Question

A ray of light traveling in water (n = 1.33) is incident at the flat surface of a block of glass (n = 1.60). If the incident ray in water makes an angle of 58.6° with the normal to the interface, what angle does the ray that is refracted into the glass make with the normal?

Answer 1

Answer:

$45.19^{\circ}$

Explanation:

We are given that

$n_1=1.33$

$n_2=1.6$

$\theta_1=58.6^{\circ}$

We have to find the angle of refraction.

By Snell's law

$n_1sin\theta_1=n_2sin\theta_2$

Substitute the values

$1.33sin58.6=1.6sin\theta_2$

$sin\theta_2=\frac{1.33sin58.6}{1.6}$

$sin\theta_2=0.7095$

$\theta_2=sin^{-1}(0.7095)=45.19^{\circ}$

Hence, the angle mad by refracted ray in to the glass with normal=$45.19^{\circ}$