Question:

A transparent solid cylindrical rod has a refractive index of 2√3. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod. The incident angle θ for which the light ray grazes along the wall of the rod is:

sin⁻¹(1/2)

sin⁻¹(1/√3).

sin⁻¹(2√3)

sin⁻¹(√3/2)

I × sin90 = 2√3sin(90 − α) ⇒ cosα = √3/2
So, sinα = √1−(√3/2)² = 1/2
Now, 1 × sinθ = 2√3sinα = 2√3 × 1/2 = √3 ⇒ θ = sin⁻¹√3