Refraction of light and Snell's law

Key ideas

Light refracts (bends) when it passes between materials because it travels at different speeds in each. Every material has a refractive index $n$ — a number that says how much it slows light down (bigger $n$ = slower light = more bending).

All angles are measured from the normal — the line at right angles to the surface.

$A light ray hitting a glass block at an angle, with the normal drawn as a dashed line, the angle of incidence in air and the smaller angle of refraction in the glass both measured from the normal.$

Snell's law

The relationship between the angles and refractive indices is Snell's law:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

$n_1$ — refractive index of the first material
$\theta_1$ — angle of incidence (in material 1, from the normal)
$n_2$ — refractive index of the second material
$\theta_2$ — angle of refraction (in material 2, from the normal)

Speed, wavelength and refractive index

The refractive index also links the speeds and wavelengths in the two materials:

$\frac{n_1}{n_2} = \frac{v_2}{v_1} = \frac{\lambda_2}{\lambda_1}$

A higher refractive index means lower speed and shorter wavelength.
Frequency is unchanged (as always across a boundary).

:::tip Watch which way the ray bends. Going into a denser material (larger $n$ ) light slows and bends toward the normal, so $\theta_2 < \theta_1$ . If your calculation gives a bigger angle in the denser material, you have swapped $n_1$ and $n_2$ — check the subscripts. :::

Worked ExampleAngle of refraction entering glass

Light travels from air ( $n_1 = 1.00$ ) into glass ( $n_2 = 1.50$ ), striking the surface at $40°$ to the normal. Find the angle of refraction in the glass.

Step 1 — Write Snell's law and substitute

$n_1 \sin\theta_1 = n_2 \sin\theta_2$ $1.00 \times \sin 40° = 1.50 \times \sin\theta_2$

Step 2 — Solve for $\sin\theta_2$

$\sin\theta_2 = \frac{1.00 \times 0.643}{1.50} = 0.4285$

Step 3 — Take the inverse sine

$\theta_2 = \sin^{-1}(0.4285) = 25.4°$

Practice question

Light passes from water ( $n = 1.33$ ) into air ( $n = 1.00$ ) at an angle of $30°$ to the normal in the water. Find the angle in the air.

Worked solution: $1.33\sin30° = 1.00\sin\theta_2$ , so $\sin\theta_2 = 0.665$ , giving $\theta_2 = 41.7°$ (bends away from the normal, into the less dense air).

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Light passes from air ( $n = 1.00$ ) into a plastic of refractive index $1.40$ at an angle of incidence of $30°$ .

Calculate the angle of refraction.

Merit

Light of wavelength $600$ nm in air ( $n = 1.00$ ) enters glass of refractive index $1.50$ .

Calculate the wavelength of the light inside the glass.

Excellence

A ray of light enters a glass block, travels through it, and exits the far parallel side back into air. Explain fully, using Snell's law, why the ray leaves parallel to the direction it entered, and why it is shifted sideways.

Refraction of light and Snell's law

Covers AS91170

Key ideas

All angles are measured from the normal — the line at right angles to the surface.

$A light ray hitting a glass block at an angle, with the normal drawn as a dashed line, the angle of incidence in air and the smaller angle of refraction in the glass both measured from the normal.$

Snell's law

The relationship between the angles and refractive indices is Snell's law:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

$n_1$ — refractive index of the first material
$\theta_1$ — angle of incidence (in material 1, from the normal)
$n_2$ — refractive index of the second material
$\theta_2$ — angle of refraction (in material 2, from the normal)

Speed, wavelength and refractive index

The refractive index also links the speeds and wavelengths in the two materials:

$\frac{n_1}{n_2} = \frac{v_2}{v_1} = \frac{\lambda_2}{\lambda_1}$

A higher refractive index means lower speed and shorter wavelength.
Frequency is unchanged (as always across a boundary).

Worked ExampleAngle of refraction entering glass

Light travels from air ( $n_1 = 1.00$ ) into glass ( $n_2 = 1.50$ ), striking the surface at $40°$ to the normal. Find the angle of refraction in the glass.

Step 1 — Write Snell's law and substitute

$n_1 \sin\theta_1 = n_2 \sin\theta_2$ $1.00 \times \sin 40° = 1.50 \times \sin\theta_2$

Step 2 — Solve for $\sin\theta_2$

$\sin\theta_2 = \frac{1.00 \times 0.643}{1.50} = 0.4285$

Step 3 — Take the inverse sine

$\theta_2 = \sin^{-1}(0.4285) = 25.4°$

Practice question

Light passes from water ( $n = 1.33$ ) into air ( $n = 1.00$ ) at an angle of $30°$ to the normal in the water. Find the angle in the air.

Worked solution: $1.33\sin30° = 1.00\sin\theta_2$ , so $\sin\theta_2 = 0.665$ , giving $\theta_2 = 41.7°$ (bends away from the normal, into the less dense air).

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Light passes from air ( $n = 1.00$ ) into a plastic of refractive index $1.40$ at an angle of incidence of $30°$ .

Calculate the angle of refraction.

Merit

Light of wavelength $600$ nm in air ( $n = 1.00$ ) enters glass of refractive index $1.50$ .

Calculate the wavelength of the light inside the glass.

Excellence

Refraction of light and Snell's law

Key ideas

Snell's law

Speed, wavelength and refractive index

Step 1 — Write Snell's law and substitute

Step 2 — Solve for sin⁡θ2\sin\theta_2sinθ2​

Step 3 — Take the inverse sine

Practice question

Test yourself

Practice by grade

Refraction of light and Snell's law

Key ideas

Snell's law

Speed, wavelength and refractive index

Step 1 — Write Snell's law and substitute

Step 2 — Solve for sin⁡θ2\sin\theta_2sinθ2​

Step 3 — Take the inverse sine

Practice question

Test yourself

Practice by grade

Step 2 — Solve for $\sin\theta_2$

Step 2 — Solve for $\sin\theta_2$