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(B) For the reflected light to be completely polarized at the liquid-glass interface,the angle of incidence $	heta$ must satisfy Brewster's law: $	a

Vedclass · Accepted Answer

$\frac{3}{\sqrt{5}}$

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(B) For the reflected light to be completely polarized at the liquid-glass interface,the angle of incidence $	heta$ must satisfy Brewster's law: $	an 	heta = \frac{n_2}{n_1} = \frac{1.5}{\mu}$.If the light is never completely polarized for any incident angle $i$,it means that the Brewster angle $	heta_B$ must be greater than or equal to the critical angle $	heta_c$ for the liquid-glass interface,so that total internal reflection occurs before the Brewster condition can be met.The critical angle $	heta_c$ is given by $\sin 	heta_c = \frac{1.5}{\mu}$ (assuming $\mu > 1.5$). However,since the light enters from air into the liquid,the maximum angle of refraction $	heta$ in the liquid is limited by the critical angle of the air-liquid interface. If the light is incident at $90^{\circ}$ from air,$\sin 	heta = \frac{1}{\mu}$.Thus,we require $	an 	heta_B \ge 	an 	heta_{max}$.Given $	an 	heta_B = \frac{1.5}{\mu}$ and $\sin 	heta_{max} = \frac{1}{\mu}$,we have $\cos 	heta_{max} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$.So,$	an 	heta_{max} = \frac{1}{\sqrt{\mu^2 - 1}}$.Setting $\frac{1.5}{\mu} \ge \frac{1}{\sqrt{\mu^2 - 1}}$,we square both sides: $\frac{2.25}{\mu^2} \ge \frac{1}{\mu^2 - 1}$.$2.25(\mu^2 - 1) \ge \mu^2 \Rightarrow 2.25\mu^2 - 2.25 \ge \mu^2 \Rightarrow 1.25\mu^2 \ge 2.25$.$\mu^2 \ge \frac{2.25}{1.25} = \frac{9}{5} \Rightarrow \mu \ge \frac{3}{\sqrt{5}}$.

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