S5 Physics Paper: Waves, Sound And Optics (100 Marks)

S5 Physics Paper: Waves, Sound and Optics (100 marks)

2026-11-11

S5 Physics – waves, sound and optics

S5 Physics paper: sound and stationary waves, refraction and critical angle, lenses, the Doppler effect, double slit interference, gratings and EM waves, with solutions.

Download　Question paper PDF　|　Answer key PDF

Paper details　S5 Physics waves sound and optics practice paper with full solutions

Question 1 Sound waves and stationary waves (18 marks)

A tuning fork vibrates in air at 680 Hz. The speed of sound in the air is 340 m/s. A pipe open at both ends resonates with the fork at its fundamental frequency.

#	Question	Answer and marking notes
1	Find the wavelength and the period of the sound wave.	Answer: the wavelength is 340 divided by 680, which gives 0.50 m; the period is 1 divided by 680, which gives 0.00147 s, about 1.5 ms. (Marking: 2 for the wavelength method, 1 for the value, 2 for the period.)
2	Find the shortest length of the open pipe.	Answer: for a pipe open at both ends the fundamental length equals half a wavelength, so 0.50 divided by 2 gives 0.25 m. (Marking: 2 for the half wavelength condition, 2 for the value.)
3	If one end of the pipe is closed with the length unchanged, how does the fundamental frequency change?	Answer: a pipe closed at one end has a fundamental length of a quarter wavelength, so with the same length the wavelength doubles and the frequency halves to 340 Hz. (Marking: 2 for the quarter wavelength condition, 2 for halving the frequency.)
4	A second fork of 684 Hz sounds together with the first fork. Find the beat frequency heard.	Answer: the beat frequency is 684 minus 680, which gives 4 Hz, so four loudness cycles are heard each second. (Marking: 3 for the subtraction, 2 for the unit and the meaning.)

Question 2 Refraction and total internal reflection (18 marks)

A beam of monochromatic light enters a glass block of refractive index 1.50 from air at an angle of incidence of 30 degrees. The speed of light in a vacuum is 3.0 times 10 to the power 8 m/s and the refractive index of water is 1.33.

#	Question	Answer and marking notes
5	Find the angle of refraction in the glass.	Answer: by Snell law, sin r equals sin 30 degrees divided by 1.50, which gives 0.333, so r is 19.5 degrees. (Marking: 2 for the law, 1 for substitution, 2 for the answer.)
6	Find the speed of light in the glass.	Answer: v equals 3.0 times 10 to the power 8 divided by 1.50, which gives 2.0 times 10 to the power 8 m/s. (Marking: 2 for the formula, 2 for the value.)
7	Find the critical angle for light travelling from the glass to air.	Answer: sin c equals 1 divided by 1.50, which gives 0.667, so c is 41.8 degrees. (Marking: 2 for the condition, 1 for the calculation, 2 for the answer.)
8	Compare the critical angles of water and glass and explain the difference.	Answer: the critical angle of water is 48.8 degrees, larger than 41.8 degrees for glass, because water has a smaller refractive index, so light bends less on leaving the water and a larger angle of incidence is needed to reach 90 degrees of refraction. (Marking: 2 for the comparison, 2 for the explanation.)

Question 3 Images formed by a convex lens (16 marks)

A convex lens has a focal length of 12 cm. Three objects of height 2.0 cm are placed 18 cm, 24 cm and 6 cm in front of the lens.

#	Question	Answer and marking notes
9	For an object distance of 18 cm, find the image distance and the magnification.	Answer: from the lens formula, 1 over v equals 1 over 12 minus 1 over 18, so v is 36 cm; the magnification is 36 divided by 18, which gives 2, and the image is real, inverted and magnified. (Marking: 2 for the formula, 1 for the image distance, 1 for the magnification.)
10	For an object distance of 24 cm, find the image distance and the magnification.	Answer: 1 over v equals 1 over 12 minus 1 over 24, so v is 24 cm; the magnification is 1 and the image is real, inverted and the same size as the object. (Marking: 2 for the image distance, 2 for the magnification and its nature.)
11	For an object distance of 6 cm, find the image distance and describe the image.	Answer: 1 over v equals 1 over 12 minus 1 over 6, giving v equal to minus 12 cm, and the negative sign means a virtual image; the magnification is 12 divided by 6, which gives 2, so the image is virtual, upright and magnified on the same side as the object. (Marking: 2 for the distance, 1 for virtual, 1 for magnified upright.)
12	Explain why the lens acts as a magnifying glass when the object distance is less than the focal length, and state one limitation.	Answer: when the object distance is less than the focal length the image is virtual, and a virtual image cannot be caught on a screen but can be seen by the eye on the other side of the lens, and its angular size is larger, so it acts as a magnifier; one limitation is that the magnification is limited by the focal length and the near point, and too much magnification makes the image blurred. (Marking: 2 for virtual image, 1 for magnification, 1 for the limitation.)

Question 4 The Doppler effect (16 marks)

A car travels along a straight road at 30 m/s and its horn emits sound of frequency 500 Hz. A stationary observer stands at the roadside and the speed of sound is 340 m/s.

#	Question	Answer and marking notes
13	Find the frequency heard by the observer as the car approaches.	Answer: f equals 500 times 340 divided by the quantity 340 minus 30, which gives 548 Hz, higher than 500 Hz. (Marking: 2 for the formula, 2 for the substitution and answer.)
14	Find the frequency heard as the car moves away.	Answer: f equals 500 times 340 divided by the quantity 340 plus 30, which gives 459 Hz, lower than 500 Hz. (Marking: 2 for using the plus sign, 2 for the answer.)
15	Find the difference between the two frequencies heard.	Answer: 548 minus 459 gives 89 Hz, so the frequency drops by about 89 Hz as the car passes. (Marking: 2 for the subtraction, 1 for the value with unit, 1 for the direction.)
16	Find the wavelength of the sound in air and explain why the frequency heard changes while the wavelength in air does not.	Answer: the wavelength is 340 divided by 500, which gives 0.68 m; as the source moves the wavefronts are compressed in the direction of travel, but the speed of the wavefronts relative to the air and their spacing are set by the medium, so the wavelength stays the same and what changes is the number of wavefronts reaching the observer each second, that is the frequency. (Marking: 2 for the wavelength, 1 for compression, 1 for the definition of frequency.)

Question 5 Double slit interference and a grating (16 marks)

In a double slit experiment the slit separation is 0.50 mm and the screen is 2.0 m from the slits. The monochromatic light has a wavelength of 600 nm. A grating has 500 lines per millimetre.

#	Question	Answer and marking notes
17	Find the spacing between adjacent bright fringes on the screen.	Answer: the spacing is 600 times 10 to the power minus 9, times 2.0, divided by 0.50 times 10 to the power minus 3, which gives 2.4 times 10 to the power minus 3 m, that is 2.4 mm. (Marking: 2 for the formula, 1 for unit conversion, 1 for the answer.)
18	If light of wavelength 400 nm is used, find the new fringe spacing.	Answer: the spacing is proportional to the wavelength, so the new spacing is 2.4 times 400 divided by 600, which gives 1.6 mm, smaller than before. (Marking: 2 for the proportionality, 2 for the value.)
19	Find the first order diffraction angle of the grating.	Answer: the grating spacing d is 1 divided by 500000, which gives 2.0 times 10 to the power minus 6 m; sin of the angle is 600 times 10 to the power minus 9 divided by 2.0 times 10 to the power minus 6, which gives 0.30, so the first order angle is 17.5 degrees. (Marking: 1 for d, 2 for the formula, 1 for the angle.)
20	Find the highest observable order of diffraction for this grating.	Answer: sin of the angle cannot exceed 1, so the highest order is the integer part of 1 divided by 0.30, which is 3; the fourth order would need sin greater than 1 and cannot occur. (Marking: 2 for the condition, 1 for the order, 1 for the explanation.)

Question 6 The electromagnetic spectrum (16 marks)

A microwave beam has a wavelength of 3.0 cm. The speed of light in a vacuum is 3.0 times 10 to the power 8 m/s and the Planck constant is 6.63 times 10 to the power minus 34 J s. Visible light ranges from about 400 nm to 700 nm.

#	Question	Answer and marking notes
21	Find the frequency of the microwave.	Answer: f equals 3.0 times 10 to the power 8 divided by 0.030, which gives 1.0 times 10 to the power 10 Hz, that is 10 GHz. (Marking: 2 for the formula, 2 for the conversion and answer.)
22	Find the energy of one microwave photon.	Answer: E equals h f, that is 6.63 times 10 to the power minus 34 multiplied by 1.0 times 10 to the power 10, which gives 6.63 times 10 to the power minus 24 J. (Marking: 2 for the formula, 2 for the value.)
23	Find the frequency of visible light of wavelength 500 nm.	Answer: f equals 3.0 times 10 to the power 8 divided by 500 times 10 to the power minus 9, which gives 6.0 times 10 to the power 14 Hz. (Marking: 1 for the conversion, 1 for the formula, 2 for the answer.)
24	Compare the photon energies of the microwave and the visible light, and give one everyday use of each.	Answer: the photon energy of 500 nm light is 6.63 times 10 to the power minus 34 multiplied by 6.0 times 10 to the power 14, which gives 3.98 times 10 to the power minus 19 J, about sixty thousand times that of a microwave photon, because energy is proportional to frequency. Microwaves are used for heating food and for wireless communication, and visible light for lighting and photography. (Marking: 2 for the comparison, 1 for proportionality, 0.5 for each use.)

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.