A prism of angle $75^{\circ}$ and refractive index $\sqrt{3}$ is coated with thin film of refractive

Question

A prism of angle $75^{\circ}$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be $\_\_\_\_$
$\left(\sin 15^{\circ}=0.25\right.$ and $\left.\sin 25^{\circ}=0.43\right)$

Professor PC Thomas Classes · Answer

For total internal reflection at the back surface, the angle of incidence $r_2$ must be greater than or equal to the critical angle: $r_2 \ge \sin^{-1}(\frac{\mu_{film}}{\mu_{prism}}) = \sin^{-1}(\frac{1.5}{\sqrt{3}}) = \sin^{-1}(\frac{\sqrt{3}}{2}) = 60^\circ$. For the prism, $r_1 + r_2 = 75^\circ$, which means $r_1 \le 75^\circ - 60^\circ = 15^\circ$. By Snell's Law at the first surface, $\sin i = \sqrt{3} \sin r_1 \le \sqrt{3} \sin 15^\circ \approx 1.732 	imes 0.258 \approx 0.44$. Therefore, the incident angle $i$ must be less than $\sin^{-1}(0.44)$, which is approximately $26^\circ$. (Note: Official key drops this question due to ambiguity in options).