Polarisation

CIE A-Level Physics· Unit 7: Waves -> Polarisation· 25 min read

1. Polarised vs Unpolarised Light★★☆☆☆⏱ 5 min

📘 Definition

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.

📘 Definition

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.

📐 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.

2. Methods of Polarisation★★☆☆☆⏱ 6 min

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

Absorption: Polaroid filters use aligned long-chain polymer molecules that absorb vibrations parallel to the chain, transmitting only vibrations perpendicular to the chain.
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.
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.

3. Malus's Law★★★☆☆⏱ 8 min

✓ Calculator OK

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.

📘 Definition

Malus's Law

I = I_0 \cos^2 \theta

Gives the intensity of plane polarised light transmitted through an analyser, where $I_{0}$ is the intensity of incident polarised light and $θ$ is the angle between the transmission axes.

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 $cos^{2} θ$ is $\frac{1}{2}$ .

📐 Worked Example

Unpolarised light of intensity $100 Wm^{- 2}$ passes through two polarisers. The transmission axis of the second polariser is at $3 0^{\circ}$ 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, $θ = 3 0^{\circ}$ :
4
$I_2 = I_1 \cos^2 30^\circ = 50 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 50 \times \frac{3}{4} = 37.5 \text{ Wm}^{-2}$

✓ Quick check

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
Reveal answer
1 —
When axes are perpendicular, $θ = 9 0^{\circ}$ , $cos^{2} 9 0^{\circ} = 0$ , so no light is transmitted.

4. Applications of Polarisation★★☆☆☆⏱ 6 min

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.

5. Common Pitfalls

Wrong move:

Claiming longitudinal waves can be polarised

Why:

Polarisation requires vibrations perpendicular to propagation, which longitudinal waves do not have

Correct move:

State only transverse waves can be polarised

Wrong move:

Applying Malus's Law directly to original intensity of unpolarised light

Why:

Malus's Law only applies to already polarised incident light; unpolarised light is halved after the first polariser

Correct move:

Always halve the intensity of unpolarised light after the first polariser before applying Malus's Law

Wrong move:

Stating reflected light is polarised perpendicular to the reflecting surface

Why:

Vibrations parallel to the surface are preferentially reflected, so polarisation is parallel to the surface

Correct move:

State reflected light is polarised parallel to the reflecting surface

Wrong move:

Writing Malus's Law as $I = I_{0} cos θ$ , forgetting the square

Why:

Intensity is proportional to the square of amplitude, so the relationship requires a square term

Correct move:

Recall Malus's Law as $I = I_{0} cos^{2} θ$

Wrong move:

Claiming any non-zero angle between polarisers gives zero intensity

Why:

Intensity only drops to zero when the angle between axes is 90° (crossed polarisers)

Correct move:

Only crossed polarisers (90° between axes) produce zero transmitted intensity

6. Quick Reference 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	$I = I_{0} cos^{2} θ$ , $I_{0}$ = 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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 13
Malus's law intensity calculation
2022 · 12
Polarisation of longitudinal waves
2021 · 22
Explain polarisation by reflection

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.