Physics - Wave Optics Question with Solution | TestHub

According to the Brewster's Law

$\Rightarrow \tan \theta =\frac{{\mu }_{\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{e}\mathrm{r}}}{{\mu }_{\mathrm{r}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{r}}}$

Or$\tan \theta =\frac{1}{\sin C_2}\times \frac{\sin C_1}{1}$

[$\because \mu =\frac{1}{\sin C},$where,$C$is the critical angle]

$\Rightarrow \tan \theta =\frac{\sin C_1}{\sin C_2}$