HKDSE Physics Practice Paper: Waves, Optics And Sound

HKDSE Physics Practice Paper: Waves, Optics and Sound

2026-11-03

HKDSE Physics: Waves and Optics

HKDSE Physics practice paper: 12 questions, 100 marks, 75 minutes, on waves, refraction, lenses, sound, diffraction and the photoelectric effect, with full worked answers.

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Paper details　Level: S4-S6 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A (Questions 1 to 4, 32 marks)

Read each question, quote the formula, substitute the values and give the unit.

#	Question	Answer and marking notes
1	A transverse wave travels along a string with frequency 50 Hz and wavelength 0.68 m. (a) Find the wave speed. (b) Find the period. (c) Find the distance travelled in 0.1 s, expressed as a number of wavelengths. (d) If the frequency is doubled while the speed stays the same, find the new wavelength.	(a) v = f x wavelength = 50 x 0.68 = 34 m/s [2]. (b) T = 1/f = 1/50 = 0.02 s [2]. (c) Distance = v t = 34 x 0.1 = 3.4 m, which is 0.1 / 0.02 = 5 wavelengths [2]. (d) The speed is unchanged, so the wavelength = 34/100 = 0.34 m: doubling the frequency halves the wavelength [2]. Marking: quote the formula before substituting; period and frequency are reciprocals.
2	Water waves pass from deep water into shallow water. The speed is 1.2 m/s in deep water and 0.8 m/s in shallow water; the source frequency is 5 Hz. (a) Find the wavelength in deep water. (b) Find the wavelength in shallow water. (c) State how the frequency, the speed and the wavelength change across the boundary. (d) If the angle of incidence is 30 degrees, find the angle of refraction.	(a) Wavelength in deep water = 1.2 / 5 = 0.24 m [2]. (b) Wavelength in shallow water = 0.8 / 5 = 0.16 m [2]. (c) The frequency is set by the source and does not change (still 5 Hz); the speed is set by the medium and is lower in shallow water; the wavelength shortens in proportion, in the ratio 3 : 2 [2]. (d) sin r = (v2/v1) sin i = (0.8/1.2) x sin 30 = 0.333, so r = 19.5 degrees and the ray bends towards the normal [2]. Marking: state why the frequency is unchanged; refract with the speed ratio.
3	A ray of light passes from air into glass of refractive index 1.5 at an angle of incidence of 45 degrees. (a) Find the angle of refraction. (b) Find the speed of light in the glass (speed in air = 3.0 x 10^8 m/s). (c) Find the critical angle for light travelling from the glass into air.	(a) n = sin i / sin r, so sin r = sin 45 / 1.5 = 0.7071/1.5 = 0.4714 and r = 28.1 degrees [3]. (b) v = c/n = 3.0 x 10^8 / 1.5 = 2.0 x 10^8 m/s [2]. (c) sin C = 1/n = 0.667, so C = 41.8 degrees; beyond this angle total internal reflection occurs [3]. Marking: refractive index is the ratio of the vacuum speed to the speed in the medium.
4	A converging lens has a focal length of 10 cm. (a) An object is placed 30 cm from the lens; find the image distance, the magnification and the nature of the image. (b) The object is moved to 5 cm from the lens; find the image distance, the magnification and the nature of the image.	(a) 1/f = 1/u + 1/v, so 1/v = 1/10 – 1/30 = 2/30 and v = 15 cm; m = v/u = 0.5; since u > 2f the image is real, inverted and diminished [4]. (b) 1/v = 1/10 – 1/5 = -1/10, so v = -10 cm on the same side as the object; m = 10/5 = 2; with the object inside the focal length the image is virtual, upright and magnified, as in a magnifying glass [4]. Marking: a negative image distance means a virtual image.

Section B (Questions 5 to 8, 36 marks)

Answer the following, showing full working and the formula used.

#	Question	Answer and marking notes
5	A person claps in an open space and hears the echo 0.6 s later. The speed of sound is 340 m/s. (a) Find the distance to the reflecting wall. (b) If the air warms up and the speed becomes 350 m/s, find the time taken for the echo. (c) Explain why sonar uses ultrasound rather than audible sound.	(a) The sound travels there and back, so d = v t / 2 = 340 x 0.6 / 2 = 102 m [3]. (b) t = 2d / v = 2 x 102 / 350 = 0.58 s [3]. (c) Ultrasound has a short wavelength and is highly directional, forming a narrow beam with little spreading; audible sound spreads widely and is masked by background noise, so the echo is hard to identify [3]. Marking: divide by two in (a); the speed rises with temperature.
6	An optical fibre has a core of refractive index 1.50 and a cladding of refractive index 1.45. (a) Find the critical angle at the core-cladding boundary. (b) Explain why the core must have the higher refractive index. (c) Find the speed of light in the core (speed in vacuum = 3.0 x 10^8 m/s).	(a) sin C = n(cladding)/n(core) = 1.45/1.50 = 0.9667, so C = 75.2 degrees [3]. (b) Total internal reflection only happens when light travels from an optically denser to a less dense medium; with the higher core index, light striking the boundary beyond the critical angle is totally reflected and stays in the core [3]. (c) v = c/n = 3.0 x 10^8 / 1.50 = 2.0 x 10^8 m/s [3]. Marking: the numerator of the critical-angle formula is the smaller index.
7	A string of length 1.2 m is fixed at both ends. Its linear density is 5.0 g/m and its fundamental frequency is 60 Hz. (a) Find the wavelength of the fundamental mode. (b) Find the wave speed on the string. (c) Find the tension in the string.	(a) With both ends fixed the fundamental has half a wavelength on the string: wavelength = 2L = 2.4 m [3]. (b) v = f x wavelength = 60 x 2.4 = 144 m/s [3]. (c) v = sqrt(T/mu), so T = mu v^2 = 0.005 x 144^2 = 103.68, about 104 N [3]. Marking: convert 5.0 g/m to 0.005 kg/m before substituting.
8	The sound intensity at a point is 1.0 x 10^-6 W/m^2 (reference intensity 1.0 x 10^-12 W/m^2). (a) Find the sound intensity level. (b) If the distance from the source is doubled, find the new intensity level. (c) Explain why sound levels are measured on a logarithmic decibel scale.	(a) Level = 10 log(I/I0) = 10 log(1.0 x 10^-6 / 1.0 x 10^-12) = 10 log 10^6 = 60 dB [3]. (b) Doubling the distance quarters the intensity to 2.5 x 10^-7 W/m^2, so the level is 10 log(2.5 x 10^5) = 54 dB, a drop of 6 dB [3]. (c) The ear responds to intensities from 10^-12 to about 1 W/m^2, a range of twelve orders of magnitude; a linear scale would be unwieldy, while the logarithmic scale compresses it into 0 to 120 dB and matches loudness perception [3]. Marking: halving the intensity is about -3 dB; doubling the distance is about -6 dB.

Section C (Questions 9 to 12, 32 marks)

Show fuller working and explanation, with units in every answer.

#	Question	Answer and marking notes
9	A diffraction grating has 500 lines per millimetre and is illuminated normally with monochromatic light of wavelength 600 nm. (a) Find the slit spacing. (b) Find the diffraction angle of the first-order maximum. (c) Find the diffraction angle of the second-order maximum.	(a) d = 1 mm / 500 = 2.0 x 10^-6 m [2]. (b) d sin(theta) = wavelength, so sin(theta) = 600 x 10^-9 / 2.0 x 10^-6 = 0.30 and theta = 17.5 degrees [3]. (c) For the second order sin(theta) = 2 x 0.30 = 0.60, so theta = 36.9 degrees [3]. Marking: d is the spacing between adjacent lines, not the width of the grating.
10	In a double-slit experiment the slit separation is 0.50 mm and the screen is 2.0 m away; the fringe spacing is 2.4 mm. (a) Find the wavelength of the light. (b) If the screen is moved to 3.0 m, find the new fringe spacing. (c) Describe how the pattern changes if white light is used.	(a) Fringe spacing = wavelength x D / a, so wavelength = a x spacing / D = (0.50 x 10^-3 x 2.4 x 10^-3) / 2.0 = 6.0 x 10^-7 m = 600 nm [4]. (b) The spacing is proportional to D, so it becomes 2.4 x 3.0/2.0 = 3.6 mm [2]. (c) White light contains many wavelengths, each producing its own spacing, so the central fringe is white and the side fringes are coloured with violet nearest the centre; higher orders overlap to give white again [2]. Marking: convert every length to metres before substituting.
11	Light of wavelength 400 nm falls on a metal of work function 2.0 eV (Planck constant 6.63 x 10^-34 J s, speed of light 3.0 x 10^8 m/s, 1 eV = 1.6 x 10^-19 J). (a) Find the frequency of the light. (b) Find the energy of one photon, in J and in eV. (c) Find the maximum kinetic energy of the emitted photoelectrons, in eV.	(a) f = c / wavelength = 3.0 x 10^8 / 400 x 10^-9 = 7.5 x 10^14 Hz [2]. (b) E = h f = 6.63 x 10^-34 x 7.5 x 10^14 = 4.97 x 10^-19 J, which is 4.97 x 10^-19 / 1.6 x 10^-19 = 3.1 eV [3]. (c) Work function = 2.0 eV = 3.2 x 10^-19 J, so maximum kinetic energy = 4.97 x 10^-19 – 3.2 x 10^-19 = 1.77 x 10^-19 J = 1.1 eV [3]. Marking: use one energy unit throughout; no emission occurs if the photon energy is below the work function.
12	A resonance tube is used to measure the speed of sound. A 500 Hz tuning fork sounds at the mouth of the tube; the first resonance occurs with an air column of 17 cm and the second with 53 cm. (a) Find the wavelength. (b) Find the speed of sound. (c) Find the end correction at the mouth of the tube.	(a) Successive resonances differ by half a wavelength, so wavelength = 2 x (0.53 – 0.17) = 0.72 m [2]. (b) v = f x wavelength = 500 x 0.72 = 360 m/s [3]. (c) At the first resonance the air column is wavelength/4 + e, so e = 0.72/4 – 0.17 = 0.18 – 0.17 = 0.01 m, that is 1 cm [3]. Marking: subtracting the two resonance lengths removes the end correction.

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