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)

Ans: C

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)