Solution:
Given,
Refractive index (μ) = 1.62,
The angle of refraction (r) =?
If θp be the polarizing angle, then
\(\tan\left(\theta_p\right)=\mu\)
Or, \(\theta_p=\tan^{-1}\left(\mu\right)\)
Or, \(\theta_p=\tan^{-1}\left(1.62\right)\)
Or, \(\theta_p=58.3^\circ\)
Again,
\(\theta_p+r=90^\circ\)
Or, \(r=90^\circ-\theta_p\)
Or, \(r=90^\circ-58.3^\circ\)
∴ r = 31.7°
Hence the required angle of refraction for the transmitted beam is r = 31.7°