The angle of polarisation for a medium is 60^ | vedclass

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(D) According to Brewster's Law,the refractive index $\mu$ of a medium is related to the angle of polarisation $	heta_p$ by the formula: $\mu = 	an

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

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(D) According to Brewster's Law,the refractive index $\mu$ of a medium is related to the angle of polarisation $	heta_p$ by the formula: $\mu = 	an 	heta_p$. Given $	heta_p = 60^o$,we have $\mu = 	an 60^o = \sqrt{3}$. The critical angle $C$ is related to the refractive index $\mu$ by the formula: $\sin C = \frac{1}{\mu}$. Substituting the value of $\mu$,we get $\sin C = \frac{1}{\sqrt{3}}$. Therefore,the critical angle is $C = \sin^{-1} \left( \frac{1}{\sqrt{3}} ight)$.

The angle of polarisation for a medium is $60^o$. What will be the critical angle for this medium?

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