IGCSE Physics 0625 · Topic 3.2

IGCSE Physics: Light Practice Questions

Light reflects so that the angle of incidence equals the angle of reflection, and refracts when it changes speed on entering a new medium. Total internal reflection occurs when light strikes a boundary from the denser side at an angle greater than the critical angle.

Cambridge IGCSE Physics (0625) · Topic 3.2: Light

Sub-topic 3.2 of Cambridge IGCSE Physics 0625 carries both a calculation and a set of ray diagram skills. Snell's law and the critical angle equation are the two calculations that recur. The questions below cover them plus the ray diagram conventions examiners expect.

What you need to know for Light

IGCSE Physics Light questions and answers

4 exam-style questions written to the 0625 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

A ray of light travels from air into a glass block. The angle of incidence in air is 42 degrees and the angle of refraction in the glass is 26 degrees. Calculate the refractive index of the glass.

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Answer: 1.5
  1. Use n = sin i divided by sin r, where i is the angle in air.
  2. sin 42 = 0.6691 and sin 26 = 0.4384.
  3. n = 0.6691 divided by 0.4384.
  4. n = 1.53, which rounds to 1.5. Refractive index has no units because it is a ratio.
How the marks are awarded. 1 mark for using n = sin i divided by sin r. 1 mark for correct substitution of both sine values. 1 mark for 1.5 with no units.
Where students lose the mark. Dividing sin r by sin i, which gives 0.66. Refractive index for light entering a denser medium is always greater than 1, so a value below 1 signals the fraction is inverted.
Question 2[4 marks]

Calculate the critical angle for a material of refractive index 1.5, and explain what happens to a ray striking the boundary from inside the material at 50 degrees.

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Answer: The critical angle is 42 degrees, so the ray at 50 degrees is totally internally reflected.
  1. Use sin c = 1 divided by n = 1 divided by 1.5 = 0.6667.
  2. c = inverse sine of 0.6667 = 41.8 degrees, which is 42 degrees to 2 significant figures.
  3. The ray strikes the boundary at 50 degrees, which is greater than the critical angle of 42 degrees.
  4. No light is refracted out. All of it is reflected back into the material, obeying the law of reflection. This is total internal reflection.
How the marks are awarded. 1 mark for using sin c = 1 divided by n. 1 mark for a critical angle of 42 degrees. 1 mark for comparing 50 degrees with the critical angle. 1 mark for stating that total internal reflection occurs.
Where students lose the mark. Using sin c = n rather than 1 divided by n. Since n is greater than 1, that would give a sine greater than 1, which is impossible.
Question 3[3 marks]

Explain why a ray of light bends towards the normal when it passes from air into glass.

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Answer: Light travels more slowly in glass, and the side of the wavefront entering first slows before the rest.
  1. Light travels more slowly in glass than in air, because glass is optically denser.
  2. When the ray meets the boundary at an angle, one side of each wavefront enters the glass and slows down before the other side does.
  3. The wavefront therefore pivots, changing the direction of travel.
  4. Because the light slows on entering, it bends towards the normal. The frequency is unchanged, so the wavelength decreases in the glass.
How the marks are awarded. 1 mark for light travelling more slowly in glass. 1 mark for one part of the wavefront slowing before the rest. 1 mark for the change of direction being towards the normal.
Where students lose the mark. Saying the light bends because the glass is thicker or heavier. The cause is the change in speed at the boundary.
Question 4[4 marks]

Explain how an optical fibre carries a light signal along a curved path, and give one advantage of using optical fibres for communication.

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Answer: Light is totally internally reflected repeatedly along the fibre. Fibres carry far more information than copper cables.
  1. The fibre has a core of higher refractive index surrounded by cladding of lower refractive index.
  2. Light entering the fibre strikes the core cladding boundary at an angle greater than the critical angle.
  3. It is therefore totally internally reflected rather than escaping, and this repeats many thousands of times along the fibre, so the light follows the fibre even around bends.
  4. An advantage is that optical fibres carry far more information per second than copper cables, with less signal loss over long distances and no interference from external electrical fields.
How the marks are awarded. 1 mark for a core of higher refractive index with cladding of lower refractive index. 1 mark for light striking the boundary above the critical angle. 1 mark for repeated total internal reflection guiding the light. 1 mark for a valid advantage.
Where students lose the mark. Saying the light reflects off the inside of the glass like a mirror. It is total internal reflection at a boundary with a less dense medium, which requires the angle to exceed the critical angle.

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Light FAQs

How do I calculate refractive index?

Divide the sine of the angle in air by the sine of the angle in the medium, using n = sin i divided by sin r. The angles are measured from the normal. Refractive index has no units and is always greater than 1 for light passing into a denser medium.

What is the critical angle?

The critical angle is the angle of incidence inside the denser medium at which the refracted ray would travel along the boundary at 90 degrees to the normal. It is calculated using sin c = 1 divided by n. Beyond that angle, no light escapes and total internal reflection occurs.

Why does light bend when it enters glass?

Light travels more slowly in glass than in air. When a ray meets the boundary at an angle, one side of each wavefront slows before the other, which pivots the wavefront and changes the direction of travel. Slowing down means the ray bends towards the normal.

How do optical fibres work?

A fibre has a core of higher refractive index surrounded by cladding of lower refractive index. Light striking that boundary at an angle greater than the critical angle is totally internally reflected. Repeating this thousands of times guides the light along the fibre, even around bends.

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Written to the published Cambridge IGCSE Physics (0625) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.