Question

A ray of light enters a rectangular glass slab of refractive index route 3 at an angle of incidence 60 degree travels a distance of 6.0 cm inside the slab before emerging out of it in. the lateral displacement of the incident ray is ?

Answer 1

angle of incidence , i = 60°

refractive index of glass slab, $\mu$ = √3

from Snell's law,

$1\times sini=\mu sinr$

or, $sin60^{\circ}=\sqrt{3}sinr$

sinr = 1/2 = sin30° => r = 30°

now using formula, $L.D=t\frac{sin(i-r)}{cosr}$

where L.D is lateral displacement and t is thickness of slab.

here, t = 6cm, i = 60° and r = 30°

so, lateral displacement = 6cm × sin(60°-30°)/cos30°

= 6cm × sin30°/cos30°

= 6cm × 1/√3 = 2√3 cm

hence,lateral displacement of the incident ray is 2√3 cm

Answer 2

Snell's law,

1\times sini=\mu sinr1×sini=μsinr

or, sin60^{\circ}=\sqrt{3}sinrsin60

∘

=

3

sinr

sinr = 1/2 = sin30° => r = 30°

now using formula, L.D=t\frac{sin(i-r)}{cosr}L.D=t

cosr

sin(i−r)

where L.D is lateral displacement and t is thickness of slab.

here, t = 6cm, i = 60° and r = 30°

so, lateral displacement = 6cm × sin(60°-30°)/cos30°

= 6cm × sin30°/cos30°

= 6cm × 1/√3 = 2√3 cm

hence,lateral displacement of the incident ray is 2√3 cm