Question

Find the critical angle for a ray of light at glass water interface if refractive indices for glass and water are $1.62$ and $1.32$ respectively.

Answer 1

From Snell's law,

${\mu }_1\sin {\theta }_1={\mu }_2\sin {\theta }_2$

The critical angle,

${\theta }_c={\sin }^{-1}\left(\frac{{\mu }_1}{{\mu }_2}\right)$

$={\sin }^{-1}\left(\frac{1.32}{1.62}\right)$

$\therefore {\theta }_c={54.57}^{\circ }$

Hence, this is the answer.