A circular beam of light having a diameter 8 cm falls on a plane glass slab n=1.5 at an angle of incidence 60o. the diameter of the refracted beam is :
Answers
Given : A circular beam of light having a diameter 8 cm falls on a plane glass slab at angle of incidence is 60° . if the refractive index of the material of slab is 1.5.
To find : The diameter of the refracted beam.
solution : see ray diagram,
let d' is the diameter of reflected beam.
here, d = PQ cos60°
and d' = PQ cos r [ r is angle of reflection ]
now d'/d = (PQ cosr)/(PQ cos60°) = 2cosr
⇒d' = 2d cosr ........(1)
from Snell's law,
1 × sini = μ × sinr
⇒sinr = sini/μ = sin60°/(1.5) = 1/√3
⇒cosr = √(1 - sin²r) = √(1 - 1/√3²) = √(2/3)
putting, cos r = √(2/3) in equation (1) we get,
d' = 2d × √(2/3)
= 2 × 8 × √(2/3) [ given d = 4cm ]
= 16√(2/3) cm
Therefore the diameter of the reflected beam is 16√(2/4) cm
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