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The angle of polarization for any medium is $$60^{\circ\ }$$, what will be critical angle for this :
At the angle of polarization $$i_p=60^\circ$$, Brewster's law gives
$$\tan i_p=\frac{n_2}{n_1}$$
Therefore, $$\frac{n_2}{n_1}=\tan60^\circ=\sqrt{3}$$
For the critical angle $$C$$, $$\sin C=\frac{n_1}{n_2}$$
Hence, $$\sin C=\frac{1}{\sqrt{3}}$$
Therefore, $${C=\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)}$$
Hence, the correct option is C.
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