Q. 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 shown in the figure. If the angle of incidence of ray is θ , the value of θ is
(a)sin-1(8/9)
(b) sin-1(13/18)
(c) sin-1(13/16)
(d) sin-1(8/13)
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Ans: (C)
Sol: Applying Snell’s Law for medium 1 & 2
1.6 sinθ =1.8 sinr , where r= angle of refraction
Sinr = 16 sinθ /18 = 8 sinθ/9 ….(i)
Applying Snell’s Law for medium 2 & 3
1.8 sinr = 1.3 sin90 , Where r= angle of incidence
Sinr = 13/18 ….(ii)
8 sinθ/9 = 13/18 from (i)
=> sinθ = 13×9/18×8 = 13/16
θ = sin-1(13/16)