In the given figure, the face A C of the equilateral prism is immersed in a liquid of refractive index 'n'. For incident angle 60° at the side A C, the refractive light beam just grazes along face A C. The refractive index of the liquid n=√x/4. The value of x is ____________.
(Given refractive index of glass =1.5 )

Solution

Physics sir · 4 steps

Grazing along AC means 90° in the liquid, so 60° is the critical angle here.

1. Read the figure. The ray travels inside the glass (μ = 1.5) and meets face AC at 60° to the normal. The liquid of index n lies on the other side of AC.
2. Grazing emergence. The refracted ray just grazes along AC, so the angle of refraction in the liquid is 90°. That makes 60° exactly the critical angle for the glass to liquid boundary.

   1.5 sin 60° = n sin 90°

3. Solve for n. n = 1.5 × √3/2 = 3√3/4 = √27/4 ≈ 1.3. This is less than 1.5, as it must be for a critical angle to exist.
4. So. Comparing with n = √x/4 gives x = 27.

The common slip: 48. Swapping the media and writing n sin 60° = 1.5 gives n = √3 = √48/4, so x = 48. The 60° is inside the glass, so it belongs with 1.5.

The idea to keep. Grazing emergence means the refraction angle is 90°, so μ_(denser) sin θ_c = μᵣₐᵣₑᵣ. Always put the index of the medium the ray is coming from with the incidence angle.

Answer from the NTA final answer key. Solution by PrepWiser.