Physics - Geometrical Optics Question with Solution | TestHub

The critical angle is that angle on incidence in the denser medium for which the angle of refraction in rarer medium is${90}^o,i.e.$
$\frac{\mathrm{s}\mathrm{i}\mathrm{n}\theta }{\mathrm{s}\mathrm{i}\mathrm{n}{90}^o}=\frac{{\mu }_y}{{\mu }_x}$
$\Rightarrow \mathrm{s}\mathrm{i}\mathrm{n}\theta =\frac{{\mu }_y}{{\mu }_x}$…. (i)
Where,$\mu =$refractive index of medium.
$Also,\mu =\frac{speedoflightinvacuum}{speedoflightinmedium}=\frac{c}{v}$… (ii)
Form Eqs. (i) and (ii), we get
$\mathrm{s}\mathrm{i}\mathrm{n}\theta \frac{c}{v_y}\times \frac{v_x}{c}=\frac{v_x}{v_y}\Rightarrow v_y=\frac{v_x}{\mathrm{s}\mathrm{i}\mathrm{n}\theta }$