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# A ray incident at a point at an angle of incidence 60∘ enters a glass sphere of μ=√3 and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :50 ∘90∘60∘40∘

A
40
B
90
C
50
D
60
Solution
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#### Using Snell's Law at point P we get:-sin 60osin r1=√3sin r1=12⟹r1=30oAs we know that r1=r2⟹r2=30oUsing Snell's Law at point Q we get:-sin 30osin i2=1√3⟹i2=60oSince reflection at point Q occurs⟹r′2=r2=60oLet angle between the refracted ray and reflected ray be αα=180o−(r′2+r2)=90oHence option (B) is correct.

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