A ray of light refracts from medium 1 into a thin layer of medium 2, crosses the layer and is incident at the critical angle on the interface between the medium 2 and 3 as shown in the figure. If the angle of incidence of ray is θ, the value of θ is
sin−1(89)
sin−1(1318)
sin−1(813)
sin−1(1316)
A
sin−1(89)
B
sin−1(1318)
C
sin−1(1316)
D
sin−1(813)
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Solution
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Applying Snell's Law at interface 2:
μ3μ2=sin(θc)......(i)
Using equality of alternate angles,
θ2=θc
Applying Snell's Law at interface 1:
μ2μ1=sin(θ)sin(θc)..........(ii)
From (i) and (ii), sin(θ)=μ3μ1=1.31.6=1316
θ=sin−1(1316)
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