IGCSE Physics Paper 4 (Theory / Extended)

Topic 3: Waves -- Mock Exam 1

Duration 1 hour 15 minutes

Total Marks 80

Questions 7

Time Remaining 75:00

Syllabus 0625 / 0972

Instructions

Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
Where required, take the speed of light = 3.0 × 10⁸ m/s and the speed of sound in air = 340 m/s unless otherwise stated.
Answer all questions first. When finished, click "Submit Exam" at the bottom.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.

Question Navigation

Question 1 -- General Wave Properties

Total: 12 marks

(a) [2]

Define the following terms:
(i) wavelength
(ii) frequency

Model Answer -- 1(a)

Wavelength is the minimum distance between two points on a wave that are in phase / the distance between two successive crests (or troughs) / the distance for one complete oscillation [1]

Frequency is the number of waves (or oscillations / cycles) passing a point per unit time (per second) [1]

Mark 1 -- correct definition of wavelength

Mark 2 -- correct definition of frequency

(b) [3]

A ripple tank in a physics lab in Bangalore produces waves with a frequency of 8.0 Hz and a wavelength of 0.025 m.
Calculate the speed of the waves.

Model Answer -- 1(b)

v = f × λ

Correct formula stated [1]

v = 8.0 × 0.025

Correct substitution [1]

v = 0.20 m/s

Correct answer with unit: 0.20 m/s [1]

Mark 1 -- correct formula (v = fλ)

Mark 2 -- correct substitution

Mark 3 -- correct answer with unit

Describe the difference between transverse waves and longitudinal waves. Give one example of each type.

Model Answer -- 1(c)

In a transverse wave, the oscillations / vibrations are perpendicular (at right angles) to the direction of energy transfer / wave travel. Example: light waves, water waves, waves on a string [1]

In a longitudinal wave, the oscillations / vibrations are parallel to the direction of energy transfer / wave travel. Example: sound waves [1]

One correct example given for each type [1]

Mark 1 -- correct description of transverse waves

Mark 2 -- correct description of longitudinal waves

Mark 3 -- one correct example for each type

(d) [4]

Describe an experiment using a ripple tank to show that waves change direction when they enter a region of different depth (refraction). Include a labelled diagram.

Model Answer -- 1(d)

Place a flat glass plate / perspex sheet in the ripple tank to create a region of shallower water [1]

Use a straight vibrating bar / dipper to generate plane (straight) wavefronts directed at an angle towards the boundary between deep and shallow water [1]

Observe that the waves change direction (refract / bend) as they cross the boundary from deep to shallow water. The wavefronts bend towards the normal because the waves travel more slowly in shallower water [1]

The wavelength decreases in the shallow region (wavefronts closer together) while the frequency remains the same. A lamp above the tank can project the wave pattern onto a screen below for clearer observation [1]

Mark 1 -- glass plate to create shallow region

Mark 2 -- straight bar to generate plane waves at angle

Mark 3 -- waves change direction at boundary

Mark 4 -- wavelength decreases / frequency constant / observation method

Question 2 -- Reflection and Refraction of Light

Total: 12 marks

(a) [2]

State the law of reflection.

Model Answer -- 2(a)

The angle of incidence is equal to the angle of reflection [1]

The incident ray, reflected ray and the normal all lie in the same plane [1]

Mark 1 -- angle of incidence = angle of reflection

Mark 2 -- all rays in the same plane

(b) [3]

A ray of light hits a plane mirror at 35° to the mirror surface.
Draw a diagram and find the angle of reflection.

Model Answer -- 2(b)

The angle to the mirror surface is 35°, so the angle of incidence (measured from the normal) = 90° − 35° = 55° [1]

By the law of reflection, the angle of reflection = the angle of incidence = 55° [1]

Correct ray diagram showing incident ray, normal, reflected ray with angles correctly labelled [1]

Mark 1 -- angle of incidence = 55° (correctly converted from surface angle)

Mark 2 -- angle of reflection = 55°

Mark 3 -- correct labelled diagram

A light ray passes from air into a glass block. The angle of incidence is 45° and the angle of refraction is 28°.
Calculate the refractive index of the glass.
Use: n = sin i / sin r

Model Answer -- 2(c)

n = sin i / sin r

Correct formula stated [1]

n = sin 45° / sin 28°

Correct substitution of angles [1]

n = 0.7071 / 0.4695

Correct evaluation of sines [1]

n = 1.51 (to 3 s.f.)

Correct answer: n = 1.51 (accept 1.5 to 1.51) [1]

Mark 1 -- correct formula stated

Mark 2 -- correct substitution

Mark 3 -- correct evaluation of sines

Mark 4 -- correct final answer (1.51)

(d) [3]

Explain why a swimming pool appears shallower than it really is, using the concept of refraction.

Model Answer -- 2(d)

Light from the bottom of the pool travels from water (optically denser medium) into air (optically less dense medium) [1]

As the light crosses the boundary, it refracts (bends) away from the normal because it speeds up when entering the less dense medium [1]

When the refracted rays are extended back (traced back in straight lines by the observer's eye/brain), they appear to come from a point that is higher up / closer to the surface than the actual bottom. This makes the pool appear shallower [1]

Mark 1 -- light travels from denser (water) to less dense (air)

Mark 2 -- light bends away from normal at boundary

Mark 3 -- image appears higher / pool looks shallower

Question 3 -- Total Internal Reflection

Total: 11 marks

(a) [2]

Define the critical angle.

Model Answer -- 3(a)

The critical angle is the angle of incidence (in the optically denser medium) [1]

for which the angle of refraction (in the less dense medium) is exactly 90° / the refracted ray travels along the boundary [1]

Mark 1 -- angle of incidence in denser medium

Mark 2 -- angle of refraction = 90° / ray along boundary

(b) [3]

The refractive index of diamond is 2.42.
Calculate the critical angle of diamond.
Use: n = 1 / sin c

Model Answer -- 3(b)

n = 1 / sin c ⇒ sin c = 1 / n

Correct rearrangement of formula [1]

sin c = 1 / 2.42 = 0.4132

Correct substitution and calculation [1]

c = sin⁻¹(0.4132) = 24.4°

Correct answer: c = 24.4° (accept 24° to 24.5°) [1]

Mark 1 -- correct rearrangement of formula

Mark 2 -- correct substitution (sin c = 1/2.42)

Mark 3 -- correct answer (24.4°)

State the two conditions required for total internal reflection to occur.

Model Answer -- 3(c)

The light must be travelling from an optically denser medium towards an optically less dense medium (e.g. from glass to air, or from water to air) [1]

The angle of incidence must be greater than the critical angle [1]

Both conditions clearly and correctly stated [1]

Mark 1 -- light travels from denser to less dense medium

Mark 2 -- angle of incidence greater than critical angle

Mark 3 -- both conditions clearly stated with correct terminology

(d) [3]

Explain how optical fibres use total internal reflection to transmit data in India's broadband network. Include a simple diagram showing TIR inside a fibre.

Model Answer -- 3(d)

An optical fibre has a core made of optically dense glass surrounded by cladding of less dense glass (or lower refractive index material). Light enters one end of the fibre and hits the core-cladding boundary at an angle greater than the critical angle [1]

Total internal reflection occurs at each bounce, so the light is continuously reflected along the length of the fibre without escaping. This allows light (carrying data as digital pulses) to travel long distances with very little signal loss [1]

In India's broadband network, optical fibres carry internet data as pulses of light (infrared) over thousands of kilometres, connecting cities like Bangalore, Mumbai and Delhi at very high speeds [1]

Mark 1 -- core/cladding structure and angle > critical angle

Mark 2 -- TIR keeps light inside / low signal loss

Mark 3 -- practical application / data as light pulses

Question 4 -- Lenses and Ray Diagrams

Total: 12 marks

(a) [2]

Define the focal length of a converging lens.

Model Answer -- 4(a)

The focal length is the distance from the centre of the lens (optical centre) [1]

to the principal focus (focal point), where rays parallel to the principal axis converge after passing through the lens [1]

Mark 1 -- distance from centre of lens / optical centre

Mark 2 -- to principal focus / where parallel rays converge

(b) [4]

Draw a ray diagram for a converging lens where the object is placed between F and 2F. State the properties of the image formed (nature, orientation, size, position).

Model Answer -- 4(b)

Correct ray 1: ray parallel to the principal axis, refracted through F on the other side of the lens [1]

Correct ray 2: ray through the optical centre passes straight through undeviated [1]

Image correctly drawn at the point where the rays converge, beyond 2F on the other side [1]

Properties of image: Real, inverted, magnified (larger than object), formed beyond 2F on the other side of the lens [1]

Mark 1 -- correct ray parallel to axis then through F

Mark 2 -- correct ray through optical centre

Mark 3 -- image drawn at correct position (beyond 2F)

Mark 4 -- image properties: real, inverted, magnified, beyond 2F

A converging lens has a focal length of 10 cm. An object is placed 15 cm from the lens.
Using the ray diagram or otherwise, describe the image formed.

Model Answer -- 4(c)

The object is at 15 cm, which is between F (10 cm) and 2F (20 cm). From the ray diagram rules, the image is formed beyond 2F on the other side of the lens (at 30 cm, using 1/f = 1/u + 1/v: 1/10 = 1/15 + 1/v, so 1/v = 1/10 - 1/15 = 1/30, v = 30 cm) [1]

The image is real and inverted (upside down) [1]

The image is magnified (larger than the object) since it is formed further from the lens than the object (magnification = v/u = 30/15 = 2, so the image is twice the size of the object) [1]

Mark 1 -- image position identified (beyond 2F / at 30 cm)

Mark 2 -- real and inverted

Mark 3 -- magnified (larger than object)

(d) [3]

State two differences between a converging lens and a diverging lens, and give one practical use of each.

Model Answer -- 4(d)

Difference 1 (Shape): A converging lens is thicker in the middle than at the edges (convex). A diverging lens is thinner in the middle than at the edges (concave) [1]

Difference 2 (Effect on light): A converging lens brings parallel rays of light together to a focus (converges them). A diverging lens spreads parallel rays of light apart (diverges them) so they appear to come from a virtual focus [1]

Practical uses: Converging lens -- used in a magnifying glass, camera, or projector. Diverging lens -- used in spectacles for correcting short-sightedness (myopia), or in peepholes in doors [1]

Mark 1 -- correct difference in shape

Mark 2 -- correct difference in effect on light

Mark 3 -- one correct practical use for each lens type

Question 5 -- Electromagnetic Spectrum

Total: 11 marks

(a) [3]

List the seven regions of the electromagnetic spectrum in order of increasing frequency.

Model Answer -- 5(a)

Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays [1]

All seven regions named correctly [1]

Correct order of increasing frequency (lowest to highest) [1]

Mark 1 -- at least 5 regions correctly named

Mark 2 -- all 7 regions correctly named

Mark 3 -- correct order of increasing frequency

(b) [2]

State two properties that are common to all electromagnetic waves.

Model Answer -- 5(b)

All electromagnetic waves travel at the same speed in a vacuum (speed of light, 3.0 × 10⁸ m/s) [1]

All electromagnetic waves are transverse waves / they can travel through a vacuum / they transfer energy [1]

Mark 1 -- travel at speed of light in vacuum

Mark 2 -- transverse / travel through vacuum / transfer energy

For each of the following, name the type of electromagnetic radiation used and explain why it is suitable:

(i) Taking an X-ray image of a broken bone at a hospital in Bangalore.
(ii) A TV remote control.
(iii) Cooking food in a microwave oven.

Model Answer -- 5(c)

(i) X-rays -- X-rays can pass through soft tissue (skin and muscle) but are absorbed by dense materials such as bone. This creates a shadow image on a detector/film, allowing doctors to see fractures [1]

(ii) Infrared -- Infrared radiation is emitted by an LED in the remote. It is suitable because it is safe, low energy, can be directed in a beam towards the TV sensor, and does not travel through walls (so it won't interfere with other devices) [1]

(iii) Microwaves -- Microwaves are absorbed by water and fat molecules in food, causing them to vibrate and heat up. This cooks the food from inside as well as outside [1]

Mark 1 -- X-rays identified with correct explanation

Mark 2 -- infrared identified with correct explanation

Mark 3 -- microwaves identified with correct explanation

(d) [3]

Explain why ultraviolet radiation is dangerous to humans and state one precaution that should be taken.

Model Answer -- 5(d)

Ultraviolet radiation can cause sunburn / damage to the surface cells of the skin [1]

Prolonged or excessive exposure to UV radiation can cause skin cancer and can also damage the eyes (e.g. cataracts) [1]

Precaution: Wear sunscreen with a high SPF / wear protective clothing / wear UV-blocking sunglasses / avoid prolonged exposure to direct sunlight, especially during peak hours [1]

Mark 1 -- causes sunburn / skin cell damage

Mark 2 -- can cause skin cancer / eye damage

Mark 3 -- appropriate precaution stated

Question 6 -- Sound Waves

Total: 12 marks

(a) [2]

State two differences between sound waves and light waves.

Model Answer -- 6(a)

Sound waves are longitudinal waves; light waves are transverse waves [1]

Sound waves require a medium to travel through (cannot travel through a vacuum); light waves can travel through a vacuum [1]

Mark 1 -- longitudinal vs transverse

Mark 2 -- sound needs medium / light can travel through vacuum

(b) [3]

Describe an experiment to show that sound cannot travel through a vacuum.

Model Answer -- 6(b)

Place a ringing electric bell (or buzzer / alarm) inside a glass bell jar connected to a vacuum pump [1]

With air inside, the sound of the bell can be clearly heard. Gradually pump the air out of the bell jar using the vacuum pump [1]

As the air is removed, the sound becomes fainter and fainter. When (nearly) all the air has been removed, the sound can no longer be heard even though the bell is still visibly ringing. This shows that sound requires a medium (air) to travel and cannot travel through a vacuum [1]

Mark 1 -- bell/buzzer inside bell jar / vacuum pump described

Mark 2 -- air pumped out gradually

Mark 3 -- sound fades / disappears, showing sound needs a medium

A student standing 680 m from a cliff claps her hands and hears an echo after 4.0 s.
Calculate the speed of sound. Show your working.

Model Answer -- 6(c)

total distance = 2 × 680 = 1360 m

Sound travels to the cliff and back, so total distance = 2 × 680 = 1360 m [1]

speed = distance / time = 1360 / 4.0

Correct substitution into speed formula [1]

speed = 340 m/s

Correct answer with unit: 340 m/s [1]

Mark 1 -- total distance = 2 × 680 = 1360 m

Mark 2 -- correct substitution (1360 / 4.0)

Mark 3 -- correct answer: 340 m/s

(d) [4]

A fishing boat near the coast of Kerala uses sonar to determine the depth of the sea. An ultrasound pulse is sent downwards and the echo is received after 0.12 s. The speed of sound in seawater is 1500 m/s.

(i) Calculate the depth of the sea at that point.
(ii) Explain why ultrasound is used rather than audible sound.

Model Answer -- 6(d)

total distance = speed × time = 1500 × 0.12 = 180 m

Correct calculation of total distance travelled by the pulse [1]

depth = total distance / 2 = 180 / 2 = 90 m

Correct depth: 90 m (dividing by 2 because sound travels down and back up) [1]

Ultrasound has a frequency above the range of human hearing (above 20 000 Hz) so it does not cause noise pollution / disturbance to those on the boat or to marine life [1]

Ultrasound can be focused into a narrow, directed beam which gives more precise/accurate depth readings. It also reflects well off surfaces and is less likely to spread out compared to lower-frequency audible sound [1]

Mark 1 -- total distance = 180 m

Mark 2 -- depth = 90 m (divided by 2)

Mark 3 -- ultrasound above human hearing / no noise disturbance

Mark 4 -- can be focused / narrow beam / more precise

Question 7 -- Waves Mixed Application

Total: 10 marks

(a) [2]

Explain what is meant by diffraction of waves.

Model Answer -- 7(a)

Diffraction is the spreading out of waves [1]

when they pass through a gap (aperture) or around an obstacle / edge [1]

Mark 1 -- spreading out of waves

Mark 2 -- through a gap / around an obstacle

(b) [3]

Describe how the wavelength of a wave affects the amount of diffraction that occurs when it passes through a gap. Include a simple diagram.

Model Answer -- 7(b)

When the wavelength is much smaller than the gap, there is very little diffraction -- the waves pass through the gap almost in straight lines with minimal spreading [1]

When the wavelength is approximately equal to the width of the gap, maximum diffraction occurs -- the waves spread out significantly into the region behind the barrier, producing circular/semicircular wavefronts [1]

Correct diagram showing two scenarios: little spreading for short wavelength relative to gap, and significant spreading for wavelength similar to gap width [1]

Mark 1 -- little diffraction when wavelength << gap

Mark 2 -- maximum diffraction when wavelength ≈ gap

Mark 3 -- correct diagram showing both scenarios

Explain the difference between analogue and digital signals.

Model Answer -- 7(c)

An analogue signal varies continuously and can take any value within a range (e.g. a smoothly varying voltage) [1]

A digital signal can only take a limited number of discrete values, typically just two states: on/off (or 1 and 0) [1]

Mark 1 -- analogue: continuous / any value

Mark 2 -- digital: discrete values / on-off / 1 and 0

(d) [3]

Explain two advantages of using digital signals rather than analogue signals for transmitting data through India's telecommunications network.

Model Answer -- 7(d)

Advantage 1 -- Noise resistance: Digital signals can be regenerated (cleaned up) during transmission. Because the signal only has two states (0 and 1), any noise picked up along the way can be easily removed. Analogue signals degrade as noise accumulates and cannot be perfectly restored [1]

Advantage 2 -- Higher data capacity: Digital signals can carry much more data in the same bandwidth. Multiple digital signals can be multiplexed (combined) and sent along the same cable or fibre at the same time. This is crucial for India's growing demand for internet, phone calls and streaming across its vast telecommunications network [1]

Clear explanation of why each advantage matters for long-distance data transmission, with reference to practical telecommunications context [1]

Mark 1 -- digital signals can be regenerated / noise removed

Mark 2 -- higher data capacity / multiplexing

Mark 3 -- clear explanation with practical context

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