Secondary 4 Pure Physics Waves Sound Light Quiz

Secondary 4 Pure Physics Quiz - Waves Sound Light (Answer Key)

Total Marks: 40 marks

Section A: Multiple Choice and Short Response (Questions 1–10)

1. B — The particles of the medium vibrate perpendicular to the direction of wave travel. [2]

Teaching note: In transverse waves, the particle displacement is perpendicular to the direction of energy transfer. Example: light, waves on a string. In longitudinal waves (sound), particles vibrate parallel to the direction of travel.

2. Speed = 340 m/s [2]

Working:

• Using $v = f \times \lambda$
• $v = 500 \times 0.68 = 340$ m/s [1 for formula, 1 for answer]

Teaching note: This is the standard wave equation. The speed of sound in air is approximately 340 m/s at room temperature, so this result is physically reasonable.

3. Any one difference: [2]

• Light waves are transverse; sound waves are longitudinal. [2]
• Light can travel through a vacuum; sound requires a medium. [2]
• Light travels much faster than sound (3×10⁸ m/s vs ~340 m/s). [2]

Teaching note: The fundamental distinction is polarization—light can be polarized (transverse property) while sound cannot.

4. (a) The speed of light decreases (or: light slows down) [1]

(b) Angle of refraction = 22.5° (accept 22° or 23° if using 1 significant figure intermediate) [1]

Working:

• Snell's law: $n_1 \sin i = n_2 \sin r$
• $1.0 \times \sin 35° = 1.5 \times \sin r$
• $\sin r = \frac{\sin 35°}{1.5} = \frac{0.574}{1.5} = 0.383$
• $r = \sin^{-1}(0.383) = 22.5°$ [1]

Common mistake: Using the refractive index ratio the wrong way round gives $r > i$, which is impossible for air→glass.

5. Speed = 340 m/s [2]

Working:

• Distance travelled by sound = $2 \times 170 = 340$ m (to cliff and back) [1]
• Speed = $\frac{\text{distance}}{\text{time}} = \frac{340}{1.0} = 340$ m/s [1]

Teaching note: Echo distance is always twice the distance to the reflecting surface. This is a standard echo-ranging principle used in sonar and ultrasound.

6. Region 1: Microwave [1] Region 2: Ultraviolet (or UV) [1]

Teaching note: Order of increasing frequency: radio waves → microwaves → infrared → visible light → ultraviolet → X-rays → gamma rays. Remember: Rattling Mice In Visible Underwear eXplode Greatly (mnemonic for order).

7. [2]

Answer: A lunar eclipse occurs when the Earth passes directly between the Sun and the Moon, casting a shadow on the Moon. [1] This alignment can only happen during a full moon because that is when the Moon is on the opposite side of Earth from the Sun. [1]

Teaching note: The three bodies must be collinear with Earth in the middle. The Moon's orbit is slightly tilted, so lunar eclipses don't occur every full moon—only when the alignment is precise.

8. Image distance = 30 cm [1]

Nature of image: Real, inverted, same size as object (accept any two characteristics for 1 mark) [1]

Working:

• Mirror equation: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
• $\frac{1}{15} = \frac{1}{30} + \frac{1}{v}$
• $\frac{1}{v} = \frac{1}{15} - \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}$
• $v = 30$ cm [1]

Teaching note: When $u = 2f$ for a concave mirror, the image forms at $v = 2f$ on the opposite side. This is the "radius of curvature" position, giving a real, inverted image the same size as the object.

9. Maximum speed = 0.314 m/s (or 0.31 m/s) [2]

Working:

• Maximum speed of particle: $v_{max} = 2\pi f A$ (or $\omega A$ where $\omega = 2\pi f$)
• $v_{max} = 2 \times 3.14 \times 2.5 \times 0.04$ [1]
• $v_{max} = 0.628 \times 2.5 \times 0.04 = 0.628 \times 0.1 = 0.0628$...

Correction: Let me recalculate: $2 \times 3.14 = 6.28$; $6.28 \times 2.5 = 15.7$; $15.7 \times 0.04 = 0.628$ m/s

Maximum speed = 0.628 m/s (or 0.63 m/s to 2 s.f.) [2]

Teaching note: The maximum speed occurs as particles pass through equilibrium position. This is $v_{max} = \omega A$ where angular frequency $\omega = 2\pi f$.

10. Any two valid applications: [2]

• Thermal imaging cameras [1]
• Remote controls [1]
• Night vision equipment [1]
• Heaters and cooking (halogen ovens) [1]
• Fibre optic communications (though this is arguably more IR/visible) [1]
• Infrared spectroscopy in chemical analysis [1]

Section B: Structured Response (Questions 11–15)

11. (a) The law of reflection: The angle of incidence equals the angle of reflection, with the incident ray, reflected ray, and normal all lying in the same plane. [1]

(b) Angle of reflection = 50° [1]

Working: Angle of incidence to normal = 90° − 40° = 50°. By law of reflection, angle of reflection = 50°.

(c) Diagram: Reflected ray drawn at 50° to normal (or 40° to mirror surface), on opposite side of normal from incident ray. [1]

Teaching note: Common error: using 40° as angle of incidence. Always measure angles from the normal, not from the surface.

12. Intensity = 3.98 × 10⁻⁵ W/m² ≈ 4.0 × 10⁻⁵ W/m² [3]

Working:

• Intensity $I = \frac{P}{4\pi r^2}$ for isotropic source [1]
• $I = \frac{2.0 \times 10^{-3}}{4 \times 3.14 \times (4.0)^2}$ [1]
• $I = \frac{2.0 \times 10^{-3}}{4 \times 3.14 \times 16} = \frac{2.0 \times 10^{-3}}{200.96} = 9.95 \times 10^{-6}$...

Recalculation with correct formula:

• $I = \frac{P}{4\pi r^2} = \frac{0.002}{4 \times \pi \times 16} = \frac{0.002}{201.06} = 9.95 \times 10^{-6}$ W/m²

Hmm, let me use $P = 2.0$ mW $= 2.0 \times 10^{-3}$ W:

• $I = \frac{2.0 \times 10^{-3}}{4 \times 3.14 \times 16} = \frac{2.0 \times 10^{-3}}{200.96} = 9.95 \times 10^{-6}$ W/m²

Wait—that seems very low. Let me recheck: 2.0 mW is indeed 0.002 W.

Actually: $\frac{0.002}{200.96} = 9.95 \times 10^{-6}$ ≈ 1.0 × 10⁻⁵ W/m² [3]

Teaching note: The inverse square law means intensity falls rapidly with distance. At double the distance, intensity is one-quarter. This is crucial for understanding radiation safety and light propagation.

13. (a) The ray enters along the normal to the curved surface (or: the ray is directed toward the centre of curvature, so angle of incidence = 0°). [1]

Teaching note: Key insight—the curved surface is radial, and rays directed at the centre meet the surface at 90°, so no bending occurs.

(b) Angle of refraction = 72° (or 71.8°) [2]

Working:

• At flat surface: light goes from glass ($n = 1.5$) to air ($n = 1.0$)
• Snell's law: $n_{glass} \sin 45° = n_{air} \sin r$ [1]
• $1.5 \times 0.707 = 1.0 \times \sin r$
• $\sin r = 1.061$...

This exceeds 1, which is impossible! Let me recalculate: 1.5 × sin(45°) = 1.5 × 0.707 = 1.06 > 1

This means total internal reflection occurs. The critical angle is $\sin^{-1}(1/1.5) = 41.8°$. Since 45° > 41.8°, no refraction occurs—light undergoes TIR.

Revised answer: Total internal reflection occurs; there is no emergent ray into air. [2]

Teaching note: This is an excellent illustration of why you must check whether the refracted angle is physically possible. The critical angle for glass-air is about 42°. Many students mechanically apply Snell's law without verifying the result.

14. (a) Distance = 0.30 m = 30 cm [2]

Working:

• Distance = $\frac{v \times t}{2}$ (divide by 2 for return journey) [1]
• Distance = $\frac{1500 \times 0.40 \times 10^{-3}}{2} = \frac{0.60}{2} = 0.30$ m [1]

(b) Any one advantage: [1]

• Ultrasound is non-ionizing / safe for fetus (X-rays are ionizing and can damage cells)
• Ultrasound can show real-time movement (X-rays give static images)
• Ultrasound can be used repeatedly without cumulative dose concerns

15. (a) Amplitude = 0.08 m [1] (read directly from graph; maximum displacement from equilibrium)

(b) Frequency = 5.0 Hz [2]

Working:

• $v = f \times \lambda$, so $f = \frac{v}{\lambda}$ [1]
• $f = \frac{2.0}{0.40} = 5.0$ Hz [1]

Teaching note: A common error is to confuse amplitude with wavelength or to use the wave speed formula incorrectly.

Section C: Extended Response (Questions 16–20)

16. (a) [2]

Answer: The student should plot a graph of sin i (y-axis) against sin r (x-axis). [1] The refractive index equals the gradient of this graph, since Snell's law gives $n = \frac{\sin i}{\sin r}$ when light travels from air to plastic (with $n_{air} ≈ 1$). [1]

Teaching note: This is the standard graphical method for determining refractive index. Using sines automatically gives a linear relationship through the origin.

(b) Speed of light in plastic = 1.97 × 10⁸ m/s ≈ 2.0 × 10⁸ m/s [2]

Working:

• $n = \frac{1}{gradient} = \frac{1}{1.52} = 0.658$...

Wait—if plotting sin i vs sin r with i in air, the gradient IS n. Let me reconsider.

From air: $n_{plastic} = \frac{\sin i}{\sin r}$ = gradient = 1.52

• $n = \frac{c}{v}$, so $v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.52}$ [1]
• $v = 1.97 \times 10^8$ m/s ≈ 1.97 × 10⁸ m/s or 2.0 × 10⁸ m/s to 2 s.f. [1]

17. (a) [2]

Answer: In shallow water, the wave interacts with the sea bed, causing friction and drag on the water particles. [1] This reduces the wave energy and hence the wave speed, while conserving energy causes the wave height (amplitude) to increase. [1] Alternatively: the wavelength decreases in shallow water, and since $v = f\lambda$ with frequency constant, speed decreases.

(b) Warning time = 5.0 hours [2]

Working:

• Time = $\frac{\text{distance}}{\text{speed}} = \frac{4000}{800}$ [1]
• Time = 5.0 hours [1]

Teaching note: Tsunami waves in deep water travel at very high speeds—comparable to jet aircraft—but slow dramatically in shallow coastal regions, allowing the dramatic height increase that causes devastation.

18. (a) [2]

Answer: The object is placed inside the focal length of the converging lens ($u < f$). [1] The refracted rays diverge on the right side of the lens; they appear to come from a point on the same side as the object, forming an image that cannot be projected on a screen—hence virtual. [1]

(b) Any two characteristics: [2]

• Upright / erect [1]
• Magnified / enlarged [1]
• Virtual [1]
• On same side of lens as object [1]

19. (a) [3]

Answer:

• Locations A and B: S-waves cannot pass through the liquid outer core of the Earth. [1] Since S-waves are transverse, they require a solid medium and cannot travel through liquid. [1] This indicates that there is liquid material between the earthquake epicentre and locations A and B.
• Location C: Both P-waves and S-waves are detected, so the path from epicentre to C passes only through solid material (crust and mantle). [1]

(b) Wavelength = 1.5 km [1]

Working:

• $\lambda = \frac{v}{f} = \frac{6}{4} = 1.5$ km [1]

20. (a) d = 2.0 × 10⁻⁶ m [1]

Working:

• $d = \frac{1}{500 \text{ lines/mm}} = \frac{10^{-3}}{500} = \frac{10^{-3}}{5 \times 10^2} = 2 \times 10^{-6}$ m [1]

(b) [2]

Working:

• Maximum order occurs when $\sin\theta = 1$ (i.e., $\theta = 90°$) [1]
• Using $d \sin\theta = n\lambda$: $n_{max} = \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}}{520 \times 10^{-9}} = \frac{2.0 \times 10^{-6}}{5.2 \times 10^{-7}} = \frac{20}{5.2} = 3.85$ [1]
• Since n must be an integer, maximum observable order is n = 3 [1]

(c) [1]

Answer: The number of observable orders increases (or more orders visible). [0.5]

Explanation: Blue light has a shorter wavelength. From $n_{max} = \frac{d}{\lambda}$, smaller $\lambda$ gives larger $n_{max}$. [0.5]

Teaching note: This is why shorter wavelength light (blue/violet) produces more widely spaced diffraction maxima and more observable orders in a diffraction grating spectrum.

END OF ANSWER KEY