Polarization, Malus's law and birefringence

Maxwell's equations make a plane wave transverse, so a wave along z has two independent field components. Write them as Eₓ = a cos(kz − ωt) and E_y = b cos(kz − ωt + δ); the wave is fully specified by a, b and δ, and the state is the curve the tip of E traces at fixed z. Eliminate the time and that curve is (x/a)² + (y/b)² − 2(x/a)(y/b) cos δ = sin²δ. For δ = 0 or π it is a line of slope ±b/a: linear polarization. For a = b and δ = ±π/2 it is a circle: circular polarization, the sign of δ fixing the handedness. Anything else is an ellipse. Put tan α = b/a; the major axis tilts at ψ with tan 2ψ = tan 2α cos δ, and the axis ratio tan χ follows from sin 2χ = sin 2α sin δ. So a = 2b with δ = 90° gives an upright ellipse of ratio ½, while a = b with δ = 60° gives one tilted at 45° with ratio tan 30° = 0.58. Writing the pair as a column of complex amplitudes, (a, b e^(iδ)), is the Jones vector; every polarizer and plate in this topic is a 2 × 2 matrix acting on it.

An ideal linear polarizer passes the field component along its transmission axis and absorbs the rest. Linearly polarized light of amplitude E₀ at angle θ to that axis leaves with amplitude E₀ cos θ, and because intensity is the time average of E² the transmitted intensity is I = I₀ cos²θ: Malus's law. Unpolarized light has no single θ. Its orientation and δ wander on femtosecond timescales, so the detector sees the average of cos²θ over every angle, (1/2π) ∫ cos²θ dθ = ½ over a full turn, and the output is polarized along the axis with half the power. Two ideal polarizers crossed at 90° pass nothing; slide a third between them at 45° and each step is a projection by cos 45°, so I₀/2 × ½ × ½ = I₀/8 of unpolarized input emerges through what was an opaque pair. Then take N polarizers each rotated by 90°/N: they pass the fraction cos^(2N)(90°/N) of already-polarized light — 0.78 for N = 10, 0.88 for N = 20, and 1 as N → ∞. Each projection discards a field component of first order in the step angle but an intensity only of second order, so the total loss, about N(π/(2N))² = π²/(4N), falls as 1/N and a fine enough stack rotates polarization without loss.

Reflected light is radiated by the dipoles that the transmitted wave drives in the second medium, and a dipole radiates nothing along its own axis. The p-component, polarized in the plane of incidence, drives dipoles perpendicular to the refracted ray; when that ray is at 90° to the reflected ray the dipoles point along the reflected direction and cannot feed it, so the reflected beam holds only the s-component, perpendicular to the plane of incidence and parallel to the surface. Impose θ_B + θₜ = 90° on Snell's law: n₁ sin θ_B = n₂ sin θₜ = n₂ cos θ_B, so tan θ_B = n₂/n₁. For air to crown glass, n = 1.52, θ_B = 56.7° and θₜ = 33.3°; from inside the glass tan θ_B = 1/1.52 gives 33.3°, the complement, as it must. Fresnel's rₚ = tan(θᵢ − θₜ)/tan(θᵢ + θₜ) says the same: its denominator diverges when the angles sum to 90°. The reflected beam is completely polarized but weak, since the s-reflectance sin²(θᵢ − θₜ)/sin²(θᵢ + θₜ) is only sin²(23.3°) = 0.157 for glass. The transmitted beam is only partially polarized; a pile of plates enriches it in p one surface at a time.

A molecule small compared with λ scatters light by the same mechanism: the incident E drives an oscillating dipole, which radiates with intensity proportional to sin²φ, φ measured from its axis. Look at a beam of unpolarized light from the side, at 90° to its direction. Dipoles driven along your line of sight radiate nothing towards you, so the light you receive comes only from dipoles perpendicular to both the beam and the line of sight: it is linearly polarized. This is why the sky 90° from the sun is strongly polarized. Partial polarization needs a number: the degree of polarization P = (Iₘₐₓ − Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ), the extremes found by rotating an analyser. For single Rayleigh scattering through angle Θ, P = sin²Θ/(1 + cos²Θ): 1 at 90°, 0.60 at 60° and 0 straight ahead. The real sky peaks near 0.75 because light is scattered more than once on the way to you, and multiple scattering is the first thing the single-dipole model leaves out. Malus's law then applies to the polarized fraction alone; the unpolarized remainder contributes its flat one half.

In calcite or quartz the index depends on the direction of E relative to the optic axis: perpendicular to the axis light sees the ordinary index nₒ, along it the extraordinary index nₑ. At 589 nm calcite has nₒ = 1.658 and nₑ = 1.486, quartz nₒ = 1.5443 and nₑ = 1.5534. Cut a plate with the optic axis in its face and the two components cross the same thickness d at different speeds, so their phases separate by δ = 2π d (nₑ − nₒ)/λ₀: no energy is lost, only phase. A quarter-wave plate has d|Δn| = λ₀/4, which is 16.2 μm of quartz at 589 nm but only 0.86 μm of calcite, so practical calcite plates are multi-order, d|Δn| = (m + ¼)λ₀. Because δ ∝ 1/λ₀, a 1.00 mm quartz plate holds 15.45 waves of delay and its retardance drifts by 9.4° per nanometre; the zero-order 16.2 μm plate drifts by 0.15° per nanometre. A half-wave plate, δ = π, reverses the sign of the slow component and so reflects a linear state about the fast axis: input at α leaves at −α, rotated through 2α with no loss. A quarter-wave plate at α = ±45° turns linear into circular; at α = 0° or 90° it leaves the state linear, and at any angle between it makes an ellipse with its axes along the plate's axes.

Send linearly polarized light through a retarder of retardance δ whose fast axis is at α to the polarization, then through an analyser at β to the original polarization, both angles measured the same way round. Project onto the plate axes, multiply the slow component by e^(iδ), project onto the analyser and square the modulus: I/I₀ = cos²β + sin 2α sin 2(β − α) sin²(δ/2). Every limit checks. δ = 0 returns Malus's law; δ = π returns cos²(β − 2α), the half-wave rotation; crossed polarizers, β = 90°, collapse it to I = I₀ sin²(2α) sin²(δ/2). At α = 45° that is a direct meter for δ: dark for a full-wave plate, half intensity for a quarter-wave plate, full transmission for a half-wave plate. Photoelastic stress analysis uses this: loaded plastic between crossed polarizers glows in fringes because its Δn is proportional to stress, and the dark fringes trace δ = 2πm. The same arrangement settles what a single polarizer cannot. Circular and unpolarized light both give a flat intensity through a rotating analyser, but a quarter-wave plate turns circular light into linear light, which the analyser extinguishes, and leaves unpolarized light untouched.