Q 12-09-146JEE MainJEE Main 2022 (26 Jul, Shift 2)Medium
In the given figure, the face $AC$ of the equilateral prism is immersed in a liquid of refractive index $n$. For incident angle $60^\circ$ at the side $AC$, the refracted light beam just grazes along face $AC$. The refractive index of the liquid $n = \dfrac{\sqrt x}{4}$. The value of $x$ is ______. (Given refractive index of glass $= 1.5$)
Numerical value type. Enter your answer.
Answer: 27
The ray inside the glass meets face $AC$ at $60^\circ$ and the refracted ray grazes the face, so $60^\circ$ is the critical angle for glass to liquid:
$$1.5\sin60^\circ = n\sin90^\circ\ \Rightarrow\ n = \frac{3\sqrt3}{4} = \frac{\sqrt{27}}{4}$$
So $x = 27$.
Solution by Sreeraj P, M.Sc Physics