Secondary 4 Physics Practice Paper: Waves, Reflection and Refraction, Convex Lenses, Sound, Interference and Beats

2026-11-13

Physics – Secondary 4 – Practice Paper

Waves, reflection and refraction, critical angle, convex lenses, sound, interference and beats; full answers and marking scheme.

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Paper details　Secondary 4 physics practice paper with full answers: wave speed and period, refractive index and critical angle, convex lens image formation, echo and ultrasound, two source interference and beat frequency.

Section A: Basic Wave Relationships (16 marks)

Data 1: a transverse wave on a string has frequency 50 Hz and wavelength 0.80 m; water waves in a ripple tank have frequency 12 Hz and wavelength 2.5 cm. Speed of sound in air is 340 m/s.

#	Question	Answer and marking notes
1	(a) Find the speed and the period of the string wave.	Answer: speed = 50 x 0.80 = 40 m/s; period = 1 / 50 = 0.020 s. (Marking: wave speed 2 marks, period 2 marks)
2	(b) Find the speed of the water waves and the time for the wave train to travel 1.5 m.	Answer: speed = 12 x 0.025 = 0.30 m/s; time = 1.5 / 0.30 = 5.0 s. (Marking: speed 2 marks, time 2 marks)
3	(c) Find the wavelengths in air of sound of 500 Hz and 17000 Hz, giving your answers in centimetres.	Answer: 340 / 500 = 0.68 m (68 cm); 340 / 17000 = 0.020 m (2.0 cm). (Marking: 2 marks each)
4	(d) A string wave travels 8.0 m along the string in 2.0 s and has frequency 5 Hz. Find its speed and wavelength.	Answer: speed = 8.0 / 2.0 = 4.0 m/s; wavelength = 4.0 / 5 = 0.80 m. (Marking: speed 2 marks, wavelength 2 marks)

Section B: Reflection and Refraction of Light (16 marks)

Data 2: a ray of light enters glass of refractive index 1.50 from air at an angle of incidence of 40 degrees. Speed of light in vacuum is 3.0 x 10^8 m/s.

#	Question	Answer and marking notes
5	(a) Find the angle of refraction in the glass, to one decimal place.	Answer: sin r = sin 40 / 1.50 = 0.4285, so r = 25.4 degrees. (Marking: formula 2 marks, value 2 marks)
6	(b) Find the critical angle at the glass air boundary.	Answer: sin c = 1 / 1.50 = 0.667, so c = 41.8 degrees. (Marking: formula 2 marks, value 2 marks)
7	(c) Find the speed of light in the glass and the wavelength in glass of light of wavelength 600 nm in air.	Answer: speed = 3.0 x 10^8 / 1.50 = 2.0 x 10^8 m/s; wavelength = 600 / 1.50 = 400 nm. (Marking: speed 2 marks, wavelength 2 marks)
8	(d) If light inside the glass meets the glass air boundary at 45 degrees, will it leave the glass? Justify with a calculation.	Answer: total internal reflection occurs and no light refracts out. Since 45 degrees exceeds the critical angle of 41.8 degrees, all the light is reflected back into the glass. (Marking: comparison 2 marks, conclusion 2 marks)

Section C: Image Formation by a Convex Lens (16 marks)

Data 3: a convex lens has focal length 10.0 cm. An object is placed 30.0 cm, 15.0 cm and 6.0 cm in front of the lens.

#	Question	Answer and marking notes
9	(a) For object distance 30.0 cm, find the image distance and magnification, and state the nature of the image.	Answer: 1/v = 1/10 – 1/30 = 1/15, so v = 15.0 cm; magnification = 15.0 / 30.0 = 0.50. The image is real, inverted and diminished. (Marking: image distance 2 marks, magnification and nature 2 marks)
10	(b) For object distance 15.0 cm, find the image distance and magnification.	Answer: 1/v = 1/10 – 1/15 = 1/30, so v = 30.0 cm; magnification = 30.0 / 15.0 = 2.0, a real inverted magnified image. (Marking: image distance 2 marks, magnification 2 marks)
11	(c) For object distance 6.0 cm, find the image distance and magnification, and state the nature of the image.	Answer: 1/v = 1/10 – 1/6 = -1/15, so v = -15.0 cm (negative means the image is on the same side as the object); magnification = 15.0 / 6.0 = 2.5. The image is virtual, upright and magnified. (Marking: image distance and sign 2 marks, magnification and nature 2 marks)
12	(d) What condition on object distance gives a magnified virtual image? Explain with the data in this question.	Answer: the object distance must be less than the focal length, that is u less than 10.0 cm. In part (c) u = 6.0 cm satisfies this and gives a virtual image magnified 2.5 times, while 15.0 cm and 30.0 cm exceed the focal length and give real images. (Marking: condition 2 marks, explanation 2 marks)

Section D: Sound (16 marks)

Data 4: speed of sound in air is 340 m/s. An ultrasound probe has frequency 40 kHz. In a resonance tube the first resonance occurs with an air column of 0.17 m and a fork of 500 Hz.

#	Question	Answer and marking notes
13	(a) A sound is sent towards a cliff and the echo is heard 1.6 s later. Find the distance to the cliff.	Answer: distance = 340 x 1.6 / 2 = 272 m. (Marking: multiplying by time 2 marks, halving 2 marks)
14	(b) Find the wavelength of this ultrasound in air, in millimetres.	Answer: wavelength = 340 / 40000 = 0.0085 m = 8.5 mm. (Marking: frequency in Hz 2 marks, wavelength 2 marks)
15	(c) Two identical sources each produce a sound level of 70 dB. Find the sound level when both sound together and explain.	Answer: 73 dB. The intensities add, so the total intensity doubles and the level rises by 10 log 2, about 3 dB. (Marking: value 2 marks, explanation 2 marks)
16	(d) At the first resonance the air column is 0.17 m long. Find the wavelength and the speed of sound obtained from the data.	Answer: the first resonance has an air column of one quarter wavelength, so wavelength = 4 x 0.17 = 0.68 m; speed = 500 x 0.68 = 340 m/s. (Marking: wavelength 2 marks, speed 2 marks)

Section E: Interference and Beats (18 marks)

Data 5: two in phase loudspeakers emit sound of wavelength 0.50 m; the path difference from the two speakers to point P is 1.25 m. Two tuning forks of 256 Hz and 260 Hz are also available.

#	Question	Answer and marking notes
17	(a) Decide whether point P shows constructive or destructive interference, showing your working.	Answer: 1.25 / 0.50 = 2.5, a path difference of 2.5 wavelengths (a half integer multiple), so the interference is destructive and the sound at P is near a minimum. (Marking: ratio 2 marks, conclusion 2 marks)
18	(b) Moving outward from the central maximum, what path difference gives the first constructive line?	Answer: path difference = 1 wavelength = 0.50 m. (Marking: value 2 marks, unit 2 marks)
19	(c) Find the beat frequency when both forks sound together. If wax is added to the 260 Hz fork so that its frequency becomes 258 Hz, how does the beat frequency change?	Answer: beat frequency = 260 – 256 = 4 Hz. After waxing, the beat frequency is 258 – 256 = 2 Hz, so the number of beats heard per second falls from 4 to 2. (Marking: original 2 marks, new value 1 mark, explanation 2 marks)
20	(d) Arrange radio waves, infrared, visible light, ultraviolet, X rays and gamma rays in order of decreasing wavelength, and state how their speeds in vacuum compare.	Answer: radio, infrared, visible, ultraviolet, X rays, gamma rays (longest to shortest wavelength). All are electromagnetic waves with the same speed in vacuum, 3.0 x 10^8 m/s, while frequency increases along the list. (Marking: order 3 marks, same speed 1 mark, frequency trend 1 mark)

Section F: Experimental Data and Errors (18 marks)

Data 6: a convex lens experiment gives (object distance u, image distance v) as (30.0 cm, 15.0 cm), (15.0 cm, 30.0 cm) and (20.0 cm, 20.0 cm). Wave speeds come from three pairs (frequency, wavelength): (5.0 Hz, 0.80 m), (4.0 Hz, 1.00 m), (8.0 Hz, 0.50 m).

#	Question	Answer and marking notes
21	(a) Use each data pair to find the focal length, then find the mean focal length.	Answer: the three focal lengths are 10.0 cm, 10.0 cm and 10.0 cm from 1/f = 1/u + 1/v, so the mean focal length is 10.0 cm. (Marking: 1 mark per pair, 1 mark for the mean)
22	(b) Find the wave speed from each of the three pairs and comment on the consistency of the data.	Answer: every pair gives 4.0 m/s (0.80 x 5.0, 1.00 x 4.0, 0.50 x 8.0). The three results agree exactly, so the data are consistent and the speed is independent of frequency, as v = f x wavelength requires. (Marking: three speeds 2 marks, comment 2 marks)
23	(c) If a wavelength is measured as 25.0 cm with an uncertainty of plus or minus 0.5 cm from the smallest scale division, find the percentage error.	Answer: percentage error = 0.5 / 25.0 x 100 per cent = 2.0 per cent. (Marking: formula 3 marks, value 2 marks)
24	(d) State two sources of error in these experiments and one improvement for each.	Answer: first, parallax when reading the scale; keep the eye level with the scale and use a set square. Second, the lens and screen may not share one principal axis, blurring the image; mount them on a bench and align them first. Third, a single reading carries random error; repeat and average. (Marking: 1 mark per source, 1 mark per improvement, 1 mark for clarity)

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