One side of an equilateral prism is painted by a transparent material of refractive index $n_2$. The refractive index of prism is $1.6$. The minimum value of $n_2$ required for total internal reflection from painted face is ______.
Answer: (D) $\dfrac{4\sqrt3}{5}$
The ray enters one face normally, so it goes straight into the prism. In an equilateral prism it then meets the painted face at an angle of incidence of $60^\circ$.
Total internal reflection at the glass–paint boundary needs $i\ge C$, where $\sin C=\dfrac{n_2}{1.6}$:
$$\sin60^\circ\ge\frac{n_2}{1.6}\ \Rightarrow\ n_2\le1.6\times\frac{\sqrt3}{2}=\frac{4\sqrt3}{5}\approx1.39$$
The limiting (critical) value of $n_2$ is $\dfrac{4\sqrt3}{5}$; TIR occurs for any paint with $n_2$ up to this value.
Solution by Sreeraj P, M.Sc Physics