🏠 Home
FractionRush AQA A-Level Physics 3

Polarisation

Understand how transverse waves can be polarised, how polarising filters work, and how Malus's law quantifies transmitted intensity.

AQA A-Level Physics · Unit 3: Waves
🌊Transverse vs longitudinal
Know why only transverse waves can be polarised
🔲Polarising filters
Explain how a polariser and analyser work together
📐Malus's law
Apply I = I₀ cos²θ to calculate transmitted intensity
📺LCD screens
Describe how polarisation is used in liquid crystal displays
🔬Stress analysis
Understand photoelastic stress analysis using polarised light
📡Real-world uses
Recall applications: sunglasses, photography, radio waves

What is Polarisation?

All electromagnetic waves are transverse waves: the electric and magnetic fields oscillate perpendicular to the direction of travel. In an unpolarised wave, the electric field vibrates in all planes simultaneously — there is no preferred direction of oscillation.

Plane polarisation restricts the oscillation of the electric field to a single plane. Once polarised, the wave's electric field vibrates in one direction only, while the magnetic field vibrates perpendicular to it (and to the direction of propagation).

Plane-polarised wave: A transverse wave in which the oscillations occur in one plane only, containing the direction of propagation.

Only transverse waves can be polarised. Sound waves, which are longitudinal (oscillations parallel to propagation), cannot be polarised because there is no transverse displacement to restrict. This is an important test: observing polarisation effects proves a wave is transverse.

Polarisation provides evidence that electromagnetic waves are transverse. Longitudinal waves (e.g. sound) cannot be polarised.

Natural light sources such as the Sun, light bulbs, and candle flames produce unpolarised light. The electric field direction changes randomly millions of times per second, covering all orientations equally.

Polarising Filters

A polarising filter (polariser) transmits only the component of the electric field aligned with its transmission axis. When unpolarised light passes through a polariser:

A second polarising filter, called an analyser, is used to investigate the polarisation state of light. When the analyser's transmission axis is parallel to the polariser, maximum light passes. When it is perpendicular (crossed polaroids), no light is transmitted — the filters are said to be crossed.

Analyser: A second polarising filter used to detect or measure the plane of polarisation of already-polarised light.

Rotating the analyser between 0° and 90° produces a smooth variation in transmitted intensity from maximum to zero. This predictable variation is described mathematically by Malus's law.

Polarisation by reflection also occurs: when light hits a non-metallic surface (e.g. water, glass) at the Brewster angle, the reflected light is completely plane-polarised. Polaroid sunglasses exploit this — they filter out horizontally polarised glare reflected from roads and water.

Malus's Law

When plane-polarised light of intensity I₀ is incident on an analyser whose transmission axis makes an angle θ with the plane of polarisation, the transmitted intensity I is given by Malus's law:

I = I₀ cos²θ
SymbolQuantityUnit
ITransmitted intensityW m⁻²
I₀Incident (polarised) intensityW m⁻²
θAngle between polarisation plane and analyser axisdegrees or radians

Key values to remember:

Malus's law applies only when the incident light is already plane-polarised. If unpolarised light hits the analyser directly, the law does not apply in this simple form.

The cos²θ dependence means intensity varies smoothly — it does not drop linearly with angle. Plotting I against θ gives a cosine-squared curve, which reaches zero only at θ = 90° and 270°.

Applications of Polarisation

LCD screens: Liquid crystal displays rely on polarisation. Two crossed polarising filters sandwich a layer of liquid crystals. In the off state, the crystals rotate the plane of polarisation by 90°, allowing light to pass through the second filter. When a voltage is applied, the crystals realign and no longer rotate the light — the pixel appears dark. Coloured sub-pixels create the full-colour display you see.

Photoelastic stress analysis: Transparent materials like glass or plastic become birefringent (doubly refracting) when mechanically stressed. When placed between two crossed polaroids and illuminated with polarised light, stressed regions rotate the plane of polarisation by different amounts, producing coloured fringes. Engineers use this technique — called photoelasticity — to visualise stress distributions in mechanical components before manufacture.

Photography: Circular polarising filters on camera lenses reduce glare from non-metallic surfaces, deepen the colour of skies, and reduce reflections on water, improving image contrast and colour saturation.

Radio and TV aerials: Radio transmitters emit polarised electromagnetic waves. Receiving aerials must be aligned with the same plane of polarisation — horizontal or vertical — to receive the maximum signal. Misalignment reduces signal strength according to Malus's law.

Polarisation is evidence that light (and all EM waves) are transverse. It has wide practical applications from LCD screens to stress testing.
Plane-polarised light of intensity 80 W m⁻² is incident on an analyser. The angle between the plane of polarisation and the analyser axis is 30°. Calculate the transmitted intensity.
1Write down Malus's law: I = I₀ cos²θ
2Identify values: I₀ = 80 W m⁻², θ = 30°
3Calculate: I = 80 × cos²(30°) = 80 × (0.866)² = 80 × 0.75
I = 60 W m⁻²
Polarised light passes through an analyser. The transmitted intensity is 25% of the incident intensity. Find the angle between the polariser and analyser axes.
1Using Malus's law: I = I₀ cos²θ, so I/I₀ = cos²θ
2Given I/I₀ = 0.25: cos²θ = 0.25
3cosθ = √0.25 = 0.5
4θ = cos⁻¹(0.5) = 60°
θ = 60°
Unpolarised light of intensity 120 W m⁻² passes through a polariser and then an analyser. The analyser axis is at 45° to the polariser axis. Calculate the final transmitted intensity.
1After the polariser, intensity = ½ × I₀ = ½ × 120 = 60 W m⁻²
2Now apply Malus's law for the analyser (θ = 45°): I = 60 × cos²(45°)
3cos(45°) = 1/√2, so cos²(45°) = 0.5
4I = 60 × 0.5 = 30 W m⁻²
I = 30 W m⁻²
An engineer uses photoelastic stress analysis. Explain the steps involved and what is observed when a stressed transparent plastic component is placed between crossed polaroids.
1Polarised light passes through the first polariser and then through the stressed plastic component.
2Stress causes the plastic to become birefringent — it rotates the plane of polarisation by an amount that depends on the local stress magnitude.
3Light emerging from different regions has had its polarisation rotated by different amounts, so different amounts pass through the second (crossed) polaroid.
Coloured fringes (isochromatic lines) appear, mapping stress distribution: closely spaced fringes indicate high stress concentration.

Q1. Which type of wave can be polarised?

Q2. Plane-polarised light of intensity 200 W m⁻² hits an analyser at θ = 60°. What is the transmitted intensity?

Q3. Two polarising filters are crossed (at 90° to each other). A third filter is inserted between them at 45°. What happens to the transmitted intensity compared to the crossed arrangement without the middle filter?

Q4. Explain in one sentence why sound waves cannot be polarised.

Q5. In an LCD pixel, what role does the voltage applied to the liquid crystals play?

Challenge Q1. Unpolarised light of intensity I₀ passes through three successive polarising filters. The first is at 0°, the second at 30°, and the third at 75° to the first. Derive the final transmitted intensity in terms of I₀.

Challenge Q2. A student claims that polarised light passing through an analyser at 89° will have almost zero intensity, but at 90° it will be exactly zero. Explain mathematically why this is correct, and discuss whether it matters in practice.

Challenge Q3. Describe an experiment to verify Malus's law using a laser, two polaroids, a light sensor, and a protractor. State what you would measure, what you would plot, and how you would verify the law.

Challenge Q4. Explain why photoelastic stress analysis produces coloured fringes rather than simply bright and dark regions when white light is used.