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Question 41

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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