Question

The polarising angle for transparent medium is ${}^{\prime }{\theta }^{\prime }{,}^{\prime }{v}^{\prime }$ is the speed of light in that medium. Then the relation between ${}^{\prime }{\theta }^{\prime }$ and ${}^{\prime }{v}^{\prime }$ is $(c=$ velocity of light in air)

• A. $\theta ={\tan }^{-1}\left(\frac{v}{c}\right)$
• B. $\theta ={\cot }^{-1}\left(\frac{v}{c}\right)$
• C. $\theta ={\sin }^{-1}\left(\frac{v}{c}\right)$
• D. $\theta ={\cos }^{-1}\left(\frac{v}{c}\right)$

Answer 1

The correct option is B $\theta ={\cot }^{-1}\left(\frac{v}{c}\right)$

We know $\tan \theta =\frac{n_2}{n_1}$ ....(1)

Speed of light in air $(n_1=1)$ is $c$.

Speed of light in a medium of refractive index $n_2$ is $v$.

Using $n_1\times c=n_2\times v$

$⟹$ $\frac{n_2}{n_1}=\frac{c}{v}$ ....(2)

From equations (1) and (2), we get $\tan \theta =\frac{c}{v}$

Or $\cot \theta =\frac{v}{c}$

$⟹$ $\theta =co{t}^{-1}(\frac{v}{c})$