S4 Physics Paper: Waves, Sound And Optics (Paper 2) (100 Marks)

S4 Physics Paper: Waves, Sound and Optics (Paper 2) (100 marks)

2026-11-11

S4 Physics – waves, sound and optics

S4 physics waves and optics practice paper: wave equation, sound and echoes, sonar, reflection, refraction, critical angle, convex lenses and interference, with full.

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Paper details　S4 physics practice paper on waves, sound and optics: wave speed, CRO frequency measurement, echoes, refraction, total internal reflection and convex lens images.

Section A – Wave properties and wave speed

Answer in full sentences; show the formula, substitution and unit in calculations.

#	Question	Answer and marking notes
1	Define wavelength and frequency, and state the relation between wave speed, frequency and wavelength.	Wavelength is the distance between two neighbouring points in phase, for example two adjacent crests, measured in m. Frequency is the number of complete waves passing a fixed point per second, measured in Hz. The wave advances one wavelength in one period, so v = f x lambda. (3 marks: wavelength 1; frequency 1; relation 1)
2	A water wave has frequency 4.0 Hz and wavelength 1.5 m. Find its speed and period.	v = f x lambda = 4.0 x 1.5 = 6.0 m/s. T = 1 / f = 1 / 4.0 = 0.25 s. (3 marks: speed 2; period 1)
3	In the same medium the frequency is raised to 8.0 Hz. What is the new wavelength? Explain.	In the same medium the speed is fixed by the medium at 6.0 m/s. lambda = v / f = 6.0 / 8.0 = 0.75 m. The frequency doubles so the wavelength halves and the speed stays the same. (3 marks: speed unchanged 1; calculation 1; explanation 1)
4	How far does the wave travel in 3.0 s? Compare the direction of vibration in transverse and longitudinal waves.	Distance = v x t = 6.0 x 3.0 = 18 m. In a transverse wave the particles vibrate perpendicular to the direction of travel, as on a rope; in a longitudinal wave they vibrate parallel to it, as in sound in air. (3 marks: distance 1; transverse 1; longitudinal 1)
5	A student says the wave carries the medium forward with it. Comment on this statement.	The statement is wrong. A wave transfers energy and the state of vibration; the particles of the medium only oscillate about their rest positions and do not move forward with the wave. A float on water, for example, only moves up and down. (3 marks: error identified 1; energy transfer 1; particles stay 1)

Section B – Measuring frequency and wavelength with a CRO

The time base gives the time per division; show how period and frequency are related.

#	Question	Answer and marking notes
6	The time base is 2.0 ms per division and one complete wave spans 4.0 divisions. Find the period and frequency.	T = 2.0 x 4.0 = 8.0 ms = 8.0 x10^-3 s. f = 1 / T = 1 / 0.0080 = 125 Hz. (4 marks: period 2; frequency 2)
7	The signal is a sound wave in air where the speed of sound is 340 m/s. Find the wavelength, and the new wavelength if the frequency doubles.	lambda = v / f = 340 / 125 = 2.72 m. After doubling, lambda = 340 / 250 = 1.36 m, half the original, because the speed is set mainly by the medium. (5 marks: wavelength 3; new value 1; explanation 1)
8	If the time base is changed to 1.0 ms per division while the trace still spans 4.0 divisions, what is the frequency? Why?	T = 1.0 x 4.0 = 4.0 ms and f = 1 / 0.0040 = 250 Hz. The signal is unchanged; only the time scale of the CRO is altered, so the same number of divisions now represents half the time and twice the frequency. (4 marks: period 1; frequency 1; explanation 2)
9	Another signal spans 5.0 divisions at 2.0 ms per division. Find its frequency and suggest how to reduce the measurement error.	T = 2.0 x 5.0 = 10.0 ms and f = 1 / 0.0100 = 100 Hz. Reduce the error by using a time base that makes the trace span more divisions, by timing several periods and averaging, and by repeating the reading with a steady triggered trace. (5 marks: frequency 2; more divisions 1; averaging 1; repeat 1)

Section C – Sound, echoes and the speed of sound

In echo questions state clearly that the sound travels twice the distance.

#	Question	Answer and marking notes
10	A ship sends a sound pulse and receives the echo after 0.80 s. If the speed of sound in water is 1500 m/s, find the depth.	The sound travels to the seabed and back, twice the depth: 2d = v x t = 1500 x 0.80 = 1200 m, so d = 600 m. (5 marks: factor of two 2; calculation 2; unit 1)
11	A person claps 102 m from a wall and hears the echo after 0.60 s. Find the speed of sound in air and explain why no echo is heard on the Moon.	v = 2d / t = (2 x 102) / 0.60 = 340 m/s. Sound is a longitudinal wave and needs a medium to carry the vibration; the Moon has almost no atmosphere, so there is no medium to return the sound and no echo is heard. (5 marks: calculation 3; needs a medium 1; no atmosphere 1)
12	State the usual order of sound speed in solids, liquids and gases, and give one everyday application.	In general sound travels fastest in solids, then liquids, then gases, because the particles are closer together and pass the vibration on more quickly. Example: putting an ear to a rail detects an approaching train earlier because sound travels faster in steel than in air; ultrasound is used for medical scanning. (5 marks: order 2; reason 2; example 1)

Section D – Reflection, refraction and total internal reflection

All angles are measured from the normal; take the refractive index as 1.50.

#	Question	Answer and marking notes
13	Light travels from air into glass with an angle of incidence of 45°. If the refractive index of the glass is 1.50, find the angle of refraction.	Using n = sin i / sin r, sin r = sin 45 / 1.50 = 0.7071 / 1.50 = 0.4714, so r = 28.1 degrees (about 28 degrees). The ray bends towards the normal because the glass is denser. (5 marks: formula 2; substitution 1; result 2)
14	Find the speed of light in this glass and explain why the frequency stays the same while the wavelength changes.	v = c / n = (3.00 x10^8) / 1.50 = 2.00 x10^8 m/s. The frequency is fixed by the source, so the number of waves arriving per second is unchanged at the boundary; the speed falls, so by v = f x lambda the wavelength becomes shorter. (5 marks: speed 2; constant frequency 2; shorter wavelength 1)
15	Find the critical angle at the glass-air boundary and state the two conditions for total internal reflection.	sin c = 1 / n = 1 / 1.50 = 0.6667, so c = 41.8 degrees. Conditions: light must travel from the denser medium (glass) towards the less dense one (air), and the angle of incidence must exceed the critical angle of 41.8 degrees. (5 marks: critical angle 3; two conditions 2)
16	Give one application of total internal reflection and explain why an optical fibre can carry a signal over a long distance.	Applications include optical fibre communication, endoscopes and prism periscopes. Inside the core the light strikes the core-cladding boundary at an angle greater than the critical angle, so it is totally internally reflected again and again with almost no light escaping, so the signal is only weakly attenuated over long distances. (5 marks: application 2; explanation 2; low loss 1)

Section E – Images formed by a convex lens

Use 1/f = 1/u + 1/v; a real image has positive v and a virtual image negative v.

#	Question	Answer and marking notes
17	A convex lens of focal length 10.0 cm has an object 30.0 cm away. Find the image distance and magnification, and describe the image.	1 / v = 1 / f – 1 / u = 1/10.0 – 1/30.0 = 2/30.0, so v = 15.0 cm. m = v / u = 15.0 / 30.0 = 0.50. The image is real, inverted and diminished, 15.0 cm on the far side of the lens. (6 marks: formula 1; image distance 2; magnification 2; nature 1)
18	The object is 4.0 cm tall. Find the image height. If the object distance becomes 5.0 cm, find the image distance, magnification and nature of the image.	Image height = 4.0 x 0.50 = 2.0 cm. At u = 5.0 cm, 1 / v = 1/10.0 – 1/5.0 = -1/10.0, so v = -10.0 cm: a virtual image 10.0 cm on the same side as the object. Magnification = 10.0 / 5.0 = 2.0 and image height = 4.0 x 2.0 = 8.0 cm. The image is virtual, upright and magnified, which is how a magnifying glass works. (6 marks: height 1; distance 2; magnification 1; nature 2)
19	Distinguish between a real image and a virtual image and explain how a screen can tell them apart.	A real image is formed where the actual rays meet, so light really passes through it; a virtual image is formed where the backward extensions of the rays meet and no light actually passes through it. Placing a screen at the image position shows a sharp image only for a real image; a virtual image cannot be caught on a screen and is seen only through the lens. (5 marks: real image 2; virtual image 2; screen test 1)

Section F – Experimental design and error analysis

Identify the source of the error and suggest a practical improvement.

#	Question	Answer and marking notes
20	In a ripple tank experiment 10 waves pass a point in 2.5 s and 5 adjacent crests span 6.0 m. Find the frequency, wavelength and speed.	f = 10 / 2.5 = 4.0 Hz. Five crests contain only four wavelengths, so lambda = 6.0 / 4 = 1.5 m. v = f x lambda = 4.0 x 1.5 = 6.0 m/s. (5 marks: frequency 2; wavelength 2; speed 1)
21	If a student wrongly treats 6.0 m as five wavelengths instead of four, is the speed too high or too low? Quantify the effect.	The wavelength would be taken as 6.0 / 5 = 1.2 m and the speed as 4.0 x 1.2 = 4.8 m/s, which is 20 percent below the true 6.0 m/s. One wavelength lies between adjacent crests, so the number of gaps is always one fewer than the number of crests; this is a systematic error. (5 marks: wrong wavelength 1; wrong speed 1; too low 1; reason 2)
22	Suggest two ways to reduce this error and explain the principle of ultrasonic distance measurement.	First, measure the total distance across many wavelengths, say ten, and divide by the number of gaps to cut the percentage error of a single reading; second, use a stroboscope or record the tank and measure crest positions frame by frame, repeating and averaging. Ultrasonic ranging sends a short pulse and times the echo t, then distance = v x t / 2 where v is the speed of sound in air. (5 marks: two methods 2; ranging principle 3)

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