Refraction, Snell's law and dispersion: VCE Physics Unit 1

Syllabus dot point

“Apply the wave model of light to refraction using Snell's law, total internal reflection and the critical angle, colour dispersion in prisms and lenses, rainbows, mirages and optical fibres”

7 min readUpdated 7 Oct 2026

Quick answer

Light refracts (changes direction) when its speed changes at a boundary. Snell's law is $n_1\sin\theta_1 = n_2\sin\theta_2$ with $n = c/v$ . Light trying to pass into a lower-index medium is totally internally reflected beyond the critical angle $\theta_c$ where $n_1\sin\theta_c = n_2\sin 90^\circ$ . Dispersion happens because each colour has a slightly different refractive index.

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What this dot point is asking
Refraction and Snell's law
Total internal reflection
Dispersion, rainbows, mirages and fibres
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What this dot point is asking

In Unit 1, Area of Study 1 (How are light and heat explained?), the VCE Physics Study Design asks you to analyse refraction with Snell's law, total internal reflection and the critical angle, colour dispersion in prisms and lenses, the formation of rainbows and mirages, and light transmission through optical fibres.

This page is a short summary. The longer treatment, with more worked examples, is on Refraction, Snell's law and dispersion, a page that keeps its older Unit 4 address. Polarisation of light, covered on Polarisation and Malus's law, is an extension to this topic and not part of the Unit 1 key knowledge.

Refraction and Snell's law

The refractive index of a medium is $n = c/v$ , where $v$ is the speed of light in the medium. Air is close to $1.00$ , water about $1.33$ and glass about $1.5$ .

When light passes from medium 1 to medium 2, its frequency stays the same (set by the source) and its speed changes, so its wavelength changes too. The ray bends according to Snell's law:

n_1\sin\theta_1 = n_2\sin\theta_2 \qquad\text{and}\qquad n_1v_1 = n_2v_2

Angles are measured from the normal. Light entering a higher-index medium bends towards the normal. Light entering a lower-index medium bends away.

Total internal reflection

Going from a higher index $n_1$ to a lower index $n_2$ , the refracted ray bends away from the normal. At the critical angle $\theta_c$ it runs along the boundary, so

n_1\sin\theta_c = n_2\sin 90^\circ \quad\Rightarrow\quad \sin\theta_c = \frac{n_2}{n_1}

For any larger angle of incidence there is no refracted ray and all the light is reflected. This is total internal reflection.

Dispersion, rainbows, mirages and fibres

Dispersion. The refractive index of a material depends slightly on frequency, and is a little higher for violet than for red. White light entering a prism is therefore split into colours, with violet bent the most. Lenses do the same, which blurs colours at the edges of images.
Rainbows. Sunlight refracts into a raindrop, is reflected off the back surface and refracts out again. Dispersion separates the colours, so each colour leaves at a slightly different angle.
Mirages. Air near a hot road is warmer and less dense, so has a lower index. Light from the sky heading towards the road is gradually bent upwards, so the sky appears as a shimmering "pool" on the ground.
Optical fibres. A glass core is surrounded by cladding of lower index. Light meets the core-cladding boundary at angles larger than $\theta_c$ and is totally internally reflected along the fibre, carrying communication signals with little loss.

Worked example

(a) Light in air strikes glass ( $n = 1.50$ ) at $40^\circ$ to the normal. Find the angle of refraction and the speed in the glass.

$\sin\theta_2 = 1.00\sin 40^\circ/1.50 = 0.4285$ , so $\theta_2 = 25^\circ$ .

$v = c/n = 3.00 \times 10^8/1.50 = 2.00 \times 10^8$ m s $^{-1}$ .

(b) An optical fibre has a core of index $1.50$ and cladding of index $1.45$ . Find the critical angle at the core-cladding boundary.

$\sin\theta_c = 1.45/1.50 = 0.967$ , so $\theta_c = 75^\circ$ . Rays hitting the boundary at more than $75^\circ$ to the normal are totally internally reflected.

Common errors

Measuring angles from the surface: Snell's law uses angles from the normal.
Using the critical angle for the wrong direction: Total internal reflection needs light to travel from the higher-index medium to the lower. It cannot happen entering glass from air.
Saying frequency changes in refraction: Speed and wavelength change. Frequency does not.

Try this

Q1. Light travels from water ( $n = 1.33$ ) to air. Calculate the critical angle. [2 marks]

Cue. $\sin\theta_c = 1.00/1.33 = 0.752$ , so $\theta_c = 48.8^\circ$ .

Q2. Explain why white light splits into colours in a glass prism. [2 marks]

Cue. The refractive index of glass is different for different frequencies (higher for violet), so each colour is refracted by a different angle.

Q3. Explain why light stays inside an optical fibre. [2 marks]

Cue. The core has a higher index than the cladding and the light meets the boundary at more than the critical angle, so it is totally internally reflected.