Question

A ray of light is incident on one face of a transparent glass slab of thickness at an angle of incidence lateral displacement of the ray on emerging from the opposite parallel face is refractive index of the material of slab is:

Answer 1

Answer:

The lateral shift is 0.03 m.

Explanation:

Given that,

Incident angle = 45°

Thickness t = 10 cm

Refractive index n = 1.5

We know that,

n =\dfrac{sin\ i}{sin\ r}

1.5 = \dfrac{sin\ 45^{\circ}}{sin\ r}

sin\ r= \dfrac{1}{\sqrt{2}\times1.5}

sin\ r= 0.471

r = sin^{-1}( 0.471)

r = 28^{\circ}

The lateral shift is

L_{s}= \dfrac{t\ sin(i-r)}{\cos r}

L_{s}= \dfrac{0.1\times\ sin(45^{\circ}-28^{\circ})}{\cos 28^{\circ}}

L_{s}=0.03\ m

Hence, The lateral shift is 0.03m