Question

A ray of light is incident on an air/water interface.

Answer 1

Answer:

$\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})$

Explanation:

Given that,

The ray makes an angle of 32 degrees with respect to the normal of the surface.

The refractive index of air, $n_1=1$

The refractive index of water, $n_2=1.33$

Snell's law is given by :

$n_1\ sin\theta_1=n_2\ sin\theta_2$

$sin\theta_2=\dfrac{n_1\ sin\theta_1}{n_2}$

$\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})$

So, option (4) is correct. Hence, this is the required solution.