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 [ $sin 9 0^{\circ} = 1$ ]

Question

TestHub · Accepted Answer

According to Snell's law for multiple parallel interfaces, the product of the refractive index and the sine of the angle made with the normal remains constant across all regions:

$n_{1} sin θ_{1} = n_{2} sin θ_{2} = n_{3} sin θ_{3} = n_{4} sin θ_{4}$

For the beam to just miss entering region IV, it must undergo total internal reflection at the boundary between region III and region IV. This means the angle of refraction in region IV is $9 0^{\circ}$ .

Equating the terms for region I and region IV, we get:

$n_{1} sin θ_{1} = n_{4} sin 9 0^{\circ}$

Given that $n_{1} = n$ , $θ_{1} = θ$ , and $n_{4} = \frac{n}{4}$ :

$n sin θ = \frac{n}{4} \times 1$

$sin θ = \frac{1}{4}$

$θ = sin^{- 1} (\frac{1}{4})$

Answer: $sin^{- 1} (\frac{1}{4})$

Physics - Geometrical Optics Question with Solution | TestHub

Options:

Doubts & Discussion