A ray of light entering from air into a denser medium of refractive index $\frac{4}{3}$, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $\theta$ should be equal to:
Answer: (A) $\sin^{-1}\frac{\sqrt{7}}{3}$
At the top surface: $\sin\theta = \frac{4}{3}\sin\theta'$.
The adjacent face is perpendicular, so the angle of incidence there is $\theta'' = 90^\circ - \theta'$. For total internal reflection:
$$\sin(90^\circ - \theta') \ge \frac{3}{4} \Rightarrow \cos\theta' \ge \frac{3}{4} \Rightarrow \sin\theta' \le \frac{\sqrt{7}}{4}$$
$$\sin\theta \le \frac{4}{3}\cdot\frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{3} \Rightarrow \theta_{max} = \sin^{-1}\frac{\sqrt{7}}{3}$$
Solution by Sreeraj P, M.Sc Physics