Q 12-09-220JEE MainJEE Main 2017 (8 Apr)Medium
Let the refractive index of a denser medium with respect to rarer medium be $n_{12}$ and its critical angle be $\theta_C$. At an angle of incidence $A$ when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is $90^\circ$. Angle $A$ is given by:
Answer: (A) $\tan^{-1}(\sin\theta_C)$
Reflected and refracted rays perpendicular means $A + r = 90^\circ$. Snell's law from the denser side:
$$n_{12}\sin A = \sin r = \cos A \;\Rightarrow\; \tan A = \frac{1}{n_{12}}$$
For the critical angle, $\sin\theta_C = \dfrac{1}{n_{12}}$. Hence
$$A = \tan^{-1}(\sin\theta_C)$$
Solution by Sreeraj P, M.Sc Physics