The angle of polarisation for any medium is 6 | vedclass

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(D) According to Brewster's law,the refractive index $\mu$ is related to the angle of polarisation $i_{p}$ as $\mu = 	an i_{p}$.Given $i_{p} = 60^{\c

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

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(D) According to Brewster's law,the refractive index $\mu$ is related to the angle of polarisation $i_{p}$ as $\mu = 	an i_{p}$.Given $i_{p} = 60^{\circ}$,we have $\mu = 	an 60^{\circ} = \sqrt{3}$.The critical angle $	heta_{c}$ is related to the refractive index by the formula $\sin 	heta_{c} = \frac{1}{\mu}$.Substituting the value of $\mu$,we get $\sin 	heta_{c} = \frac{1}{\sqrt{3}}$.Therefore,the critical angle is $	heta_{c} = \sin^{-1} \left( \frac{1}{\sqrt{3}} ight)$.

The angle of polarisation for any medium is $60^{\circ}$. What will be the critical angle for this medium?

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