A circular beam of light of diameter d=2cm falls on a plane surface of glass. The angle of incidence is 60∘ and refractive index of glass is μ=3/2. The diameter of the refracted beam is
A
4.00cm
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B
3.0cm
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C
3.26cm
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D
2.52cm
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Solution
The correct option is C3.26cm Let d′ be the diameter of refracted beam. Then, d=PQcos60∘ and d′=PQcosr i.e. d′d=cosrcos60∘=2cosr or d′=2dcosr sinr=siniμ=√3/23/2=1√3 ∴cosr=√1−sin2r=√23 ∴d′=(2)(2)√23 =4√23cm≈3.26cm.