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 :
A
50∘
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B
90∘
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C
60∘
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D
40∘
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Solution
The correct option is B90∘
Using Snell's Law at point P we get:-
sin60osinr1=√3
sinr1=12⟹r1=30o
As we know that r1=r2⟹r2=30o
Using Snell's Law at point Q we get:-
sin30osini2=1√3⟹i2=60o
Since reflection at point Q occurs
⟹r′2=r2=60o
Let angle between the refracted ray and reflected ray be α