Q 12-09-110JEE MainJEE Main 2023 (25 Jan, Shift 1)Medium
A ray of light is incident from air on a glass plate having thickness $\sqrt3\ \text{cm}$ and refractive index $\sqrt2$. The angle of incidence of a ray is equal to the critical angle for glass–air interface. The lateral displacement of the ray when it passes through the plate is ______ $\times10^{-2}\ \text{cm}$. (given $\sin15^\circ=0.26$)
Numerical value type. Enter your answer.
Answer: 52
Critical angle: $\sin C=\dfrac1{\sqrt2}\Rightarrow i=45^\circ$. Refraction: $\sin r=\dfrac{\sin45^\circ}{\sqrt2}=\dfrac12\Rightarrow r=30^\circ$.
$$d=\frac{t\sin(i-r)}{\cos r}=\frac{\sqrt3\times0.26}{\sqrt3/2}=0.52\ \text{cm}=52\times10^{-2}\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics