KinexiaLEARN PHYSICS

Snell's Law — Fiber Optic Cladding (Worked Example)

Hard
Problem

Light travels in a glass fibre core (n = 1.5) and strikes the core–cladding boundary (cladding n ≈ 1.0) at 50° to the normal. Does it stay in the fibre? What is the critical angle?

Given

  • n₁ = 1.5

  • n₂ = 1.0

  • θ₁ = 50°

Find

Critical angle and whether TIR occurs.

Principle

Fibre optics work by total internal reflection: rays steeper than θc = arcsin(n₂/n₁) cannot escape.

Formula

Solution

  1. Critical angle: θc=arcsin⁡(1.0/1.5)=arcsin⁡(0.667)=41.8°\theta_c = \arcsin(1.0/1.5) = \arcsin(0.667) = 41.8°

  2. Check: sin⁡θ2=1.5×sin⁡50°=1.5×0.766=1.149>1\sin\theta_2 = 1.5 \times \sin50° = 1.5 \times 0.766 = 1.149 > 1

  3. No refraction possible — the ray is totally internally reflected

  4. Since 50° > 41.8°, the light stays trapped in the core

Trapped by TIR (critical angle 41.8°)

Unit check

Degrees throughout — consistent.

What this means physically

Trapped — the ray zigzags down the fibre with almost no loss, carrying internet data as light pulses. This is why your broadband works.

Common mistake

Measuring the angle from the surface. Snell's law uses the angle from the normal — 50° to the surface would be 40° to the normal, a different answer.