S5 Physics Practice Paper: Waves, Refraction And Lenses (12 Questions, 100 Marks, 75 Minutes) With Answers

S5 Physics Practice Paper: Waves, Refraction and Lenses (12 Questions, 100 Marks, 75 Minutes) with Answers

2026-11-09

HKDSE Physics: Waves, Refraction and Lenses

S5 Physics practice paper with 12 questions, 100 marks and 75 minutes. Covers wave speed, refraction, total internal reflection, lens images and two-slit interference. Answers included.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show the formula and the law you use, and give the unit with every answer. For lens problems, take the image distance as positive for a real image and negative for a virtual image.

Give numerical answers to three significant figures unless stated otherwise.

Spend about 20 minutes on Section A, 30 minutes on Section B and the last 25 minutes on Section C.

Section A — Basic Properties of Waves

Show the formula, substitute the values and give the unit for each question.

#	Question	Answer and marking notes
1	A transverse wave travels along a string with frequency 50 Hz and wavelength 0.68 m. Find (a) the wave speed; (b) the period; (c) the new wavelength if the frequency is raised to 100 Hz while the wave speed stays the same.	Answer: (a) speed = frequency × wavelength = 50 × 0.68 = 34 m/s. (b) period T = 1 ÷ frequency = 1 ÷ 50 = 0.020 s. (c) The speed is fixed by the medium, so doubling the frequency halves the wavelength: new wavelength = 34 ÷ 100 = 0.34 m. Marking: stating v = f times lambda earns 1 mark; correct substitution and unit 1 mark; reciprocal relation in (b) 2 marks; stating that the speed is unchanged 1 mark; final value with unit 1 mark.
2	Water waves travel from deep water into shallow water. The speed in deep water is 0.60 m/s and the source frequency is 3.0 Hz. (a) Find the wavelength in deep water; (b) find the wavelength in shallow water where the speed is 0.36 m/s; (c) explain why the frequency is the same in both regions.	Answer: (a) wavelength = 0.60 ÷ 3.0 = 0.20 m. (b) wavelength = 0.36 ÷ 3.0 = 0.12 m. (c) The frequency is set by the source. At the boundary the same number of crests passes through every second, so the frequency cannot change; only the speed and hence the wavelength change. Marking: (a) 2 marks; (b) 2 marks; (c) 1 mark for frequency fixed by the source, 1 mark for the continuous passage of crests at the boundary.
3	A sound wave in air has frequency 440 Hz and speed 330 m/s. (a) Find its wavelength in air; (b) find its wavelength in water where the speed is 1500 m/s; (c) state which of frequency, wavelength and speed stays unchanged when the wave enters water.	Answer: (a) wavelength = 330 ÷ 440 = 0.75 m. (b) wavelength = 1500 ÷ 440 = 3.41 m (three significant figures). (c) The frequency stays unchanged; both the speed and the wavelength increase. Marking: (a) 2 marks; (b) 2 marks; (c) 1 mark for frequency only, 1 mark for noting that speed and wavelength change together.
4	A transverse wave on a string has wavelength 0.20 m. Points P and Q on the string are 0.30 m apart and vibrate in phase. The wave speed is 4.0 m/s. Find (a) the phase difference between P and Q in terms of pi; (b) the frequency; (c) the shortest time for P to be in phase with Q again.	Answer: (a) phase difference = 2 pi × (0.30 ÷ 0.20) = 3 pi radians. (b) frequency = 4.0 ÷ 0.20 = 20 Hz. (c) period = 1 ÷ 20 = 0.050 s; the in-phase state repeats once per period, so the shortest time is 0.050 s. Marking: (a) 2 marks; (b) 2 marks; (c) 1 mark for the period, 1 mark for the periodic argument.

Section B — Refraction, Total Internal Reflection and Lenses

State the law used and check whether your answer is reasonable.

#	Question	Answer and marking notes
5	A ray of light travels from glass of refractive index 1.50 towards air. (a) Find the critical angle at the glass-air boundary; (b) decide whether total internal reflection occurs at an angle of incidence of 45 degrees, giving a reason; (c) state whether the critical angle becomes larger or smaller if the glass is placed in water of refractive index 1.33.	Answer: (a) sin C = 1 ÷ 1.50 = 0.667, so C = 41.8 degrees. (b) Yes. The angle of incidence 45 degrees is greater than the critical angle 41.8 degrees, so no light refracts out and all of it is reflected back into the glass. (c) It becomes larger, because the refractive index gap between glass and water is smaller than that between glass and air: sin C = 1.33 ÷ 1.50 = 0.887, giving C = 62.5 degrees. Marking: (a) formula and value 2 marks; (b) decision 1 mark and reason 1 mark; (c) direction 2 marks and value 2 marks.
6	A ray of light enters water of refractive index 1.33 from air at an angle of incidence of 45 degrees. (a) Find the angle of refraction; (b) the same ray then travels from water to air at an angle of incidence of 50 degrees: find the angle of refraction or explain what happens; (c) name the law used.	Answer: (a) 1.00 × sin 45 = 1.33 × sin r, so sin r = 0.707 ÷ 1.33 = 0.532 and r = 32.1 degrees. (b) For the water-air boundary, sin C = 1 ÷ 1.33 = 0.752, so C = 48.8 degrees. The angle of incidence 50 degrees exceeds the critical angle, so total internal reflection occurs and no light refracts into the air. (c) Snell law, that is n1 sin theta1 = n2 sin theta2. Marking: (a) formula 2 marks and value 3 marks; (b) critical angle 3 marks and conclusion 1 mark; (c) 1 mark.
7	An object is placed 15.0 cm in front of a convex lens of focal length 10.0 cm. (a) Use the lens formula to find the image distance; (b) find the magnification; (c) describe the image (real or virtual, upright or inverted, magnified or diminished).	Answer: (a) 1 ÷ f = 1 ÷ u + 1 ÷ v, so 1 ÷ 10.0 = 1 ÷ 15.0 + 1 ÷ v and 1 ÷ v = 0.100 − 0.0667 = 0.0333, giving v = 30.0 cm on the far side of the lens. (b) magnification = v ÷ u = 30.0 ÷ 15.0 = 2.0. (c) The image is real, inverted and magnified twice. Marking: (a) formula 2 marks, value and position 3 marks; (b) 3 marks; (c) 1 mark for each of the three properties.
8	With the same convex lens of focal length 10.0 cm, the object is moved to 30.0 cm in front of the lens. (a) Find the image distance and the magnification; (b) describe the image; (c) compare with the previous question and state how the image distance and magnification change as the object distance increases.	Answer: (a) 1 ÷ v = 1 ÷ 10.0 − 1 ÷ 30.0 = 0.0667, so v = 15.0 cm and magnification = 15.0 ÷ 30.0 = 0.50. (b) The image is real, inverted and half the size of the object. (c) When the object distance rises from 15.0 cm to 30.0 cm, the image distance falls from 30.0 cm to 15.0 cm and the magnification falls from 2.0 to 0.50; the further the object, the closer the image to the focal point and the smaller it is. Marking: (a) 4 marks; (b) 2 marks; (c) 2 marks.

Section C — Experiments and Applications

State the principle used. Give numerical answers to three significant figures.

#	Question	Answer and marking notes
9	In a two-slit interference experiment the slit separation is 0.50 mm, the screen is 1.50 m from the slits and the spacing between adjacent bright fringes is 1.80 mm. (a) Find the wavelength of the monochromatic light in nm; (b) find the new fringe spacing if a 450 nm source is used; (c) explain why the fringe spacing depends on the wavelength.	Answer: (a) fringe spacing x = lambda D ÷ d, so lambda = x d ÷ D = (1.80 × 10^-3 m)(0.50 × 10^-3 m) ÷ 1.50 m = 6.00 × 10^-7 m = 600 nm. (b) x = (450 × 10^-9 m)(1.50 m) ÷ (0.50 × 10^-3 m) = 1.35 × 10^-3 m = 1.35 mm. (c) The fringe spacing is directly proportional to the wavelength: a shorter wavelength gives a smaller angle for the same path difference, so the bright fringes are packed closer together. Marking: (a) formula 2 marks, substitution 3 marks, answer 2 marks; (b) 2 marks; (c) 1 mark.
10	The speed of sound is measured with a resonance tube. A 500 Hz tuning fork gives a first resonance at an air-column length of 0.16 m and a second resonance at 0.50 m. (a) Find the wavelength; (b) find the speed of sound; (c) find the end correction of the tube.	Answer: (a) The length difference between successive resonances is half a wavelength, so lambda = 2 × (0.50 − 0.16) = 0.68 m. (b) speed = frequency × wavelength = 500 × 0.68 = 340 m/s. (c) At the first resonance the column is about a quarter wavelength minus the end correction e: lambda ÷ 4 = 0.17 m, so e = 0.17 − 0.16 = 0.01 m, that is 1.0 cm. Marking: (a) 2 marks for the half-wavelength argument and 1 mark for the value; (b) 3 marks; (c) 2 marks for the formula and 2 marks for the value.
11	A beam of white light is dispersed into a spectrum by a triangular prism. The refractive index of the glass is 1.52 for violet light and 1.50 for red light. (a) Explain why red light is deviated least; (b) find the critical angle at the glass-air boundary for violet and for red light, and the difference between them; (c) explain why a prism is used instead of a parallel-sided glass block.	Answer: (a) The refractive index increases with frequency. Red light has the lowest frequency and therefore the smallest refractive index, so it is deviated least. (b) Violet: sin C = 1 ÷ 1.52 = 0.658, so C = 41.1 degrees. Red: sin C = 1 ÷ 1.50 = 0.667, so C = 41.8 degrees. The difference is about 0.7 degrees. (c) The two faces of a parallel-sided block are parallel, so the two refractions cancel and the emergent ray is parallel to the incident ray; the colours are only displaced, not separated. The faces of a prism are not parallel, so different colours are deviated by different amounts and a spectrum is formed. Marking: (a) 3 marks; (b) 2 marks each and 1 mark for the difference; (c) 2 marks for the parallel-block argument and 2 marks for the prism.
12	A ship measures the depth of the sea bed with ultrasound of frequency 40 kHz. The speed of sound in sea water is 1500 m/s and the echo returns 0.024 s after transmission. (a) Find the depth of the sea bed; (b) find the wavelength in water; (c) explain why ultrasound rather than audible sound is used for ranging.	Answer: (a) The pulse travels to the sea bed and back, so depth = speed × time ÷ 2 = 1500 × 0.024 ÷ 2 = 18.0 m. (b) wavelength = 1500 ÷ 40000 = 0.0375 m, that is 3.75 cm. (c) Ultrasound has a high frequency and a short wavelength, so the beam spreads very little, stays narrow and carries its energy in a well defined direction; it is also inaudible, so it does not disturb the crew. Marking: (a) 3 marks for dividing by two and 2 marks for the value; (b) 2 marks for the formula and 1 mark for the value; (c) 1 mark for direction and short wavelength, 1 mark for avoiding disturbance.

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