KinexiaLEARN PHYSICS

Snell's Law — Water to Air (Worked Example)

Medium
Problem

A diver's torch beam underwater (n = 1.33) hits the water surface at 40° to the normal. At what angle does it emerge into the air?

Given

  • n₁ = 1.33 (water)

  • n₂ = 1.00 (air)

  • θ₁ = 40°

Find

The refracted angle θ₂ in air.

Principle

Going to a rarer medium bends the ray away from the normal — θ₂ > θ₁.

Formula

Solution

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

  2. Substitute: sin⁡θ2=1.33×sin⁡40°=1.33×0.6428=0.8549\sin\theta_2 = 1.33 \times \sin40° = 1.33 \times 0.6428 = 0.8549

  3. Arcsin: θ2=arcsin⁡(0.8549)=58.8°\theta_2 = \arcsin(0.8549) = 58.8°

58.8° from the normal

Unit check

Degrees throughout — consistent.

What this means physically

58.8° — spread well away from the normal. Past the critical angle (48.8°) this beam would not escape at all, which is why the water surface looks mirror-like from below at shallow angles.

Common mistake

Getting θ₂ < θ₁. Leaving a dense medium always bends away from the normal — if your answer is smaller, recheck.