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Critical angle and total internal reflectionEdexcel International A Level Physics: Revision notes

Section 1

Total internal reflection

Total internal reflection (TIR) is the complete reflection of light at a boundary, with no light refracted out. It occurs only when both conditions are met:

  • the light is travelling from a medium of higher refractive index towards one of lower refractive index, for example glass to air
  • the angle of incidence is greater than the critical angle

TIR can never occur when light travels towards a medium of higher refractive index.

Key termstotal internal reflection
Common mistake

Total internal reflection does not occur for light going from air into glass, whatever the angle.

Section 2

The critical angle

The critical angle C is the angle of incidence in the denser medium for which the angle of refraction is 90°, so the refracted ray travels along the boundary. Putting θ₂ = 90° in Snell's law gives n₁ sin C = n₂. For a boundary with air (n₂ = 1):

sin C = 1/n

The larger the refractive index, the smaller the critical angle. For a general boundary between two materials, sin C = n₂/n₁ with n₁ > n₂.

Worked example. Glass n = 1.52: sin C = 1/1.52 = 0.658, so C = 41°. Diamond n = 2.42: C = 24°.

Key termscritical angle
Exam tip

Use sin C = 1/n only for a boundary with air. For glass in water use sin C = n₂/n₁.

Section 3

Three cases at a boundary

For light travelling from a higher to a lower refractive index:

  • angle of incidence < C: most light refracts away from the normal, with a weak partial reflection
  • angle of incidence = C: the refracted ray emerges along the boundary at 90°
  • angle of incidence > C: total internal reflection, and the angle of reflection equals the angle of incidence

To predict whether TIR occurs: work out C for the boundary, check that n decreases across it, then compare the angle of incidence with C.

Key termspartial reflection

Section 4

Applications

TIR is used in optical fibres, where light signals are guided along a glass core by repeated total internal reflection at the core wall, with very little loss. It is also used in prisms in binoculars and periscopes, and in the sparkle of cut diamond: its high refractive index gives a small critical angle (24°), so a large fraction of the light inside is totally internally reflected and emerges from the front face.

Key termsoptical fibre
Exam tip

In explanation questions, state the two conditions for TIR and then apply them to the numbers in the question.

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Exam questions on Critical angle and total internal reflection

  1. A ray of light travels inside a rectangular block of glass of refractive index 1.60 and strikes a flat glass–air boundary from inside the glass. Take the refractive index of air as 1.00.
    The block is now immersed in water of refractive index 1.33. Calculate the critical angle for the glass–water boundary.2 marks
  2. A bare optical fibre made of glass with refractive index 1.52 is surrounded by air. Light signals travel along the fibre by repeatedly meeting the side wall of the fibre, which is a glass–air boundary.
    A ray inside the fibre meets the side wall at an angle of incidence of 38°. Determine whether the ray is totally internally reflected.2 marks
  3. A small lamp is fixed at the bottom of a swimming pool, 2.0 m below the water surface, and shines light in all directions upwards. The refractive index of the water is 1.33 and that of air is 1.00. A swimmer looks at the surface from above.
    Calculate the critical angle for the water–air surface. Explain why light from the lamp that reaches the surface at an angle of incidence of 60° does not emerge into the air.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).