Critical angle and total internal reflection

Critical angle

Going from a higher to a lower refractive index, the refracted ray bends away from the normal. At the critical angle C the angle of refraction is 90°. For a material in air: sin C = 1/n. For a boundary between two materials (n1 > n2): sin C = n2 / n1.

Total internal reflection

Total internal reflection (TIR) happens when both:

• light travels towards a boundary with a material of lower refractive index, and
• the angle of incidence is greater than the critical angle.

All the light is then reflected. Optical fibres use a core surrounded by a cladding of lower refractive index, so light is totally internally reflected along the core.

Measuring the refractive index of a solid

• Place a rectangular glass block on paper and draw round it. Shine a ray from a ray box into one face.
• Mark the incident and emergent rays, join them through the block and draw normals.
• Measure angles of incidence i and refraction r for several values of i.
• Plot sin i (y-axis) against sin r (x-axis): the gradient is n.

Alternatively, use a semicircular block to find the critical angle and use n = 1 / sin C.

Worked example

Find the critical angle for glass of refractive index 1.50 in air.

1. sin C = 1 / 1.50 = 0.667

Answer: C = 41.8°

Check yourself

1. Find the critical angle for water (n = 1.33) in air, in degrees, to 3 s.f.
   Show answer

   48.8 °. sin C = 1 / 1.33 = 0.752, so C = 48.8°.

2. A plastic has a critical angle of 42° in air. Find its refractive index, to 3 s.f.
   Show answer

   1.49. n = 1 / sin 42° = 1 / 0.669 = 1.49.

3. Which conditions together cause total internal reflection?
   • Light travels towards a higher refractive index at any angle
   • The angle of incidence is less than the critical angle
   • Light travels along the normal
   • Light travels towards a lower refractive index and the angle of incidence is greater than the critical angle
   Show answer

   Light travels towards a lower refractive index and the angle of incidence is greater than the critical angle. Both conditions are needed.

4. Light in glass (n = 1.50, critical angle 41.8°) hits the glass-air boundary at 45° to the normal. What happens?
   • It is totally internally reflected
   • It refracts out into the air
   • It travels along the boundary
   • It passes straight through
   Show answer

   It is totally internally reflected. 45° is greater than the critical angle, so TIR occurs.

5. An optical fibre has a core of refractive index 1.50 and cladding of 1.45. Find the critical angle at the core-cladding boundary in degrees, to 3 s.f.
   Show answer

   75.2 °. sin C = 1.45 / 1.50 = 0.967, so C = 75.2°.

6. In the glass block experiment, which graph gives the refractive index as its gradient?
   • sin i against sin r
   • i against r
   • sin r against sin i
   • i against sin r
   Show answer

   sin i against sin r. sin i = n sin r, so the gradient of sin i against sin r is n.

7. Why does an optical fibre have a cladding of lower refractive index than its core?
   • To make the light travel faster in the core
   • To absorb stray light only
   • So that total internal reflection happens at the core-cladding boundary
   • To increase the critical angle to 90°
   Show answer

   So that total internal reflection happens at the core-cladding boundary. TIR needs light travelling towards a lower refractive index.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)

Last updated 7 November 2026. Spotted a mistake? Tell us.