Question

refractive index of water with respect to air is root2 .A light is incident on the surface at an angle 60 degrees travelling through water . The deviation of light ray is

Answer 1

Given:

Light is incident on water-air interface at an angle of 60° with the surface. Refractive Index of water is √2.

To find:

Net deviation of light ray.

Calculation:

We shall apply Snell's Law to find out the angle of refraction. With that angle , we can calculate the net deviation suffered by the light ray.

Whenever light ray passes from a denser medium like that of water to rarer medium like air , it is refracted in such a way that the light ray gets deviated away from the normal.

We need to find this deviation. (Refer to the diagram)

Since angle with surface is 60° , hence the angle of incidence will be (90°-60°) = 30°

Applying Snell's Law:

$\mu \times \sin( \angle i) = 1 \times \sin( \angle r)$

$= > \sqrt{2} \times \sin(30 \degree) = 1 \times \sin( \angle \: r)$

$= > \sqrt{2} \times \dfrac{1}{2} = \sin( \angle \: r)$

$= > \sin( \angle \: r) = \dfrac{1}{ \sqrt{2} }$

$= > \angle r = 45 \degree$

Now , let deviation be $\delta$

$\therefore \: \delta = \angle r - \angle i$

$= > \delta = 45 \degree - 30 \degree$

$= > \delta = 15 \degree$

So final answer :

Net deviation of light ray is 15°