If the polarizing angle for a medium is 60^ci | vedclass

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(D) According to Brewster's law,the refractive index $\mu$ is given by $\mu = 	an(i_p)$,where $i_p$ is the polarizing angle.Given $i_p = 60^\circ$,we

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$\sin^{-1} \frac{1}{\sqrt{3}}$

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(D) According to Brewster's law,the refractive index $\mu$ is given by $\mu = 	an(i_p)$,where $i_p$ is the polarizing angle.Given $i_p = 60^\circ$,we have $\mu = 	an(60^\circ) = \sqrt{3}$.For a medium,the relationship between the refractive index $\mu$ and the critical angle $i_c$ is $\mu = \frac{1}{\sin(i_c)}$.Substituting the value of $\mu$,we get $\sqrt{3} = \frac{1}{\sin(i_c)}$.Therefore,$\sin(i_c) = \frac{1}{\sqrt{3}}$,which implies $i_c = \sin^{-1}(\frac{1}{\sqrt{3}})$.Thus,the correct option is $D$.

If the polarizing angle for a medium is $60^\circ$,then the critical angle for that medium is . . . . . .

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