Physics - Wave Optics Question with Solution | TestHub

Brewster's angle for light propagating from medium 1 to medium 2 is given by $\tan {\theta }_B=\frac{{\mu }_2}{{\mu }_1}$.

For light propagating from air to glass, $\tan \theta =\frac{\mu }{1}=\mu$.

For light propagating from glass to air, let the Brewster's angle be ${\theta }^{\prime }$.

$\tan {\theta }^{\prime }=\frac{1}{\mu }$

$\tan {\theta }^{\prime }=\frac{1}{\tan \theta }=\cot \theta =\tan \left(\frac{\pi }{2}-\theta \right)$

${\theta }^{\prime }=\frac{\pi }{2}-\theta$

Thus, if Brewster's angle for light propagating from air to glass is $\theta$, then Brewster's angle for light propagating from glass to air is $\left(\frac{\pi }{2}-\theta \right)$.

Answer: If the Brewster's angle for the light propagation from air to glass is ' $\theta$ ', then Brewster's angle for the light propagating from glass to air is $\left(\frac{\pi }{2}-\theta \right)$.