If velocity of light in air is $3\times {10}^8$ m/s and that in water is$2\times {10}^{8}m/s$ , then what would be the critical angle?

${}^a{\mu }_w=\frac{\text{velocity of light in air }}{\text{velocity of light in water}}$

$=\frac{3\times {10}^8}{2\times {10}^8}=\frac{3}{2}$

$\sin C=\frac{1}{{}^a{\mu }_w}$        [where c is the critical angle ]

$\therefore$ $\sin C=\frac{1}{3/2}=\left(\frac{2}{3}\right)$  or $C={\sin }^{-1}\left(\frac{2}{3}\right)$