Question

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

Answer 1

Snell's Law is $\mu_1 \sin \theta_1=\mu_2 \sin \theta_2$.

For a ray going from water to glass,

If $\mu_1 <\mu_2$, the critical angle is

$\theta_c=\sin^{-1}(\frac{ \mu_1}{\mu_2} )\\ \theta_c=\sin^{-1}(\frac{ 1.32}{1.62} )\\ \theta_c=54.57^o$

Answer 2

Hello mate,

◆ Critical angle-
The angle at which the incident ray after refraction grazes is called the critical angle.

◆ Answer-
Critical angle = 54.57°

◆ Explaination-
# Given-
μi = 1.62
μr = 1.32

# Solution-
According to Snell's law for refraction,
μi.sini = μr.sinr

At critical angle of incidence, i = c, r = 90°,
μi.sinc = μr.sin90
sinc = μr / μi
sinc = 1.32 / 1.62
sinc = 0.8148

Taking sin inverse,
c = 54.57°

Hope this helps...