A transparent cube of side $d$, made of a material of refractive index $\mu_2$, is immersed in a liquid of refractive index $\mu_1$ $(\mu_1 < \mu_2)$. A ray is incident on the face AB at an angle $\theta$ (shown in the figure). Total internal reflection takes place at the point E on the face BC. Then, $\theta$ must satisfy
Answer: (A) $\theta < \sin^{-1}\sqrt{\dfrac{\mu_2^2}{\mu_1^2} - 1}$
At AB: $\mu_1\sin\theta = \mu_2\sin r$. The angle of incidence at BC is $90^\circ - r$, and TIR needs
$$\sin(90^\circ - r) = \cos r > \frac{\mu_1}{\mu_2}$$
$$1 - \frac{\mu_1^2\sin^2\theta}{\mu_2^2} > \frac{\mu_1^2}{\mu_2^2} \Rightarrow \sin^2\theta < \frac{\mu_2^2}{\mu_1^2} - 1$$
$$\theta < \sin^{-1}\sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$$
Solution by Sreeraj P, M.Sc Physics