Question

The polarising angle for a medium is found to be $60^{\circ}$. The critical angle of the medium is

• A. $\sin^{-1}\left(\frac{1}{2}\right)$
• B. $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• C. $\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. $\sin^{-1}\left(\frac{1}{4}\right)$

Answer 1

Answer: (C)

$\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer 2

From Brewster's law
The refractive index of the medium is given by
$n=\tan i_{p}$
where $i_{p}=$ polarising angle
$\Rightarrow n=\tan 60^{\circ}$
$\Rightarrow n=\sqrt{3}$
If $C$ be the critical angle for the medium, then
$n =\frac{1}{\sin C}$
$\sin C =\frac{1}{n}=\frac{1}{\sqrt{3}}$
$C =\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$