The polarising angle for a transparent medium is '$\theta$' and 'V' is the speed of light in that medium,…
- $\theta=\sin ^{-1}\left(\frac{\mathrm{V}}{\mathrm{c}}\right)$
- $\theta=\tan ^{-1}\left(\frac{\mathrm{V}}{\mathrm{c}}\right)$
- $\theta=\cot ^{-1}\left(\frac{\mathrm{V}}{\mathrm{c}}\right)$
- $\theta=\cos ^{-1}\left(\frac{\mathrm{V}}{\mathrm{c}}\right)$
Solution
We know \(\tan \theta=\frac{n_2}{n_1}\) ...(1)
Speed of light in air \(\left(n_1=1\right)\) is \(c\)
Speed of light in a medium of refractive index \(n_2\) is \(v\).
Using \(n_1 \times c=n_2 \times v\)
\(\Rightarrow \frac{n_2}{n_1}=\frac{c}{v}\)
From equations (1) and (2), we get \(\tan \theta=\frac{c}{v}\)
Or \(\cot \theta=\frac{v}{c}\)
\(\Rightarrow \theta=\cot ^{-1}\left(\frac{v}{c}\right)\)
Asked in: MHT CET 2020 (15 Oct Shift 2)