Refraction and Snell's Law Revision Notes | A Level AQA Physics

1. Why Light Bends at a Boundary

Picture a ray of light travelling through air and then entering a glass block at an angle. The moment it crosses the surface, it changes direction — bending towards the normal as it slows down. That bending is refraction, and understanding it precisely is what this lesson is about.

The key physical reason is straightforward: light travels at different speeds in different materials. When a wavefront hits a boundary at an angle, one side of the wavefront slows down before the other, causing the whole wavefront to pivot. The result is a change in direction.

This is directly analogous to a car driving from tarmac onto mud at an angle — the wheel that hits the mud first slows down, and the car turns. The physics of refraction follows exactly the same geometry.

2. Refractive Index — Quantifying How Much Light Slows

To describe how much a material slows light, physicists use the refractive index, given the symbol $n$. It is defined as the ratio of the speed of light in a vacuum, $c$, to the speed of light in the substance, $c_s$.

Because light can never travel faster than $c$, the value of $c_s$ is always less than or equal to $c$. This means $n$ is always greater than or equal to 1. A higher refractive index means light travels more slowly in that material and bends more when it enters.

This approximation matters enormously in calculations. When one medium is air, Snell's law simplifies considerably — you will see this in action shortly. For now, commit this to memory: $n_{\text{air}} \approx 1$.

Some typical values worth knowing: glass has $n \approx 1.5$, water has $n \approx 1.33$, and diamond has $n \approx 2.4$. These numbers tell you that light travels roughly half as fast in diamond as it does in a vacuum.

3. Snell's Law — The Equation That Governs Refraction

Knowing that light bends at a boundary is useful, but to actually calculate angles and indices you need Snell's law. It connects the refractive indices of two media to the angles the ray makes with the normal on each side of the boundary.

The equation tells you something elegant: the product $n \sin \theta$ is conserved across a boundary. If $n_2 > n_1$ (light entering a denser medium), then $\sin \theta_2 < \sin \theta_1$, so $\theta_2 < \theta_1$ — the ray bends towards the normal. If $n_2 < n_1$ (light entering a less dense medium), the ray bends away from the normal.

When one medium is air ($n_1 = 1$), Snell's law becomes $\sin \theta_1 = n_2 \sin \theta_2$. This is the form you will use most often in AQA questions, so recognise it quickly.

4. Worked Calculations Using Snell's Law

Theory only becomes useful when you can execute it under exam conditions. The following worked examples cover the two most common question types: finding an angle of refraction and finding a refractive index. Work through each step before revealing the next — that active engagement is what builds exam fluency.

Worked example: Finding the Angle of Refraction (Air to Glass)

A ray of light travels from air into a glass block. The angle of incidence is 40°. The refractive index of the glass is 1.50. Calculate the angle of refraction. [3 marks]

1. $n_1 \sin \theta_1 = n_2 \sin \theta_2$ — [M1] Stating the correct relationship earns the method mark.
2. $n_1 = 1$ (air), $\theta_1 = 40°$, $n_2 = 1.50$, $\theta_2 =?$ — Label each quantity before substituting. This prevents sign and index errors.
3. $1 \times \sin 40° = 1.50 \times \sin \theta_2$ — [M1] Correct substitution with n_1 = 1 applied.
4. $\sin \theta_2 = \frac{\sin 40°}{1.50} = \frac{0.6428}{1.50} = 0.4285$ — sin 40° = 0.6428 (to 4 s.f.). Dividing by 1.50 gives 0.4285.
5. $\theta_2 = \sin^{-1}(0.4285) = 25.4°$ — [A1] Answer to 3 significant figures, consistent with the data given.

Answer: $\theta_2 = 25.4°$ (the ray bends towards the normal, as expected when entering a denser medium)

Notice the sanity check built into that answer: the refracted angle (25.4°) is smaller than the incident angle (40°), which is exactly what you expect when light enters a medium with a higher refractive index. If your answer gave a larger angle, that would signal an error in the rearrangement.

Worked example: Finding the Refractive Index of a Substance

A ray of light passes from air into an unknown liquid. The angle of incidence is 55° and the angle of refraction is 38°. Calculate the refractive index of the liquid. [3 marks]

1. $n_1 \sin \theta_1 = n_2 \sin \theta_2$ — [M1] Always start by stating the relationship.
2. $n_1 = 1$ (air), $\theta_1 = 55°$, $\theta_2 = 38°$, $n_2 =?$ — The ray travels from air into the liquid, so medium 1 is air.
3. $n_2 = \frac{n_1 \sin \theta_1}{\sin \theta_2} = \frac{1 \times \sin 55°}{\sin 38°}$ — [M1] Rearranging correctly for n_2.
4. $\sin 55° = 0.8192$, $\quad \sin 38° = 0.6157$ — Compute each sine value separately to avoid arithmetic slips.
5. $n_2 = \frac{0.8192}{0.6157} = 1.33$ — [A1] The result is dimensionless and greater than 1, consistent with a refractive index. This value matches water closely.

Answer: $n_2 = 1.33$ (dimensionless)

5. Snell's Law Between Two Non-Air Media

So far, one medium has always been air. AQA questions can also involve a boundary between two materials where neither is air — for example, light passing from glass into water. The full form $n_1 \sin \theta_1 = n_2 \sin \theta_2$ handles this without modification; you substitute both refractive indices directly.

Worked example: Refraction at a Glass–Water Boundary

Light travels from glass (n = 1.50) into water (n = 1.33). The angle of incidence in the glass is 45°. Calculate the angle of refraction in the water. [3 marks]

1. $n_1 \sin \theta_1 = n_2 \sin \theta_2$ — [M1] The same equation applies regardless of whether air is involved.
2. $n_1 = 1.50$, $\theta_1 = 45°$, $n_2 = 1.33$, $\theta_2 =?$ — Medium 1 is glass (where the ray starts); medium 2 is water (where it is going).
3. $1.50 \times \sin 45° = 1.33 \times \sin \theta_2$ — [M1] Both refractive indices are substituted correctly.
4. $\sin 45° = 0.7071$, so $1.50 \times 0.7071 = 1.0607$ — 1.50 multiplied by 0.7071: 1.5 × 0.7071 = 1.0607.
5. $\sin \theta_2 = \frac{1.0607}{1.33} = 0.7974$ — Dividing 1.0607 by 1.33 gives 0.7974.
6. $\theta_2 = \sin^{-1}(0.7974) = 52.9°$ — [A1] The angle is larger than 45°, confirming the ray bends away from the normal when moving from glass to water (lower n).

Answer: $\theta_2 = 52.9°$

This result connects back to the earlier principle: moving from a higher to a lower refractive index always increases the angle. The full form of Snell's law handles every case — air-to-glass, glass-to-water, or any other pairing — with identical method.

6. Required Practical — Measuring the Refractive Index of Glass

The refractive index equation and Snell's law are not just theoretical tools — they can be verified directly in the laboratory. The standard AQA approach uses a glass block, a ray box, and a protractor to measure angles of incidence and refraction, then applies Snell's law to determine $n$.

The graphical method is more reliable than using a single pair of angles because it averages out random errors across multiple measurements. A best-fit straight line through the origin reduces the effect of any one anomalous reading.

7. Exam Technique and Common Mistakes

Refraction questions are reliable mark-scorers once you know the traps. The following guidance consolidates the most common errors seen in AQA mark schemes and explains exactly how to avoid them.

For 'explain' questions on refraction, the mark scheme consistently rewards a two-part answer: state that the speed of light changes at the boundary, then link this to the change in direction. A response that only says 'light bends' without explaining why will not access the higher marks.

Refraction and Snell's law underpin several further topics in the AQA specification, including total internal reflection and optical fibres. The core skill you have built here — correctly applying $n_1 \sin \theta_1 = n_2 \sin \theta_2$ with careful labelling of media and angles — transfers directly to those topics without modification.