Q. A transparent solid cylindrical rod has a refractive index of 2/√3. It is surrounded by air. A light ray is incident at the midpoint of one end of the rod as shown in the figure. The incident angle θ for which the light ray grazes along the wall of the rod is

(a) sin^{-1}(√3/2)

(b) sin^{-1}(2/√3)

(c) sin^{-1}(1/√3)

(d) sin^{-1}(1/2)

Ans:(C)

Sol:Applying Snell’s Law for Air & medium inside the cylinder

1× sinθ = (2/√3)×sinr

Sinr = (√3/2 )sinθ —-(i)

Again by applying snell’s law for medium inside cylinder & Air

(2/√3 )sin(90-r) = sin90

cosr = √3/2 => r = 30

From (i) sin30 = √3/2 sinθ

sinθ = 1/ √3 => θ= sin^{-1}(1/√3 )