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]

⚠ If you missed marks here: "The distance of one wave" is too loose to earn the wavelength mark — name two points in phase, or two successive crests. For frequency, "how fast the wave moves" describes speed and scores nothing; the mark needs waves (or oscillations) passing a point per second.

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]

⚠ If you missed marks here: Dividing the wrong way round, 8.0 / 0.025, gives 320 m/s instead of 0.20 m/s — a ripple-tank answer should come out in tens of centimetres per second, not hundreds of metres. The unit carries the third mark, so 0.20 written on its own does not score it.

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]

⚠ If you missed marks here: "Transverse waves move up and down and longitudinal waves move side to side" earns nothing, because neither is compared with the direction of energy transfer — you need perpendicular and parallel to the direction the wave travels. Sound is the safe longitudinal example; offering sound as transverse, or light as longitudinal, loses the example mark.

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]

⚠ If you missed marks here: Describing a ray of light through a glass block answers a different question — these marks are for a ripple tank, so the glass plate that creates the shallow region has to be named. Saying "the frequency drops in the shallow water" is wrong: the frequency is fixed by the dipper, and it is the wavelength that shortens.

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]

⚠ If you missed marks here: Most students write only "angle of incidence = angle of reflection" and stop, throwing away the second mark, which is for the incident ray, reflected ray and normal all lying in the same plane. Both angles must be understood as measured from the normal, never from the mirror surface.

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)

Your diagram should look like this:

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]

⚠ If you missed marks here: Answering 35° is the standard trap: 35° is measured from the mirror, and the law of reflection uses angles from the normal, so i = 90° − 35° = 55°. A diagram drawn without a normal also loses the third mark, even when the arithmetic is right.

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]

⚠ If you missed marks here: Turning the fraction upside down gives sin 28° / sin 45° = 0.66, and a refractive index below 1 is impossible for glass — that is your check. An answer near 3.1 means the calculator is in radian mode; switch it to degrees before going any further.

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]

⚠ If you missed marks here: "Because the light bends" is not enough for any of the three marks — you must say which way. Travelling from water into air the ray bends away from the normal; writing "towards the normal" describes light entering the water instead. The last mark needs the eye tracing the emerging rays back in straight lines to a point higher than the real floor.

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]

⚠ If you missed marks here: "The angle where total internal reflection starts" describes the effect rather than defining the angle, and scores at most one mark. The examiner wants the angle of incidence in the denser medium that produces an angle of refraction of exactly 90°, with the refracted ray grazing along the boundary.

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]

⚠ If you missed marks here: Stopping at sin c = 0.4132 and offering that as the answer is the usual loss — you still owe the inverse sine that turns it into 24.4°. Working out sin⁻¹(2.42) instead gives a calculator error, which is the signal that you used n = 1 / sin c without rearranging it first.

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]

⚠ If you missed marks here: Reversing the media is fatal: light going from air into glass can never be totally internally reflected, however large the angle. "Greater than or equal to the critical angle" is also marked wrong — at exactly c the light grazes along the boundary, so the condition is strictly greater than.

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]

⚠ If you missed marks here: "The light bounces off the inside of the fibre" reads as ordinary reflection and misses the first mark — name the dense core and the less dense cladding, and say the light meets that boundary above the critical angle. A ray drawn straight down the middle of the fibre shows no total internal reflection at all.

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]

⚠ If you missed marks here: Defining focal length as "the distance from the lens to the image" loses both marks, because the image moves whenever the object moves while the focal length does not. The measurement runs from the optical centre to the principal focus, the point where rays that arrived parallel to the principal axis are brought together.

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)

Your diagram should look like this:

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]

⚠ If you missed marks here: Rays change direction once, at the vertical lens line — drawing them curving through the glass shape costs the ray marks. An upright image is a contradiction here: two real rays crossing on the far side always give a real, inverted image, and with the object between F and 2F it comes out magnified and beyond 2F. "Diminished" belongs to the object-beyond-2F case.

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). Drawing the two standard construction rays — one parallel to the axis that refracts through F, and one straight through the centre of the lens — they meet beyond 2F on the far side of the lens [1]

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

The image is magnified (larger than the object), because the object lies between F and 2F [1]

⚠ If you missed marks here: Do not reach for a lens formula — 0625 does not use one, and algebra earns nothing here. Compare 15 cm with f = 10 cm and 2f = 20 cm: the object sits between F and 2F, so the image is real, inverted and magnified. "Virtual and upright" is what you get by wrongly treating 15 cm as being inside the focal length.

Mark 1 -- image position identified as beyond 2F on the far side

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]

⚠ If you missed marks here: "One is convex and one is concave" restates the names without describing the shape, so it scores nothing — say thicker in the middle against thinner in the middle. The last mark goes if you offer a magnifying glass for both; a diverging lens corrects short sight (myopia), not long sight.

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]

⚠ If you missed marks here: Listing the spectrum from gamma rays down to radio waves answers the reverse of the question — increasing frequency starts at radio. The regions most often dropped are infrared and ultraviolet, and "light" on its own is not accepted for the visible band. Six correct names out of seven caps you at the first mark.

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]

⚠ If you missed marks here: "They are all waves" followed by "they all carry energy" is one idea stretched over two lines and earns one mark at best. Choose two genuinely separate properties: the same speed of 3.0 × 10⁸ m/s in a vacuum, and being transverse (or able to cross a vacuum). "They all have the same wavelength" is false — wavelength is what separates the regions.

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 the food, making them vibrate faster, so the food heats up. They penetrate a few centimetres into the food, and the heat then spreads inwards by conduction, so the food cooks faster than under a grill [1]

⚠ If you missed marks here: Naming the radiation is half of each mark; the question says "explain why it is suitable", so "X-rays" with no mention of bone absorbing them scores nothing. The reversed version, "X-rays pass through bone but are stopped by flesh", is a common and costly slip. Answering "radio waves" for the remote control is also wrong: the emitter is an infrared LED.

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]

⚠ If you missed marks here: One hazard written twice, "sunburn" and then "burns the skin", gives one mark rather than two — the second mark needs the longer-term damage, skin cancer or damage to the eyes. A vague precaution such as "be careful in the sun" is not creditable; name sunscreen, covering up, or UV-blocking sunglasses.

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]

⚠ If you missed marks here: "Sound is longitudinal" and then "sound vibrates parallel to the wave" is the same point in two costumes and earns one mark. The second difference that always scores is the medium: sound needs particles and cannot cross a vacuum, whereas light can. Expecting to hear an explosion in space is the misconception behind that lost mark.

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]

⚠ If you missed marks here: The observation mark is earned only if you note the bell is still seen to be ringing after the sound has gone — without that, the fading could be blamed on the bell running down. Omitting the vacuum pump, or claiming the bell stops vibrating, loses the apparatus mark.

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]

⚠ If you missed marks here: Forgetting that the sound travels out and back gives 680 / 4.0 = 170 m/s instead of 340 m/s, and 170 m/s should look wrong to you for sound in air. The doubling is a mark in its own right, so write the line 2 × 680 = 1360 m rather than hiding it inside the division.

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]

⚠ If you missed marks here: Offering 180 m as the depth is the classic error — that is the there-and-back distance, and the sea floor is half of it, 90 m. In part (ii), "ultrasound is louder" or "ultrasound is faster" scores nothing; credit goes to a frequency above 20 000 Hz, so it is inaudible, and to a narrow beam giving a more precise reading.

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]

⚠ If you missed marks here: "Waves bending" is the wrong verb and reads as refraction — diffraction is spreading out. The second mark needs the cause: passing through a gap or round the edge of an obstacle. Any answer that mentions a change of speed has described refraction instead.

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]

⚠ If you missed marks here: "A narrower gap gives more diffraction" is half the physics and does not score, because the effect depends on the gap compared with the wavelength — maximum spreading comes when the two are about equal. Never write that the wavelength changes as the wave diffracts; wavelength, frequency and speed all stay the same.

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]

⚠ If you missed marks here: "Digital is clearer than analogue" is an advantage, not a difference, and answers nothing here. The two marks are for continuous variation across a range set against a small number of discrete values, typically the two states 1 and 0.

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]

⚠ If you missed marks here: "Digital signals are better quality" repeats the claim instead of explaining it — the mark is for regeneration, that noise can be stripped out because only two states have to be recognised. "Digital travels faster" is false: both are carried at the same speed along the same cable. The second advantage should be capacity, more signals multiplexed onto one fibre.

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

Mark 2 -- higher data capacity / multiplexing

Mark 3 -- clear explanation with practical context

Self-Assessment

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