{0.8} = \frac{1}{4}$.
Phase difference = $\frac{1}{4} \times 2\pi = \frac{\pi}{2}$ rad (or 90°).
Mark breakdown: 1 mark for correct fraction of wavelength, 1 mark for correct phase difference in rad or degrees.
Alternative: Q is at $x = 0.2 \text{ m}$ which is $\lambda/4$ ahead. Since wave travels right, Q leads P by 90° (Q reaches maximum first).

(d) [2]
Marking points for displacement-time graph:

• 1 mark: Correct sinusoidal shape starting at origin (0,0) and going positive (upwards).
• 1 mark: Period = 0.2 s shown (two complete oscillations = 0.4 s on x-axis), amplitude = 0.05 m.
  Expected graph: Sine wave crossing at $t = 0, 0.1, 0.2, 0.3, 0.4 \text{ s}$; peaks at $t = 0.05, 0.25 \text{ s}$ (+0.05 m); troughs at $t = 0.15, 0.35 \text{ s}$ (-0.05 m).

End of Answer Key