Question

A light ray Is Incident from medium-1 to medium-2. If the refractive indices of medium-1 and medium-2 are 1.5 and 1.36 respectively then determine the angle of refraction for an angle of incidence of 30°.​

Answer 1

Answer:

Refractive index of medium 1 = n1 = 1.5

Refractive index of medium 2 = n2 = 1.36

Angle of incidence = ∅1 = 30°

Angle of refraction = ∅2 = ?

We will use Snell's law in this question.

Snell's law states that the ration of sines of angles of incidence and refraction is constant.

The formula related to this law is : n1sin∅1 = n2sin∅2.

Using the same formula,

⇒ n1sin∅1 = n2sin∅2

Rearranging,

⇒ sin∅2 = (n1sin∅1)/n2

Putting the given values,

⇒ sin∅2 = (1.5×sin30°)/1.36

⇒ sin∅2 = (1.5×½)/1.36

⇒ sin∅2 = 0.75/1.36

⇒ sin∅2 = 0.55147

We know that, ½ = sin30°. Now, as the value we found is slightly greater than 0.5, the angle will be also slightly greater than 30°.

Hence, Angle of refraction = ∅2 = 33°

Answer 2

Given :-

A light ray Is Incident from medium-1 to medium-2. If the refractive indices of medium-1 and medium-2 are 1.5 and 1.36 respectively

To Find :-

The angle of refraction for an angle of incidence of 30°.​

Solution :-

By using snell law

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

$\sf 1.5 \times \sin 30^{\circ} = 1.36 \times \sin\theta_2$

$\sf 1.5\times \dfrac{1}{2} = 1.36\times\sin\theta_2$

$\sf 0.75 = 1.36 \times \sin\theta_2$

$\sf\dfrac{0.75}{1.36}=\sin\theta_2$

$\sf 0.55 = \sin\theta_{2}$

$\sf 34^{\circ} = \sin\theta_{2}$