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The critical angle for a medium of refractive index n (with respect to vacuum) is:

  1. (a) sin⁻¹(n)
  2. (b) sin⁻¹(1/n)
  3. (c) cos⁻¹(n)
  4. (d) tan⁻¹(n)
Answer and solution

Answer (b) sin⁻¹(1/n)

Solution

Physics sir · 3 steps

At the critical angle the refracted ray just grazes the surface, so put r = 90° in Snell's law.

  1. Set up Snell's law. Light goes from the denser medium (index n) into vacuum (index 1). At the critical angle C, the refracted ray travels along the surface, so the angle of refraction is 90°.

    n sin C = 1 × sin 90° = 1

  2. Solve for C. This gives sin C = 1/n. Since n > 1, the value 1/n is less than 1, so a real angle C exists.
  3. So. C = sin⁻¹(1/n), which is option (b).

The idea to keep. Critical angle always means refraction angle 90°. From denser to rarer, sin C = nᵣₐᵣₑᵣ/n_(denser), which is always less than 1.

Answer from the NTA final answer key. Solution by PrepWiser.

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