A beam of light is travelling from region I to region IV. The refractive indices in regions I, II, III and IV are n, n/2, n/3 and n/4 respectively. The angle of incidence for which the beam just misses entering region IV is [ sin 90° = 1 ] (a) sin ⁻¹(1) (b) sin ⁻¹(1/2) (c) sin ⁻¹(1/3) (d) sin ⁻¹(1/4)

Answer: (d). For parallel layers, nsin θ stays the same in every region. Just missing region IV means the ray grazes the III-IV boundary: nsin θ = n/4sin 90°. sin θ = 1/4. Regions II and III do not change the result. So the answer is (d) sin ⁻¹(1/4).

A beam of light is travelling from region I to region IV. The… | MHT CET 2026 (20 May, Shift 1)

Q5MHT CET 2026 · 20 MAY · SHIFT 1Medium

Total Internal Reflection

A beam of light is travelling from region I to region IV. The refractive indices in regions I, II, III and IV are $n,\ \frac{n}{2},\ \frac{n}{3}$ and $\frac{n}{4}$ respectively. The angle of incidence for which the beam just misses entering region IV is [ $\sin 90^\circ = 1$ ]

Show answer and solution

Answer: (d)

For parallel layers, $n\sin\theta$ stays the same in every region.
Just missing region IV means the ray grazes the III-IV boundary: $n\sin\theta = \frac{n}{4}\sin90^\circ$ .
$\sin\theta = \frac{1}{4}$ . Regions II and III do not change the result.
So the answer is (d) $\sin^{-1}\left(\frac{1}{4}\right)$ .

Report a mistake

Total Internal Reflection question from MHT CET 2026 (20 May, Shift 1)

More questions on Total Internal Reflection