A transparent block A having refractive index $\mu = 1.25$ is surrounded by another medium of refractive index $\mu = 1.0$ as shown in the figure. A light ray is incident on the flat (vertical) face of the block with incident angle $\theta$. What is the maximum value of $\theta$ for which the light suffers total internal reflection at the top surface of the block?
Answer: (C) $\sin^{-1}(3/4)$
Let the refraction angle at the side face be $r$. The two faces are perpendicular, so the ray meets the top face at an angle of incidence $90^\circ - r$.
TIR at the top needs $90^\circ - r \ge \theta_c$, where $\sin\theta_c = \dfrac{1}{1.25} = 0.8$, so $\cos\theta_c = 0.6$. The limit is $r = 90^\circ - \theta_c$, i.e. $\sin r = \cos\theta_c = 0.6$.
Snell's law at the side face:
$$\sin\theta = 1.25\sin r = 1.25\times0.6 = 0.75$$
So $\theta_{\max} = \sin^{-1}(3/4)$; larger $\theta$ makes $r$ larger and the ray escapes through the top.
Solution by Sreeraj P, M.Sc Physics