# Polarisation

> CIE A-Level Physics · Unit 7: Waves
> Source: https://www.owlsprep.com/study/cie-9702-u7-polarisation/

This module covers polarisation of transverse waves, distinguishes polarised and unpolarised light, covers methods of polarisation, Malus's law calculations, and common applications, aligned to CIE A-Level Physics 9702 requirements.

**Prerequisites:** [Transverse and longitudinal waves](https://www.owlsprep.com/study/cie-9702-u7-transverse-longitudinal-waves/); [Wave intensity](https://www.owlsprep.com/study/cie-9702-u7-wave-intensity/)

## Learning objectives

- Distinguish between polarised and unpolarised transverse waves
- Explain different methods of polarisation of light
- Apply Malus's Law to calculate transmitted intensity through polarisers
- Describe common practical applications of polarisation

## Polarised vs Unpolarised Light

**Unpolarised light** — Light consisting of transverse vibrations that occur in all planes perpendicular to the direction of wave propagation.

*Example:* Light from the Sun or an incandescent bulb is unpolarised.

**Polarised light** — Light where all transverse vibrations are restricted to a single plane perpendicular to the direction of wave propagation.

*Example:* Light passed through a polarising filter is plane polarised.

Only transverse waves can be polarised. Longitudinal waves (like sound) cannot be polarised, because their vibrations are parallel to the direction of travel, so there is no plane perpendicular to restrict vibrations to.

> **warning**
>
> A very common exam trap is asking if longitudinal waves can be polarised — the answer is always no.

**Worked example:** Can a sound wave travelling through air be polarised? Explain your answer.

1. Polarisation requires vibrations to be perpendicular to the direction of wave propagation.
2. Sound waves in air are longitudinal, so their vibrations are parallel to the direction of travel.
3. No, sound waves cannot be polarised, because longitudinal waves cannot have their vibrations restricted to a single plane perpendicular to propagation.

## Methods of Polarisation

CIE A-Level requires you to know three common methods of polarisation:

1. **Absorption**: Polaroid filters use aligned long-chain polymer molecules that absorb vibrations parallel to the chain, transmitting only vibrations perpendicular to the chain.
2. **Reflection**: Unpolarised light reflecting off a non-metallic surface (water, glass, asphalt) becomes partially polarised parallel to the reflecting surface. At Brewster's angle, it becomes fully polarised.
3. **Refraction**: Unpolarised light entering a birefringent crystal splits into two separate polarised rays that refract at different angles.

**Worked example:** Explain why sunlight reflected off a lake is partially polarised.

1. Incoming sunlight is unpolarised, with vibrations in all planes perpendicular to its direction of travel.
2. Vibrations parallel to the lake (horizontal) surface are preferentially reflected, while vibrations perpendicular to the surface are mostly refracted into the water.
3. The reflected light is therefore partially polarised parallel to the lake surface.

> **Exam tip:** Always specify that reflected light is polarised parallel to the reflecting surface — this is a common marking point.

## Malus's Law

When plane polarised light passes through a second polariser (analyser), the intensity of transmitted light depends on the angle between the transmission axes of the two polarisers.

**Malus's Law** — Gives the intensity of plane polarised light transmitted through an analyser, where $I_0$ is the intensity of incident polarised light and $\theta$ is the angle between the transmission axes.

*Notation:* I = I_0 \cos^2 \theta

$$I = I_0 \cos^2 \theta$$

If unpolarised light hits the first polariser, the intensity after the first polariser is always half the original intensity. This is because unpolarised light averages out over all angles, so the average value of $\cos^2 \theta$ is $\frac{1}{2}$.

**Worked example:** Unpolarised light of intensity $100 \text{ Wm}^{-2}$ passes through two polarisers. The transmission axis of the second polariser is at $30^\circ$ to the first. Calculate the transmitted intensity.

1. After the first polariser, unpolarised light becomes polarised with intensity:
2. $$I_1 = \frac{I_{original}}{2} = \frac{100}{2} = 50 \text{ Wm}^{-2}$$
3. Apply Malus's Law to the second polariser, $\theta = 30^\circ$:
4. $$I_2 = I_1 \cos^2 30^\circ = 50 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 50 \times \frac{3}{4} = 37.5 \text{ Wm}^{-2}$$

**Check your understanding**

1. Unpolarised light passes through two polarisers with perpendicular transmission axes. What is the final transmitted intensity?

   - Half the original intensity
   - Zero
   - Quarter the original intensity
   - Same as original

   *Answer:* Zero

   *Why:* When axes are perpendicular, $\theta = 90^\circ$, $\cos^2 90^\circ = 0$, so no light is transmitted.

*Calculator:* allowed

## Applications of Polarisation

CIE often asks to explain how polarisation is used in common applications. The most frequently tested applications are:

- Polaroid sunglasses: Block horizontally polarised glare reflected from roads and water
- Liquid Crystal Displays (LCDs): Use polarisers and liquid crystals that rotate polarisation to control light output for pixels
- Stress analysis: Stressed transparent plastic rotates polarisation to reveal high-stress areas
- Sugar concentration measurement: Sugar solutions rotate the plane of polarisation, angle of rotation gives concentration

**Worked example:** Explain how polaroid sunglasses reduce glare from a horizontal road surface.

1. Light reflected from the road is partially polarised parallel (horizontal) to the road surface.
2. Polaroid sunglasses have a vertical transmission axis, so they absorb the horizontally polarised glare.
3. Only a fraction of unpolarised ambient light is transmitted, reducing overall glare intensity.

## Common pitfalls

- **Wrong:** Claiming longitudinal waves can be polarised
  - Why it fails: Polarisation requires vibrations perpendicular to propagation, which longitudinal waves do not have
  - Correct: State only transverse waves can be polarised
- **Wrong:** Applying Malus's Law directly to original intensity of unpolarised light
  - Why it fails: Malus's Law only applies to already polarised incident light; unpolarised light is halved after the first polariser
  - Correct: Always halve the intensity of unpolarised light after the first polariser before applying Malus's Law
- **Wrong:** Stating reflected light is polarised perpendicular to the reflecting surface
  - Why it fails: Vibrations parallel to the surface are preferentially reflected, so polarisation is parallel to the surface
  - Correct: State reflected light is polarised parallel to the reflecting surface
- **Wrong:** Writing Malus's Law as $I = I_0 \cos \theta$, forgetting the square
  - Why it fails: Intensity is proportional to the square of amplitude, so the relationship requires a square term
  - Correct: Recall Malus's Law as $I = I_0 \cos^2 \theta$
- **Wrong:** Claiming any non-zero angle between polarisers gives zero intensity
  - Why it fails: Intensity only drops to zero when the angle between axes is 90° (crossed polarisers)
  - Correct: Only crossed polarisers (90° between axes) produce zero transmitted intensity

## Cheatsheet

| Key Concept | Key Fact/Formula |
| --- | --- |
| Polarisation possible? | Only transverse waves, no longitudinal |
| Unpolarised → 1 polariser | Intensity = $I_0/2$, output polarised |
| Malus's Law | $I = I_0 \cos^2 \theta$, $I_0$ = incident polarised intensity |
| Reflected light polarisation | Parallel to the reflecting surface |
| Polaroid sunglasses | Vertical axis blocks horizontal glare |
| Common applications | Sunglasses, LCDs, stress analysis, sugar measurement |

## What's next

Polarisation confirms the transverse wave nature of light, and is a core concept for optics in CIE A-Level Physics. It regularly appears in both multiple choice and structured questions, so mastering Malus's Law and explanations of polarisation is key for exam success. This topic builds on your foundational knowledge of wave properties, and paves the way for understanding interference and diffraction, the next key topics in Unit 7. Polarisation is also fundamental for advanced study of optics and quantum physics in higher education.

- [Electromagnetic Spectrum](https://www.owlsprep.com/study/cie-9702-u7-electromagnetic-spectrum/)
- [Superposition](https://www.owlsprep.com/study/cie-9702-u8-overview/)
- [Principle of superposition](https://www.owlsprep.com/study/cie-9702-u8-principle-of-superposition/)

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