IGCSE Physics 0625 · Topic 3.2.2

IGCSE Physics Refraction of Light Questions and Worked Solutions

50 topic-wise practice questions on Refraction of Light (Unit 3: Waves), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions18 Core22 Extended10 Challenge135 marksTextbook §3.2 · pp. 136–159

Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Refraction of Light

Light changes speed and direction when it passes between materials. It bends towards the normal entering a denser medium and away from it on leaving. Above the critical angle, light is totally internally reflected — the basis of optical fibres.

Key formulas

n = sin i ÷ sin r	refractive index n, angle of incidence i, angle of refraction r (in air → medium)	Extended
n = 1 ÷ sin c	critical angle c	Extended
n = speed in air (vacuum) ÷ speed in medium	definition of refractive index	Extended

Prerequisite knowledge

Reflection and the normal (3.2.1)
Wave speed (3.1)
Using sin and sin⁻¹ on a calculator (Extended)

Exam tips & common mistakes

Calculator must be in degrees
n is always greater than 1
State BOTH conditions for total internal reflection

Key facts from the syllabus (3.2.2)

Normal, angle of incidence and angle of refraction are measured from the normal
Experiment: trace rays through rectangular, triangular and semicircular glass blocks
Air → glass: slows down, bends towards the normal; glass → air: speeds up, bends away from the normal
Light along the normal does not change direction
Critical angle: the angle of incidence for which the angle of refraction is 90°
Total internal reflection: light travelling from a denser to a less dense medium with i greater than the critical angle (prisms, optical fibres)
Extended optical fibres carry signals by repeated total internal reflection, especially in telecommunications

Textbook: §3.2 Light, pp. 136–159

Core[1 mark]

State what is meant by refraction.

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Method: Change of direction due to change in speed.

The change in direction of a wave when it changes speed passing from one medium to another

Core[2 marks]

Light passes from air into glass at an angle. State what happens to its speed and direction.

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Method: Slows down, bends towards the normal.

It slows down and bends towards the normal

Core[2 marks]

Light passes from glass into air at an angle. State what happens to its speed and direction.

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Method: Speeds up, bends away from the normal.

It speeds up and bends away from the normal

Core[1 mark]

Light enters a glass block along the normal. Describe its path.

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Method: No change in direction.

It passes straight through without changing direction (but slows down)

Core[3 marks]

Describe an experiment to show refraction of light through a rectangular glass block.

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Method: Trace block, ray box, mark rays, join.

Draw round the block; shine a ray from a ray box at an angle; mark the incident and emerging rays with crosses; remove the block and join the points; draw normals and measure angles

Core[1 mark]

State the meaning of critical angle.

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Method: Angle of incidence for 90° refraction.

The angle of incidence in the denser medium at which the angle of refraction is 90°

Core[2 marks]

State the two conditions for total internal reflection.

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Method: Denser to less dense; i > c.

Light travelling from a denser to a less dense medium, and the angle of incidence greater than the critical angle

Core[2 marks]

Give two everyday examples of total internal reflection.

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Method: Prisms, optical fibres.

Optical fibres; prisms in periscopes/binoculars (also diamonds sparkling)

Core[2 marks]

A swimming pool looks shallower than it really is. Explain why.

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Method: Refraction at the surface.

Light from the bottom bends away from the normal as it leaves the water, so it appears to come from a shallower point

Core[1 mark]

Define the angle of refraction.

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Method: Between refracted ray and normal.

The angle between the refracted ray and the normal

Core[2 marks]

Explain why a straw in a glass of water appears bent.

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Method: Refraction at surface.

Light from the underwater part refracts as it leaves the water, so it appears to come from a different position

Core[2 marks]

State what happens to the speed of light as it passes from glass into water.

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Method: Glass denser.

It increases

Core[3 marks]

A ray enters a rectangular glass block at an angle. Describe the direction of the ray when it leaves the block.

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Method: Parallel to original.

It leaves parallel to the incident ray but displaced sideways

Core[2 marks]

State what happens to light hitting a glass–air boundary from inside the glass at an angle smaller than the critical angle.

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Method: Refracts out, some reflects.

Most is refracted out of the glass, bending away from the normal; a little is reflected

Core[2 marks]

State what happens at an angle of incidence exactly equal to the critical angle.

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Method: 90° refraction.

The refracted ray travels along the boundary (angle of refraction 90°)

Core[2 marks]

Describe how a 45° right-angled prism can turn light through 90°.

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Method: Total internal reflection.

Light enters one face normally, hits the hypotenuse at 45° (greater than the critical angle of about 42°) and is totally internally reflected

Core[2 marks]

Explain why diamonds sparkle.

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Method: Small critical angle.

Diamond has a small critical angle, so light undergoes many total internal reflections before leaving

Core[2 marks]

State one use of total internal reflection in medicine.

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Method: Endoscope.

Endoscopes (optical fibres to see inside the body)

Extended[1 mark]

Define refractive index in terms of speed.

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Method: n = speed in vacuum/air ÷ speed in medium.

The ratio of the speed of light in air (vacuum) to the speed of light in the medium

Extended[2 marks]

Light travels at 3.0 × 10⁸ m/s in air and 2.0 × 10⁸ m/s in glass. Calculate the refractive index of the glass.

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Method: n = c ÷ v.

n = 3.0×10⁸ ÷ 2.0×10⁸ = 1.50

Extended[3 marks]

Light enters glass (n = 1.5) at an angle of incidence of 40°. Calculate the angle of refraction.

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Method: sin r = sin i ÷ n.

sin r = sin 40° ÷ 1.5 = 0.429; r = 25.4°

Extended[3 marks]

Light enters water at 50° and refracts at 35°. Calculate the refractive index of water.

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Method: n = sin i ÷ sin r.

n = sin 50° ÷ sin 35° = 1.34

Extended[3 marks]

Calculate the critical angle for glass of refractive index 1.5.

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Method: sin c = 1/n.

c = sin⁻¹(1/1.5) = 41.8°

Extended[3 marks]

The critical angle for a material is 24.4°. Calculate its refractive index.

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Method: n = 1 ÷ sin c.

n = 1 ÷ sin 24.4° = 2.42

Extended[3 marks]

Calculate the speed of light in a diamond of refractive index 2.4 (c = 3.0 × 10⁸ m/s).

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Method: v = c ÷ n.

v = 3.0×10⁸ ÷ 2.4 = 1.25 × 10⁸ m/s

Extended[3 marks]

Light inside glass (n = 1.5) hits the glass–air surface at 45°. State and explain what happens.

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Method: Compare with c = 41.8°.

Critical angle = 41.8°; 45° is greater, so total internal reflection occurs

Extended[3 marks]

Describe how optical fibres are used in telecommunications.

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Method: TIR along fibre.

Light (or infrared) signals travel along thin glass fibres by repeated total internal reflection, carrying large amounts of data over long distances with little loss

Extended[2 marks]

State two advantages of optical fibres over copper cables for communication.

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Method: Data rate, interference.

Carry much more data at higher rates; less signal loss/no electrical interference; lighter

Extended[3 marks]

Light enters a block of refractive index 1.33 at an angle of incidence of 60°. Calculate the angle of refraction.

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Method: sin r = sin i ÷ n.

r = sin⁻¹(sin 60° ÷ 1.33) = 40.6°

Extended[3 marks]

Light enters water (n = 1.33) at an angle of incidence of 30°. Calculate the angle of refraction.

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Method: sin r = sin i / n.

r = 22.1°

Extended[3 marks]

A ray passes from air into a material at 55° and refracts at 33°. Calculate the refractive index.

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Method: n = sin i / sin r.

n = 1.50

Extended[3 marks]

Calculate the critical angle for water (n = 1.33).

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Method: sin c = 1/n.

c = 48.8°

Extended[3 marks]

Calculate the critical angle for diamond (n = 2.42).

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Method: sin c = 1/n.

c = 24.4°

Extended[3 marks]

Light travels at 2.25 × 10⁸ m/s in water. Calculate the refractive index of water (c = 3.0 × 10⁸ m/s).

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Method: n = c/v.

n = 1.33

Extended[3 marks]

A glass has refractive index 1.60. Calculate the speed of light in it.

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Method: v = c/n.

v = 1.88 × 10⁸ m/s

Extended[3 marks]

Calculate the angle of incidence in air needed to give an angle of refraction of 20° in glass (n = 1.50).

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Method: sin i = n sin r.

i = 30.9°

Extended[3 marks]

A ray inside water (n = 1.33) meets the surface at 52°. State and explain whether it escapes.

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Method: Compare with c.

Critical angle = 48.8°; 52° is greater, so it is totally internally reflected and does not escape

Extended[2 marks]

Explain why the cladding of an optical fibre has a lower refractive index than the core.

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Method: TIR at boundary.

So that light in the core can be totally internally reflected at the core–cladding boundary

Extended[2 marks]

State one advantage of using optical fibres for broadband over radio waves.

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Method: Data rate, security.

Higher data rates (and signals are secure and less affected by interference)

Extended[3 marks]

Light enters a glass block (n = 1.52) at 0° incidence. State the angle of refraction and the speed of light in the glass.

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Method: sin 0 = 0.

0°; v = 3.0×10⁸ ÷ 1.52 = 1.97 × 10⁸ m/s

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

ChallengeCore[4 marks]

Describe and explain the path of a ray of light through a rectangular glass block when it enters at 40° to the normal.

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Method: Refracts in and out.

It bends towards the normal on entering (slows), travels straight, then bends away from the normal on leaving (speeds up), emerging parallel to the incident ray but displaced

ChallengeCore[4 marks]

Explain how an optical fibre carries light around bends using total internal reflection, and state two conditions needed.

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Method: TIR.

Light hits the core boundary at angles greater than the critical angle and is totally internally reflected repeatedly; conditions: travelling from higher to lower refractive index, angle of incidence > critical angle

ChallengeCore[3 marks]

Explain why a coin at the bottom of a cup can become visible when water is poured in, though your eye position is unchanged.

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Method: Refraction at surface.

Light from the coin refracts away from the normal at the water surface, bending over the cup's rim into your eye

ChallengeExtended[4 marks]

Light enters glass (n = 1.52) at 60°. Calculate the angle of refraction and the speed of light in the glass.

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Method: Snell and n = c/v.

r = 34.7°; v = 1.97 × 10⁸ m/s

ChallengeExtended[5 marks]

Light travels from glass (n = 1.50) into air. Calculate the critical angle and the angle of refraction in air for an angle of incidence of 30° in the glass.

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Method: sin c = 1/n; sin r = n sin i.

c = 41.8°; r = 48.6°

ChallengeExtended[5 marks]

A ray enters one face of a right-angled 45° glass prism (n = 1.50) along the normal. Calculate what happens at the hypotenuse and the total deviation.

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Method: i = 45° > c.

c = 41.8°; 45° > c so total internal reflection occurs; the ray is turned through 90°

ChallengeExtended[4 marks]

A diamond has critical angle 24.4°. Calculate its refractive index and the speed of light in it, and explain why it sparkles.

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Method: n = 1/sin c.

n = 2.42; v = 1.24 × 10⁸ m/s; the small critical angle causes many total internal reflections

ChallengeExtended[4 marks]

Light in water (n = 1.33) hits the surface at 45°. Calculate the angle of refraction in air, if any, and state what happens at 55°.

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Method: Compare with c = 48.8°.

r = 70.1°; at 55° (> 48.8°) total internal reflection occurs

ChallengeExtended[4 marks]

A ray passes from water (n = 1.33) into glass (n = 1.50) at 40° to the normal. Calculate the angle of refraction. (Use n₁ sin θ₁ = n₂ sin θ₂ by going via air.)

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Method: Same sin × n.

sin θ₂ = 1.33 sin 40° ÷ 1.50; θ₂ = 34.7°

ChallengeExtended[4 marks]

Explain why the cladding of an optical fibre must have a lower refractive index than the core, and why scratches on an unclad fibre would cause loss of signal.

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Method: TIR condition.

TIR only occurs going from higher to lower refractive index; without cladding, scratches or contact change the boundary so light escapes

All IGCSE Physics topics ← 3.2.1 Reflection of Light 3.2.3 Thin Lenses →

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