A transparent block A having refractive index μ = 1.25 is surrounded by another medium of refractive index μ = 1.0 as shown in the figure. A light ray is incident on the flat face of the block with incident angle θ as shown in the figure. What is the maximum value of θ for which light suffers total internal reflection at the top surface of the block?

Solution

Physics sir · 4 steps

The two faces are at right angles, so the ray's angle at the top face is just 90° minus its refraction angle.

1. Critical angle. At the top face, light goes from the block (μ = 1.25) to the outside (μ = 1). So sin C = 1/1.25 = 4/5, which gives cos C = 3/5.
2. Link the two faces. Let the ray refract into the block at angle r at the side face. That face is perpendicular to the top face, so the angle of incidence at the top is 90° - r. A larger θ gives a larger r, and so a smaller angle at the top.
3. Condition for TIR. Total internal reflection needs 90° - r ≥ C. So r ≤ 90° - C, which means sin r ≤ cos C = 3/5.

   sin θ = 1.25 sin r ≤ 5/4 × 3/5 = 3/4

4. So. The largest angle that still gives total internal reflection is θₘₐₓ = sin⁻¹(3/4), option (b).

The idea to keep. When a ray enters one face and meets a perpendicular face, its angles at the two faces add up to 90°. Write the TIR condition at the second face, then use Snell's law at the first face.

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