Question

A ray of light is incident at an angle of 60o on one surface of a glass plate
4 cm thick. Find the later displacement between the incident ray and
emergent ray. (refractive index of glass = 1.5)

Answer 1

The lateral displacement is 0.513 cm

Explanation:

Given:

The refractive index of glass n = 1.5

A ray of light is incident at an angle of 60° on the glass plate 4 cm thick

To find out:

The lateral displacement between the incident ray and the emergent ray

Solution:

Here

$i=60^\circ$

$n=1.5$

$t=4$ cm

From Snell's law

$n=\frac{\sin i}{\sin r}$

$\implies 1.5=\frac{\sin 60^\circ}{\sin r}$

$\implies \sin r=\frac{\sqrt{3}}{2\times 1.5}$

$\implies \sin r=\frac{\sqrt{3}}{3}$

$\implies \sin r=\frac{1}{\sqrt{3}}$

$\implies \sinr=0.577$

$\implies r=\sin^{-1}0.577\approx 35.24^\circ$

We know that for a t thickness of the plate the lateral shift d is given by

$d=t\frac{\sin(i-r)}{\cos r}$

Therefor, the lateral shift

$d=4\times\frac{\sin (60^\circ-35.24^\circ}{\cos 35.24^\circ}$

$\implies d=4\times\frac{\sin 24.76^\circ}{\cos 35.24^\circ}$

$\implies d=4\times\frac{0.419}{0.817}$

$\implies d=0.513$ cm

Hope this answer is helpful.

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