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HKDSE Physics Practice Paper: Waves, Light and Sound (12 Questions, 100 Marks, 75 Minutes, with Answers)
HKDSE Physics: Waves, Light and Sound
S4 physics waves, light and sound practice paper: 12 questions on wave speed, echoes, refraction, lenses, standing waves, interference, ultrasound and resonance tubes, with full answers.
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Paper details Level: S4 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Basic Wave Properties and Sound (32 marks)
Answer all questions in Section A.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (Wave quantities) A sound wave in air has a frequency of 250 Hz and a wavelength of 1.36 m. (a) Find the wave speed (3 marks). (b) Find the period in milliseconds (3 marks). (c) The wave enters another medium where the wavelength becomes 0.68 m while the frequency stays the same; find the new speed and explain why the frequency does not change (2 marks). | Answer: (a) v = f x wavelength = 250 x 1.36 = 340.0 m/s. (b) T = 1 / f = 1 / 250 = 4.0 ms. (c) v = 250 x 0.68 = 170.0 m/s; the frequency is set by the source, so the medium changes only the speed and the wavelength. Marking: 1 mark for unit conversion; the period is the reciprocal of the frequency. |
| 2 | (Echo ranging) A person shouts towards a cliff and hears the echo 0.80 s later; the speed of sound in air is 340 m/s. (a) Find the distance to the cliff (3 marks). (b) On a cold night the speed falls to 320 m/s; how long does the echo take then (3 marks). (c) Explain why the distance travelled is halved when finding the distance to the cliff (2 marks). | Answer: (a) s = vt / 2 = 340 x 0.80 / 2 = 136.0 m. (b) t = 2s / v = 2 x 136.0 / 320 = 0.85 s. (c) The sound travels to the cliff and back, so the total path is twice the distance and must be halved. Marking: half marks if the factor of two is forgotten. |
| 3 | (Refraction and total internal reflection) A ray of light enters glass of refractive index 1.5 from air at an angle of incidence of 30 degrees; the speed of light in a vacuum is 3.0 x 10^8 m/s. (a) Use Snell law to find the angle of refraction (3 marks). (b) Find the speed of light in the glass (3 marks). (c) Find the critical angle at the glass-air boundary (2 marks). | Answer: (a) sin r = sin 30 / 1.5 = 0.333, giving r = 19.5 degrees. (b) v = c / n = 3.0 x 10^8 / 1.5 = 2.0 × 10⁸ m/s. (c) sin c = 1 / 1.5 = 0.667, so c = 41.8 degrees. Marking: angles are measured from the normal; a critical angle exists only when light goes from a denser to a less dense medium. |
| 4 | (Converging lens) An object is placed 15 cm in front of a converging lens of focal length 10 cm. (a) Use the lens formula to find the image distance (3 marks). (b) Find the magnification (3 marks). (c) Describe the nature of the image (2 marks). | Answer: (a) 1/f = 1/u + 1/v, so 1/10 = 1/15 + 1/v and 1/v = 1/30, giving v = 30.0 cm. (b) m = v / u = 30.0 / 15 = 2.0. (c) The object lies between one and two focal lengths, so the image is inverted, magnified and real and can be caught on a screen. Marking: correct signs in the formula earn 1 mark. |
Section B: Wave Phenomena, Standing Waves and Interference (36 marks)
Answer all questions in Section B.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (Standing waves on a string) A string 1.2 m long, fixed at both ends, vibrates at 120 Hz and forms 3 antinodes. (a) Find the wavelength (3 marks). (b) Find the wave speed on the string (3 marks). (c) Find the fundamental frequency of this string (3 marks). | Answer: (a) wavelength = 2L / n = 2 x 1.2 / 3 = 0.8 m. (b) v = f x wavelength = 120 x 0.8 = 96.0 m/s. (c) The fundamental has n = 1, so f1 = f3 / 3 = 40.0 Hz. Marking: the number of antinodes equals the harmonic number. |
| 6 | (Electromagnetic waves) A radio station transmits at 100 MHz; the speed of light in a vacuum is 3.0 x 10^8 m/s. (a) Find the wavelength (3 marks). (b) Find the period in nanoseconds (3 marks). (c) Compare radio waves with X-rays in frequency, energy and speed in a vacuum (3 marks). | Answer: (a) wavelength = c / f = 3.0 x 10^8 / 100 x 10^6 = 3.0 m. (b) T = 1 / f = 10.0 ns. (c) Radio waves have a lower frequency, lower photon energy and longer wavelength, while X-rays have a higher frequency, higher energy and shorter wavelength; both travel at 3.0 x 10^8 m/s in a vacuum. Marking: the equal speed is the key point. |
| 7 | (Double-slit interference) Monochromatic light of wavelength 600 nm passes through two slits 0.30 mm apart and fringes are observed on a screen 1.0 m away. (a) Find the spacing between adjacent bright fringes in mm (3 marks). (b) Find the new spacing if the slit separation is doubled (3 marks). (c) Find the spacing if light of wavelength 400 nm is used with the original slit separation (3 marks). | Answer: (a) x = wavelength x D / a = 600 x 10^-9 x 1.0 / 0.30 x 10^-3 = 2.0 mm. (b) The spacing is inversely proportional to the slit separation, so doubling it gives 1.0 mm. (c) The spacing is proportional to the wavelength, giving 1.33 mm. Marking: the 10^-9 conversion from nm to m is essential. |
| 8 | (Ultrasound) A bat navigates with ultrasound of 40 kHz; the speed of sound in air is 340 m/s. (a) Find the wavelength in mm (3 marks). (b) The upper limit of human hearing is about 20 kHz; find the corresponding wavelength (3 marks). (c) Give two reasons why ultrasound is suitable for medical imaging (3 marks). | Answer: (a) wavelength = v / f = 340 / 40000 = 8.5 mm. (b) 340 / 20000 = 17.0 mm. (c) Ultrasound has a short wavelength and a narrow beam, so it resolves fine structure, and it is inaudible so it causes no noise nuisance. Marking: 1.5 marks per sound reason. |
Section C: Optics, Media and Experimental Analysis (32 marks)
Answer all questions in Section C.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | (Refraction of water waves) Water waves of frequency 5.0 Hz travel at 0.30 m/s in deep water and slow to 0.20 m/s in shallow water. (a) Find the wavelength in deep water (3 marks). (b) Find the wavelength in shallow water and explain why the frequency stays the same (3 marks). (c) State whether the rays bend towards or away from the normal in shallow water (2 marks). | Answer: (a) wavelength = v / f = 0.30 / 5.0 = 0.06 m. (b) 0.20 / 5.0 = 0.04 m; the frequency is fixed by the source and does not change with depth. (c) The waves slow down in shallow water, so the rays bend towards the normal. Marking: constant frequency is the basis for the direction of bending. |
| 10 | (Resonance tube) A tube 0.25 m long and closed at one end resonates at its fundamental frequency in air; the speed of sound is 340 m/s. (a) Find the fundamental wavelength (3 marks). (b) Find the fundamental frequency (3 marks). (c) Find the fundamental frequency if the tube were open at both ends (2 marks). | Answer: (a) For a closed pipe the fundamental has wavelength = 4L = 4 x 0.25 = 1.0 m. (b) f = v / wavelength = 340 / 1.0 = 340.0 Hz. (c) An open pipe has wavelength = 2L, so the frequency doubles to 680.0 Hz. Marking: mixing up 4L and 2L is the most common error. |
| 11 | (Light and media) Light of wavelength 600 nm travels in air at 3.0 x 10^8 m/s and enters glass of refractive index 1.5. (a) Find the frequency of the light (3 marks). (b) Find the wavelength in the glass (3 marks). (c) State whether the frequency or the wavelength stays unchanged and explain why (2 marks). | Answer: (a) f = v / wavelength = 3.0 x 10^8 / 600 x 10^-9 = 5.0 × 10¹⁴ Hz. (b) wavelength = 600 / 1.5 = 400.0 nm. (c) The frequency stays the same because it is set by the source; the speed and the wavelength both fall, keeping their ratio equal to the frequency. Marking: find the frequency first, then the wavelength in the medium. |
| 12 | (Measuring the speed of sound) A student uses a resonance tube and a 680 Hz tuning fork. The air column resonates first at 0.18 m and again at 0.43 m. (a) Find the wavelength (3 marks). (b) Find the speed of sound in air from these data (3 marks). (c) Suggest two ways to reduce the experimental error (2 marks). | Answer: (a) The difference between successive resonances is half a wavelength, so wavelength = 2 x (0.43 – 0.18) = 0.5 m. (b) v = f x wavelength = 680 x 0.5 = 340.0 m/s. (c) Use a finer scale to read the water level, repeat the readings and average them, and keep the tube away from draughts. Marking: the difference of the two lengths must be used. |
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