Q 12-09-085JEE MainJEE Main 2024 (4 Apr, Shift 2)Medium
A light ray is incident on a glass slab of thickness $4\sqrt3\ \text{cm}$ and refractive index $\sqrt2$. The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of the ray after passing through the glass slab is ______ cm. (Given $\sin15^\circ = 0.25$)
Numerical value type. Enter your answer.
Answer: 2
Critical angle: $\sin C = \dfrac{1}{\sqrt2} \Rightarrow i = 45^\circ$.
Refraction: $\sin r = \dfrac{\sin45^\circ}{\sqrt2} = \dfrac12 \Rightarrow r = 30^\circ$.
$$d = \frac{t\sin(i - r)}{\cos r} = \frac{4\sqrt3\times\sin15^\circ}{\sqrt3/2} = 8\times0.25 = 2\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics