Question

The polarising angle for a medium is found to be ${60}^o$. 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. ${\sin }^{-1}\left(\frac{1}{4}\right)$
• E. ${\sin }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

Answer: (C)

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

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