Question

10.
A ray of light enters a rectangular glass slab of refractive index √3 at an angle of
incidence 60°'. It travels a distance of 5 cm inside slab and emerges out of slab. What is
I distance between incident and emergent ray?​

Answer 1

Answer:

angle of incidence , i = 60°

refractive index of glass slab, \muμ = √3

from 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

Explanation:

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