Question

A parallel sided glass slab of
thickness 4cm is made of a
material of refractive index is
v3.When light is incident on
one of parallel face at an
angle of 60', it emerge from
the other parallel face. Find
the lateral displacement of
the emergent beam in cm.​

Answer 1

Given :

The thickness of glass slab = t = 4 cm

The refractive index = $\mu$ = √3

The angle of incident = i = 60°

Light emerge from  the other parallel face

To Find :

The lateral displacement

Solution :

Let The Lateral displacement = L cm

Let The angle of incident = i°

The angle of reflection = r°

∵ Sin r = $\dfrac{Sin i}{\mu}$

          = $\dfrac{Sin60^{\circ}}{\sqrt{3} }$

          = $\dfrac{\sqrt{3} }{\sqrt{3} }$  = 1

Or,  r = $Sin^{-1}$ 1

         = 45°

∵ Lateral displacement = $\dfrac{t Sin ( i - r )}{Cos r}$

                                       = $\dfrac{4 \times Sin (60^{\circ}-45^{\circ})}{Cos45^{\circ}}$

                                       = $\dfrac{4 \times Sin (15^{\circ})}{Cos45^{\circ}}$

                                       = $\dfrac{4 \times 0.25}{0.707}$

                                       = $\dfrac{1}{0.707}$ =  $\sqrt{2}$ cm

i.e Lateral displacement = L = 1.414 = $\sqrt{2}$ cm

Hence, The lateral displacement is  $\sqrt{2}$ cm   . Answer