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Polarisation is the restriction of the oscillations of a transverse wave to a single plane. It provides definitive evidence that light is a transverse wave and has numerous practical applications.
Spec mapping: This lesson sits under AQA 7408 section 3.3.1 and covers polarisation as a property of transverse waves only, the action of Polaroid filters, the I = I₀/2 result for unpolarised light passing through a single filter, Malus's law I = I₀ cos² θ for two filters with a relative angle θ, the use of polarisation in everyday applications (sunglasses, LCD screens, radio/TV aerial alignment), and the use of polarisation as definitive evidence that light is a transverse wave. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.3.1 (transverse vs longitudinal waves): polarisation is the experimental test that distinguishes transverse from longitudinal waves — the conceptual prerequisite is the first lesson of this course.
- Section 3.3.1 (electromagnetic spectrum): all EM waves are transverse and can be polarised, but the wavelength-dependence of polariser materials varies (Polaroid film for visible; wire-grid polarisers for microwave; thin-film polarisers for UV).
- Section 3.2.2 (photon model): an unpolarised beam of light can be described either classically (random oscillation directions) or quantum-mechanically (each photon in a superposition of two orthogonal polarisation states); the I = I₀/2 result applies in both pictures.
An unpolarised transverse wave oscillates in all planes perpendicular to the direction of propagation. For example, light from a bulb has electric field oscillations in every direction perpendicular to the ray.
A polarised wave has oscillations restricted to a single plane containing the direction of propagation. This plane is called the plane of polarisation.
Longitudinal waves (such as sound) cannot be polarised because their oscillations are already in a single direction — along the direction of propagation. The fact that light can be polarised is direct evidence that light is a transverse wave.
Exam Tip: If asked for evidence that light is a transverse wave, state that light can be polarised, and only transverse waves can be polarised. This is the definitive proof.
A Polaroid filter (such as Polaroid film) contains long-chain molecules aligned in one direction. These molecules absorb the component of the electric field oscillating parallel to the chains, transmitting only the component oscillating perpendicular to the chains. The transmitted direction is called the transmission axis.
When unpolarised light passes through a single Polaroid:
When light reflects off a non-metallic surface (e.g., glass, water) at a certain angle, the reflected light is partially polarised. At the Brewster angle, the reflected light is completely polarised.
Light scattered by small particles (e.g., molecules in the atmosphere) is partially polarised. This is why polarising sunglasses reduce glare from the sky.
If a second Polaroid filter (the analyser) is placed after the first (the polariser):
Rotate two Polaroid filters relative to each other while looking through both. The transmitted intensity varies from a maximum (axes parallel) to zero (axes crossed, at 90°). This can only be explained if light is a transverse wave.
When plane-polarised light of intensity I₀ is incident on a polarising filter (analyser) whose transmission axis makes an angle θ with the plane of polarisation, the transmitted intensity is:
I = I₀ cos² θ
This is Malus's law.
| Angle θ | cos θ | cos² θ | Transmitted Intensity |
|---|---|---|---|
| 0° | 1.00 | 1.00 | I₀ (maximum) |
| 30° | 0.866 | 0.750 | 0.75 I₀ |
| 45° | 0.707 | 0.500 | 0.50 I₀ |
| 60° | 0.500 | 0.250 | 0.25 I₀ |
| 90° | 0.000 | 0.000 | 0 (no transmission) |
Worked Example 1 — Plane-polarised light of intensity 8.0 W m⁻² passes through a polaroid filter whose transmission axis is at 35° to the plane of polarisation. Calculate the transmitted intensity.
I = I₀ cos² θ = 8.0 × cos² 35°
cos 35° = 0.8192
cos² 35° = 0.6711
I = 8.0 × 0.6711 = 5.4 W m⁻²
Worked Example 2 — Unpolarised light of intensity 12.0 W m⁻² passes through two polaroid filters. The first polarises the light, and the second has its transmission axis at 60° to the first. Calculate the intensity after each filter.
After the 1st filter (polariser): I₁ = I₀/2 = 12.0/2 = 6.0 W m⁻² (the light is now plane polarised)
After the 2nd filter (analyser at 60°): I₂ = I₁ cos² 60° = 6.0 × cos² 60° = 6.0 × (0.500)² = 6.0 × 0.250 = 1.5 W m⁻²
Worked Example 3 — Light passes through a polariser and then an analyser. The transmitted intensity is 40% of the intensity between the two filters. What is the angle between the transmission axes?
I = I₀ cos² θ
0.40 I₀ = I₀ cos² θ
cos² θ = 0.40
cos θ = √0.40 = 0.6325
θ = arccos(0.6325) = 50.8° ≈ 51°
Worked Example 4 — Three polaroid filters are arranged in series. The first and third have their transmission axes at 90° to each other. The second (middle) filter has its axis at 45° to both the first and third. Unpolarised light of intensity I₀ is incident. Calculate the final transmitted intensity.
After filter 1: I₁ = I₀/2 (polarised at 0°)
After filter 2 (at 45° to filter 1): I₂ = I₁ cos² 45° = (I₀/2) × (1/√2)² = (I₀/2) × (1/2) = I₀/4
The light is now polarised at 45° (aligned with filter 2's axis).
After filter 3 (at 90° to filter 1 = 45° to filter 2): I₃ = I₂ cos² 45° = (I₀/4) × (1/2) = I₀/8
Without the middle filter, no light would pass (crossed polaroids). With the middle filter, the final intensity is I₀/8 — the intermediate filter rotates the plane of polarisation, allowing some light through.
Reflected glare from water, roads, and glass surfaces is partially polarised (mostly horizontal). Polarising sunglasses have vertical transmission axes, blocking the horizontally polarised glare while transmitting other light. This reduces glare and improves visibility.
Liquid crystal displays use polarised light. The liquid crystal layer rotates the plane of polarisation of light passing through it. By applying a voltage, the rotation is controlled, determining whether light passes through the front polariser — this is how pixels are switched on and off.
When transparent materials (e.g., plastic, glass) are placed between crossed polaroids and subjected to stress, coloured patterns appear. These patterns reveal the stress distribution within the material, which is useful in engineering design.
Radio and TV aerials must be aligned with the polarisation of the transmitted signal (horizontal or vertical) for maximum reception. This is why TV aerials are mounted either horizontally or vertically depending on the local transmitter.
A metal grille can polarise microwaves (λ ~ 3 cm). When the grille bars are parallel to the electric field oscillation, the microwaves are absorbed. When the bars are perpendicular, the microwaves pass through.
By rotating the grille, the detected intensity varies, confirming that microwaves are transverse waves and can be polarised.
Exam Tip: When describing polarisation experiments, always state what is observed (e.g., intensity variation, extinction at 90°) and explain why (only transverse waves can be polarised; when transmission axis is perpendicular to the oscillation plane, no energy passes through).
When unpolarised light reflects from a non-metallic surface (glass, water, polished wood), the reflected light is partially polarised. At a specific angle of incidence — Brewster's angle, θ_B — the reflected light is completely plane-polarised in the plane perpendicular to the plane of incidence. The condition is:
tan θ_B = n₂/n₁
where n₁ is the refractive index of the medium light is coming from and n₂ is the refractive index of the medium it is partially entering. At Brewster's angle, the reflected and refracted rays are exactly 90° apart, and the component of the electric field oscillating in the plane of incidence cannot be reflected (the dipoles set up in the glass that would re-radiate this component are oriented along the reflected ray direction, which is the direction along which a dipole cannot radiate).
tan θ_B = 1.33/1.00 = 1.33 → θ_B = arctan(1.33) = 53.1°
So sunlight striking a calm lake at 53.1° from the normal produces fully polarised reflected light, with the electric field oscillating horizontally (perpendicular to the plane of incidence, which is vertical). Polarising sunglasses with vertical transmission axes block this reflected glare almost completely — a notable example of polarisation being deployed for practical optical engineering.
tan θ_B = 1.52 → θ_B = 56.7°
Brewster windows in laser cavities exploit this geometry: optical components angled at θ_B reflect zero of one polarisation component, allowing high-purity polarisation selection inside the cavity. Many helium–neon lasers and gas-discharge lasers use Brewster-angle windows for precisely this reason.
When sunlight is scattered by air molecules in the upper atmosphere, the scattered light is partially polarised — most strongly so when the scattering angle is exactly 90°. The mechanism is dipole radiation: each air molecule oscillates as an electric dipole driven by the incident E-field, and a dipole cannot radiate along its own oscillation axis. The radiation scattered perpendicular to both the propagation direction and the dipole-oscillation direction is fully polarised.
This is why a polarising filter held up to the blue sky at 90° from the Sun shows strong intensity variation as the filter is rotated — the sky light is significantly polarised, and most strongly so along the band 90° from the solar position. Photographers exploit this with polarising filters to darken blue skies for dramatic landscape shots; geographically, the principle was even exploited by the Vikings, who reputedly used iolite crystals (a natural polariser) to navigate under overcast skies by detecting the polarisation pattern of the obscured Sun.
The scattered light is most strongly polarised at exactly 90° scattering angle — the direction perpendicular to the line from the Sun. For ideal Rayleigh scattering (point dipoles in vacuum), the degree of polarisation at 90° is theoretically 100%; in practice atmospheric depolarising effects (multiple scattering, particle size effects) reduce this to approximately 75% on a clear day.
The classical "three-filter trick" — placing a middle filter at 45° between two crossed (90°) filters and observing light pass through — has an enlightening quantum interpretation. In the photon picture, each photon arriving at a filter has two possible outcomes: pass through (probability cos² θ where θ is the angle between the photon's polarisation and the filter's transmission axis) or be absorbed (probability sin² θ).
When the middle filter is removed, photons arriving from the first filter (polarised at 0°) face a filter at 90°, with passage probability cos² 90° = 0. With the middle filter at 45° inserted, the photons now face two probabilistic gates: (i) cos² 45° = 0.5 to pass the middle filter, (ii) cos² 45° = 0.5 to pass the third filter. Total probability: 0.25, multiplied by the initial half-intensity gives I₀/8 as derived classically. The same answer, two different interpretations — both fully valid.
The three-filter result is sometimes presented as a "quantum mystery", but classically it is just Malus's law applied twice, with the middle filter re-polarising the light to 45° before it reaches the third filter. The photon-picture version is a useful illustration of how probabilities multiply in successive measurements — a foundational concept in quantum mechanics.
After filter 1 (at 0°): I₁ = I₀/2 (polarised at 0°)
After filter 2 (at 30° to filter 1): I₂ = I₁ × cos² 30° = (I₀/2) × 0.75 = 0.375 I₀ (polarised at 30°)
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