KinexiaLEARN PHYSICS

Snell's Law — Light Leaving Water (Worked Example)

Medium
Problem

A ray of light travels upward through water (n₁ = 1.33) and hits the water–air surface at 30° to the normal. At what angle does it emerge into the air (n₂ = 1.00)?

Given

  • n₁ = 1.33 (water)

  • n₂ = 1.00 (air)

  • θ₁ = 30° from the normal

Find

Refraction angle θ₂ in air.

Principle

Snell's law: when light crosses a boundary, n·sinθ is conserved. Moving to a lower index bends the ray away from the normal.

Formula

Solution

  1. Rearrange: sin⁡θ2=(n1/n2)sin⁡θ1\sin\theta_2 = (n_1/n_2)\sin\theta_1

  2. Substitute: sin⁡θ2=(1.33/1.00)×sin⁡30°=1.33×0.5\sin\theta_2 = (1.33/1.00) \times \sin30° = 1.33 \times 0.5

  3. So sin⁡θ2=0.665\sin\theta_2 = 0.665

  4. Inverse sine: θ2=sin⁡−1(0.665)=41.7°\theta_2 = \sin^{-1}(0.665) = 41.7°

≈ 41.7° from the normal

Unit check

Refractive indices are unitless and sine is unitless, so θ₂ is a pure angle — consistent.

What this means physically

The ray bends away from the normal, from 30° to 41.7°. This bending is why a straw in a glass looks kinked at the surface — and past the critical angle (~48.6°), the light cannot escape at all.

Common mistake

Measuring the angle from the surface instead of the normal. Snell's law always uses angles from the normal — the 30° here is already correct.