Consider a tank made of glass (refractive index $1.5$) with a thick bottom. It is filled with a liquid of refractive index $\mu$. A student finds that, irrespective of what the incident angle $i$ (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of $\mu$ is
Answer: (A) $\dfrac{3}{\sqrt5}$
Reflected light at the liquid-glass interface is completely polarised at Brewster's angle $\theta_B$, where
$$\tan\theta_B = \frac{1.5}{\mu} \Rightarrow \sin\theta_B = \frac{1.5}{\sqrt{\mu^2 + 2.25}}$$
Light entering from air travels in the liquid at an angle $r$ with $\sin r = \dfrac{\sin i}{\mu} < \dfrac1\mu$. The reflection is never polarised if even the largest possible $r$ is below $\theta_B$:
$$\frac1\mu \le \frac{1.5}{\sqrt{\mu^2+2.25}} \Rightarrow \mu^2 + 2.25 \le 2.25\mu^2 \Rightarrow \mu^2 \ge \frac{2.25}{1.25} = \frac95$$
$$\mu_{\min} = \frac{3}{\sqrt5}$$
Solution by Sreeraj P, M.Sc Physics