The polarising angle of a transparent medium | vedclass

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(C) According to Brewster's Law,the refractive index $\mu$ of a medium is given by $\mu = 	an 	heta$,where $	heta$ is the polarising angle.We also

Vedclass · Accepted Answer

$	heta=\cot ^{-1}\left(\frac{v}{c}ight)$

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(C) According to Brewster's Law,the refractive index $\mu$ of a medium is given by $\mu = 	an 	heta$,where $	heta$ is the polarising angle.We also know that the refractive index $\mu$ is the ratio of the speed of light in air $(c)$ to the speed of light in the medium $(v)$,so $\mu = \frac{c}{v}$.Equating the two expressions for $\mu$,we get $	an 	heta = \frac{c}{v}$.Therefore,$	heta = 	an^{-1}\left(\frac{c}{v}ight)$.However,looking at the options provided,there seems to be a reciprocal relationship intended. If we define the refractive index as $\mu = \frac{1}{	an 	heta}$ or if the question implies the angle of incidence relative to the surface,we re-evaluate: $	an 	heta = \frac{c}{v} \implies \cot 	heta = \frac{v}{c}$.Thus,$	heta = \cot^{-1}\left(\frac{v}{c}ight)$.

The polarising angle of a transparent medium is $\theta$. Let the speed of light in the medium be $v$. Then the relation between $\theta$ and $v$ is [where $c$ is the velocity of light in air].

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