Q 12-09-134JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
A ray of light is incident at an angle of incidence $60^\circ$ on a glass slab of refractive index $\sqrt3$. After refraction, the light ray emerges out from the other parallel face and the lateral shift between incident ray and emergent ray is $4\sqrt3\ \text{cm}$. The thickness of the glass slab is ______ cm.
Numerical value type. Enter your answer.
Answer: 12
$\sin r = \dfrac{\sin60^\circ}{\sqrt3} = \dfrac12\Rightarrow r = 30^\circ$.
$$d = \frac{t\sin(i-r)}{\cos r} = \frac{t\sin30^\circ}{\cos30^\circ} = \frac{t}{\sqrt3}$$
$$\frac{t}{\sqrt3} = 4\sqrt3\ \Rightarrow\ t = 12\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics