Question

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

Answer 1

Given,

Refrective indices of glass and water are $1.62,1.32$

Snell's Law is ${\mu }_1\sin {\theta }_1={\mu }_2\sin {\theta }_2$.

For a ray going from water to glass,

the critical angle is

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

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

$={54.57}^0$