Question

An equilateral prism of glass of refractive index 1.45 is immersed in water of refractive index 1.33. A narrow beam of light falls normally on one face of the prism. Find the angle of emergence of the beam.

Answer 1

Given:

Refractive index of glass, μ_g = 1.45

Refractive index of water, μ_w = 1.33

Angle of prism A = 60° (since prism is equilateral)

Light incident normally on first face ⇒ i₁ = 0°

To Find:

Angle of emergence e

Calculation:

Step 1: Refraction at first face:

Since light falls normally, it goes undeviated:

r₁ = 0°

Use the prism angle relation:

r₁ + r₂ = A

r₂ = A = 60°

Step 2: Use Snell’s law at the second face (glass to water):

μ_g sin r₂ = μ_w sin e

1.45 sin (60°) = 1.33 sin e

1.45 × 0.866 = 1.33 sin e

sin e = `1.2557/1.33`

sin e = 0.9444

Step 3: Find angle of emergence:

e = sin⁻¹ (0.9444)

e = 70.8°