Polarisation

Fast start (what to remember first)

1. Light from an ordinary source is randomly polarised: its polarisation direction changes rapidly and unpredictably.
2. An ideal polariser transmits half the intensity of randomly polarised light: I₁ = Iᵢₙ/2.
3. For linearly polarised light reaching an ideal analyser, Malus’s law gives I = I₀ cos² θ, where I₀ is the intensity reaching that analyser and θ is the angle between the light’s polarisation direction and the analyser’s transmission axis.
4. At the Brewster angle θ_B, with tan θ_B = n₂/n₁, the reflected and refracted rays are perpendicular and the reflected light is linearly polarised perpendicular to the plane of incidence.

1. Randomly polarised light and polarisers

An ordinary light source contains a very large number of atomic emitters, each radiating a short polarised wave train (roughly 10⁻⁸ s long) in its own direction. The overall polarisation therefore changes far faster than any detector can follow, so the light behaves as randomly polarised. A polariser transmits only the electric-field component along its transmission axis. Averaged over all the random directions, the transmitted intensity is half the incident intensity, and the light leaving the polariser is linearly polarised along that axis.

2. Malus’s law and polariser chains

Only the field component along the analyser axis passes: an amplitude E₀ becomes E₀ cos θ. Intensity is proportional to amplitude squared, so

I = I₀ cos² θ.

Apply the law one element at a time. After each polariser the light is polarised along that polariser’s axis, and its intensity becomes the I₀ for the next element.

Worked chain. Randomly polarised light of intensity 80 W m⁻² passes through three ideal polarisers with transmission axes at 0°, 30° and 90° to the vertical.

1. First polariser: I₁ = 80/2 = 40 W m⁻², polarised at 0°.
2. Second: the angle between the arriving polarisation (0°) and the axis (30°) is 30°, so I₂ = 40 cos² 30° = 30 W m⁻², now polarised at 30°.
3. Third: the angle is 90° - 30° = 60°, so I₃ = 30 cos² 60° = 7.5 W m⁻².

Without the middle polariser, the first and last are crossed and transmit nothing. Inserting a polariser between them increases the output because it changes the polarisation reaching the last one.

Try it first. Randomly polarised light of intensity 120 W m⁻² passes through ideal polarisers with axes at 0°, 45° and 90°. Find the intensity after each one, then the output if the 45° polariser is removed.

3. Polarisation by reflection: the Brewster angle

Resolve the incident field into a component perpendicular to the plane of incidence (dots in the figure) and a component in the plane of incidence (double arrows). For light travelling in medium 1 (index n₁) towards medium 2 (index n₂), there is one angle of incidence at which the in-plane component is not reflected at all. This is the Brewster angle, θ_B.

At θ_B the reflected and refracted rays are perpendicular, so θ_B + θₜ = 90°. Snell’s law then gives

tan θ_B = n₂/n₁.

Why the in-plane component vanishes. The refracted wave drives electrons in medium 2 to oscillate along its field. For the in-plane component, that oscillation lies in the plane of incidence, perpendicular to the refracted ray. When the reflected and refracted rays are perpendicular, this oscillation direction points along the reflected ray. An oscillating dipole radiates nothing along its own axis, so the reflected wave receives no in-plane component. The perpendicular component drives oscillations normal to the page, which do radiate along the reflected direction.

Consequences:

• The reflected light at θ_B is linearly polarised perpendicular to the plane of incidence. For reflection from a horizontal surface, that is horizontal. Polarising sunglasses with vertical transmission axes block this glare.
• The refracted light still carries both components, so it is only partly polarised.
• θ_B depends on which medium the light starts in. Air to glass ( n = 1.50): θ_B = tan⁻¹ 1.50 = 56.3°. Glass to air: θ_B = tan⁻¹ (1/1.50) = 33.7°.
• θ_B is an angle of incidence measured from the normal. Do not confuse it with the analyser angle θ in Malus’s law, which is measured between a polarisation direction and a transmission axis.

Try it first. Light travelling in water ( n = 1.33) reflects from a glass block ( n = 1.50). Find the Brewster angle and the angle of the refracted ray, and state the polarisation of the reflected light.

4. Birefringence

In an optically anisotropic crystal such as calcite, the refractive index depends on the direction of the light’s electric field relative to a special direction in the crystal, the optic axis. A ray entering such a crystal generally splits into two rays with perpendicular polarisations:

• the ordinary ray (o-ray), whose field is perpendicular to the plane containing the optic axis and the ray. It has the same index nₒ in every direction and obeys Snell’s law;
• the extraordinary ray (e-ray), whose field has a component in that plane. Its index varies with propagation direction, between nₒ and a limiting value nₑ.

Which ray is which depends on the crystal’s optic-axis orientation relative to the ray, not on a fixed propagation direction. Light travelling exactly along the optic axis does not split, because its field is then perpendicular to the axis whatever its polarisation. For calcite at 589 nm, nₒ = 1.658 and nₑ = 1.486. The birefringence is Δ n = |nₑ - nₒ| = 0.172.

Because the two polarisations travel at different speeds, a plate of thickness d introduces a phase difference δ = 2π Δ n d/λ between them. This changes the polarisation state of the light that emerges.

Using birefringence between crossed polarisers

Crossed polarisers normally transmit no light. A birefringent sample placed between them changes the polarisation, so some light gets through.

• Photoelasticity: glass and many plastics become birefringent under mechanical stress. A plastic model of a component viewed between crossed polarisers shows coloured fringes that map the stress distribution.
• Liquid-crystal displays (LCDs): liquid-crystal molecules are themselves birefringent, and their orientation can be controlled. In a twisted-nematic cell between crossed polarisers, the molecules twist through 90° across the cell and guide the polarisation round, so light passes. An applied voltage turns the molecules to line up with the field, removing the twist, so the second polariser blocks the light. The display works by electrically reorienting birefringent molecules. It is not an example of the Kerr effect, which is birefringence induced in an otherwise isotropic material by an electric field.

Learn next

• Optical Fibres
• Compton Shift
• Back To Photonics