Question

Chapter 6: Refraction of Light
With a neat labelled diagram, prove that if the angle of
incidence and angle of emergence of a light ray falling on a glass
slab are i and e respectively, then i = e.
(3 marks)
llowing figure, SRPQ and NM is the refracted ray.​

Answer 1

Explanation:

hope you understand and this helps you

Answer 2

Given

SRQP is a rectangular glass slab

NM is the refracted ray

From the given figure

let the Refractive index of air be n₁ and that of glass be n₂

Ray AN is the incident ray at N with angle of incidence as ∠i

Ray NM is a refracted ray from N with angle of refraction as ∠r₁

SRQP is a rectangle, so PS || QR

Since normal to the surface is 90° i.e perpendicular

so, XY || X’Y’ ( Alternate angles are equal)

If XY || X’Y’

then, ∠r₁ = ∠r₂ ( Alternate interior angles)

Consider refraction at N

$\mathsf{\frac{sin ∠i}{sin ∠r_{1}} = \frac{n_{2}}{n_{1}} \:\rightarrow (1)}$

Consider refraction at M

$\mathsf{\frac{sin ∠r_{2}}{sin ∠e} = \frac{n_{1}}{n_{2}} \:\rightarrow (2)}$

Mulitiply (1) and (2):

$\mathsf{\implies\:\frac{sin ∠i}{sin ∠r_{1}} × \frac{sin ∠r_{2}}{sin ∠e}}$= $\mathsf{\frac{n_2}{n_1} × \frac{n_1}{n_2}}$

$\mathsf{\implies\:\frac{sin ∠i}{sin ∠r_{1}} × \frac{sin ∠r_{1}}{sin ∠e}}$= $\mathsf{\frac{n_2}{n_1} × \frac{n_1}{n_2}}$

Cancelling r₁ and n₁; we get

$\mathsf{\implies\:\frac{sin ∠i}{sin ∠e} = 1}$

$\mathsf{\implies\:sin ∠i = sin ∠e }$

$\fbox{\mathsf{\implies\: ∠i = ∠e }}$

We have proved that angle of incidence is equal to the angle of emergence