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Refraction and Snell's law

Waves, sound and light · level 7, lesson 1 of 6

A question to start

An optical fibre carries light round corners, for kilometres, with almost no loss. How can light be trapped inside a thin strand of glass?

The idea

When light crosses from a medium of refractive index \(n_1\) into one of index \(n_2\), the angles to the normal obey Snell's law:

\[n_1 \sin\theta_1 = n_2 \sin\theta_2.\]

Going from a slower medium (bigger \(n\)) to a faster one, the ray bends away from the normal. At one angle, the critical angle \(\theta_c\), the refracted ray runs along the surface:

\[\sin\theta_c = \frac{n_2}{n_1}.\]

Beyond the critical angle, no light gets out: it is all reflected. This is total internal reflection, and it traps light inside an optical fibre.

airglassrefracted outcritical angletotally reflected

Worked examples

1. Air into water. A ray hits water at 40° to the normal: \(\sin\theta_2 = \sin 40°/1.33 = 0.483\), so \(\theta_2 = 28.9°\).

2. Critical angle of glass (\(n = 1.5\)) into air: \(\sin\theta_c = 1/1.5 = 0.667\), so \(\theta_c = 41.8°\).

3. Water into air: \(\sin\theta_c = 1/1.33\), so \(\theta_c = 48.8°\). A fish looking up sees the outside world squeezed into a circle overhead.

Try it

  1. State Snell's law.
  2. Light goes from air into glass (\(n = 1.5\)) at 60° to the normal. Find the angle in the glass.
  3. Find the critical angle for diamond, \(n = 2.42\).
  4. Why can total internal reflection only happen going from a slower medium into a faster one?
Answers
  1. \(n_1 \sin\theta_1 = n_2 \sin\theta_2\).
  2. \(\sin\theta_2 = 0.866/1.5 = 0.577\): \(\theta_2 = 35.3°\).
  3. \(\sin\theta_c = 1/2.42 = 0.413\): \(\theta_c = 24.4°\).
  4. Only then does the ray bend away from the normal, so it can reach 90° and beyond.

Common mistake

Using the angle to the surface. Snell's law uses angles to the normal. A ray 30° above the surface is at 60° to the normal.

Where it leads