Polarization of Light

We take one fact from electromagnetism and build a whole chapter on it. The companion article aq-em-polarization-em-waves establishes that Maxwell's equations confine the electric field of a plane wave to a two-component transverse plane — Gauss's law kills any component along the propagation direction, leaving exactly two independent field directions in the wavefront. Polarization is that surviving freedom. Everything else in this chapter — linear, circular, elliptical states; polarizers and Malus's law; wave plates; the blue of the sky and the glare off a wet road — is the physics of how those two components share amplitude and phase. Stand at a fixed point and watch the tip of the electric-field vector over one optical period. If the two transverse components rise and fall in step, the tip runs back and forth along a line: linear polarization. If they are equal and a quarter-cycle out of step, the tip sweeps a circle: circular polarization. Every other case is an ellipse. These are not three separate phenomena but one two-parameter family — the amplitude ratio and the relative phase δ — read at different settings. Ordinary light shows none of this at first glance because a hot filament or the Sun emits an incoherent churn of every state at once: unpolarized light, whose degree of polarization is zero. Nature and the laboratory both edit that churn. Selective absorption (a Polaroid sheet), reflection near Brewster's angle, right-angle scattering, and birefringent crystals each imprint a preferred direction, which is why sunglasses cut glare, why the daytime sky is polarized, and why a stressed windshield glows in colour between crossed filters. This article is the hub of the polarization chapter: it fixes the vocabulary, defines the one number that grades any beam, and surveys the four ways to make polarized light — handing each mechanism off to the article that owns its details.

Key takeaways

Polarization is the surviving two-component transverse freedom of light's electric field: amplitude ratio and relative phase δ pick one state out of the line–ellipse–circle family.
Linear (δ = 0 or π), circular (equal amplitudes, δ = ±π/2), and elliptical (everything else) are one geometry read at different relative phases.
Unpolarized light is a fast random mixture of states; a single ideal polarizer transmits exactly half of it, since ⟨cos²θ⟩ = ½.
The degree of polarization V = (I_max − I_min)/(I_max + I_min) grades any beam from 0 (unpolarized) to 1 (fully polarized) via a rotating analyzer.
Four mechanisms make polarized light: dichroism, reflection at Brewster's angle (tan θ_B = n₂/n₁), 90° scattering, and birefringence.
Key idea: light's electric field owns two transverse components; their relative amplitude and phase are the polarization state, and every polarizer, every crystal, and every reflection is a way of reading or rewriting those two numbers.

The core insight

Light's electric field owns exactly two transverse components. Their relative amplitude and phase δ fix the polarization state — line, circle, or ellipse. Producing polarized light means imprinting a preferred component; analyzing it means projecting onto one. Every result in this chapter is bookkeeping on those two numbers.

Where this lives today

Polarization is a working control knob across science and industry. Every LCD screen gates each pixel between crossed polarizers, converting a voltage-controlled birefringence into brightness. Photographers and drivers cut glare with polarizing filters aligned against Brewster-reflected light. Field biologists and navigators exploit the sky's polarization pattern — the same pattern Vikings may have read with "sunstones" and that desert ants and bees use to steer — now reproduced in polarization-based navigation sensors. Engineers map stress in transparent parts by photoelasticity (colour fringes between crossed polarizers), and remote-sensing satellites use polarimetry to distinguish clouds, aerosols, and ocean-surface roughness. In the laboratory the degree of polarization is a routine diagnostic, and the polarization of emitted photons reveals the angular momentum exchanged in atomic transitions — the bridge to optical pumping and laser cooling.

The physical picture

Freeze a plane wave travelling along $\hat{\mathbf z}$ and look end-on at the electric field. Because the wave is transverse, $\mathbf E$ lives entirely in the $xy$ -plane; write it as two oscillations, $E_x = E_1\cos(kz-\omega t)$ and $E_y = E_2\cos(kz-\omega t+\delta)$ . Only two knobs matter: the amplitude ratio $E_2/E_1$ and the relative phase $\delta$ (an overall amplitude just sets brightness, an overall phase just resets the clock). Turn those two knobs and the tip of $\mathbf E$ traces every polarization state there is. If $\delta = 0$ the components peak together and the tip runs along a fixed line; if $\delta = \pi$ it runs along the mirror line. Give the components equal amplitude and a quarter-cycle offset ( $\delta = \pm\pi/2$ ) and the tip's two perpendicular oscillations are in quadrature — it traces a circle. Anything in between is an ellipse. This is why the whole subject has three named states and no more: they are the degenerate, the generic, and the symmetric readings of one two-component object.

Figure

Four ways to imprint a direction on light — dichroism, Brewster reflection, right-angle scattering, birefringence — the mechanisms this chapter hands off to its member articles

A single number grades how thoroughly a beam has been polarized. Send it through a rotating analyzer and record the transmitted intensity: a fully polarized beam gives a curve that dips to zero once per half-turn, an unpolarized beam gives a flat line, and a partial beam lies between. The contrast of that curve is the degree of polarization — the punchline figure below reads it straight off the trace, and pins the Brewster angle at which a reflection makes polarized light for free.

Figure

Punchline — the rotating-analyzer trace reads the degree of polarization V straight off the contrast; the inset fixes Brewster's angle at 56.7° for air→glass

The core equations

A plane wave travelling along $\hat{\mathbf z}$ carries two transverse field components. Written with an explicit relative phase, they are the seed of every polarization state.

\mathbf E(z,t) = \hat{\mathbf x}\,E_1\cos(kz-\omega t) + \hat{\mathbf y}\,E_2\cos(kz-\omega t+\delta) \qquad (1)

where: $E_1, E_2 \ge 0$ — the two transverse amplitudes; $\delta = \phi_y - \phi_x$ — the relative phase, the only phase that matters; $k = 2\pi/\lambda$ — wavenumber; $\omega$ — angular frequency. Read it physically: four real numbers describe the field, but overall amplitude sets brightness and overall phase sets timing, so the shape of the polarization is the two-parameter family $(E_2/E_1,\ \delta)$ . The transversality that makes this a two-component (not three-component) object is inherited, not derived here: $\mathbf k \cdot \mathbf E = 0$ .

\mathbf k \cdot \mathbf E = 0 \quad\Longrightarrow\quad \text{two independent components in the wavefront plane} \qquad (2)

where: $\mathbf k$ — wavevector along propagation; the identity is Gauss's law for a source-free plane wave, owned by aq-em-polarization-em-waves. Optics takes it as given and asks what the two survivors do. Any real beam is graded by one measurable number, obtained by rotating an analyzer and watching the transmitted intensity swing between extremes.

V = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} \qquad (3)

where: $I_{\max}, I_{\min}$ — the largest and smallest transmitted intensities as the analyzer turns; $V=1$ is fully polarized (the analyzer can extinguish the beam), $V=0$ is unpolarized (no orientation changes the reading). This is the degree of polarization — the modulation depth of the rotating-analyzer trace. Of the four ways to make polarized light, reflection carries the cleanest number: at one special incidence angle the reflected beam is completely linearly polarized.

\tan\theta_B = \frac{n_2}{n_1} \qquad (3\text{-Brewster})

where: $\theta_B$ — Brewster's angle; $n_1, n_2$ — indices of the incidence and transmission media. At $\theta_B$ the reflected and refracted rays are perpendicular and only the component perpendicular to the plane of incidence reflects. Why the parallel component vanishes — the Fresnel amplitude going to zero — is derived by fresnel-equations (1472) and brewster-angle (1473); here we take the condition and its number.

The physical picture in equations

Derivation AWhy light has two polarization states, not three

The goal of this argument is to show that polarization is a two-state freedom — no more, no less — and to locate exactly where that "two" comes from, so that the entire chapter rests on a single inherited fact rather than a re-derivation. The transversality of light is an electromagnetic result; optics quotes it and then counts degrees of freedom. Watch how a three-component vector field collapses to a two-component one the instant we impose the source-free plane-wave condition, and why that leftover pair is irreducible.

Step 1. Start with a plane wave $\mathbf E = \mathbf E_0\,e^{i(\mathbf k\cdot\mathbf r-\omega t)}$ , a full three-component vector amplitude $\mathbf E_0$ . Why start general: nothing so far forbids a longitudinal component; if one survived, light would have three polarization states and this chapter would be different.

Step 2. Impose Gauss's law in a source-free region, $\nabla\cdot\mathbf E = 0$ . For the plane wave $\nabla \to i\mathbf k$ , so $i\,\mathbf k\cdot\mathbf E_0 = 0$ , i.e. $\mathbf k\cdot\mathbf E_0 = 0$ . Why this is the whole story: the constraint is one scalar equation, and it removes exactly the component of $\mathbf E_0$ along $\mathbf k$ — the longitudinal one.

Step 3. Choose $\mathbf k = k\hat{\mathbf z}$ . The surviving amplitude lies in the $xy$ -plane: $\mathbf E_0 = E_1\hat{\mathbf x} + E_2 e^{i\delta}\hat{\mathbf y}$ , two complex components, i.e. two amplitudes and one relative phase after discarding an overall phase. Why irreducible: the two transverse directions are physically inequivalent to a detector aligned with either, so neither can be dropped — the freedom is genuinely two-dimensional. This transversality is owned by aq-em-polarization-em-waves; optics inherits "two components in the wavefront plane" and spends the rest of the chapter on what they can do. Verified: the constraint-counting and the collapse to a 2-component complex vector are checked symbolically in verify_polarization-of-light.py.

Go deeper · Derivation B · 1 minThe degree of polarization as the contrast of a rotating analyzer

Derivation BThe degree of polarization as the contrast of a rotating analyzer

The single most useful measurement on a beam of unknown polarization is to spin a linear analyzer in front of a detector and log intensity against angle. This derivation shows why the contrast of that trace — not its shape, not its average — is the physically meaningful grade, and why it runs cleanly from 0 to 1. The result, $V = (I_{\max}-I_{\min})/(I_{\max}+I_{\min})$ , is the working definition of the degree of polarization and the number every later article quietly assumes.

Step 1. Model any beam as a fully polarized part of intensity $I_p$ plus an unpolarized part of intensity $I_u$ ; total $I_0 = I_p + I_u$ . Why this split is always possible: the Stokes description (owned by stokes-parameters-poincare-sphere, 1469) proves any beam decomposes uniquely this way; we use only the conclusion.

Step 2. Through an analyzer at angle θ, the polarized part obeys Malus, $I_p\cos^2(\theta-\theta_0)$ , while the unpolarized part always gives half: $I_u/2$ . So $I(\theta) = I_u/2 + I_p\cos^2(\theta-\theta_0)$ . Why the unpolarized half is orientation-blind: averaging $\cos^2$ over random input angles gives ½ regardless of the analyzer setting.

Step 3. The extremes are $I_{\max} = I_u/2 + I_p$ (aligned) and $I_{\min} = I_u/2$ (crossed). Their contrast is

V = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} = \frac{I_p}{I_p + I_u} = \frac{I_p}{I_0}.

Why this is exactly the polarized fraction: the algebra collapses the analyzer geometry and leaves the ratio of polarized to total intensity — a property of the beam alone. So $V$ is both the measured contrast and the intrinsic polarized fraction. Verified: the collapse $V = I_p/I_0$ and the worked value $V = 0.64$ for $(I_{\max},I_{\min})=(8.2,1.8)$ are checked in verify_polarization-of-light.py.

Go deeper · Derivation C · 1 minBrewster's condition tan θ_B = n₂/n₁ from geometry, without Fresnel

Derivation CBrewster's condition tan θ_B = n₂/n₁ from geometry, without Fresnel

Reflection makes polarized light, and it does so most sharply at one angle. This short derivation reaches Brewster's condition from a purely geometric fact — that at the polarizing angle the reflected and refracted rays are perpendicular — combined with Snell's law. It deliberately stops short of the amplitude question (why the parallel reflection vanishes), which belongs to the Fresnel equations; the aim is the angle and its number, which the hub needs to survey reflection as a production mechanism.

Step 1. Take as the defining geometric fact that at the polarizing angle the reflected ray is perpendicular to the refracted ray: $\theta_B + \theta_t = 90^\circ$ , so $\theta_t = 90^\circ - \theta_B$ . Why this is the right condition: an oscillating dipole radiates nothing along its own axis; the refracted-beam dipoles driven parallel to the plane of incidence point exactly toward where the reflected ray would go, so that reflected component cannot be radiated — the mechanism owned by fresnel-equations (1472).

Step 2. Snell's law across the interface: $n_1\sin\theta_B = n_2\sin\theta_t = n_2\sin(90^\circ-\theta_B) = n_2\cos\theta_B$ . Why substitute the complement: it converts the second angle into the first, leaving a single unknown.

Step 3. Divide by $\cos\theta_B$ : $n_1\tan\theta_B = n_2$ , hence

\tan\theta_B = \frac{n_2}{n_1}.

Why this matters for the hub: it gives the production angle for every dielectric — 56.7° for air→glass, 53.1° for air→water — with no reference to reflection coefficients. The strength of the effect (that $R_\parallel \to 0$ exactly, not just weakens) is the Fresnel-equation result carried by brewster-angle (1473). Verified: tan θ_B = n₂/n₁ evaluated for glass and water in verify_polarization-of-light.py.

Limiting cases

Case	Condition	Meaning
Linear	δ = 0 or π	Components in step; tip on a fixed line at azimuth ψ = arctan(E₂/E₁). Owned by linear-polarization (1460).
Circular	E₁ = E₂, δ = ±π/2	Equal amplitudes in quadrature; tip on a circle. Handedness fixed by circular-polarization (1461).
Elliptical	generic E₂/E₁, δ	The general fully polarized state; tilt and shape from both parameters. Owned by elliptical-polarization (1462).
Unpolarized	random δ, time-averaged	V = 0; no fixed ellipse — thermal/natural light.
Partially polarized	0 < V < 1	Polarized fraction V rides on an unpolarized background.

Common trapUnpolarized light is not "light with no field direction" — it is a direction that reshuffles faster than you can measure

The commonest error is to picture unpolarized light as an electric field pointing nowhere, or as a static 45° blur. At every instant the field points somewhere definite; what makes the beam unpolarized is that the direction and relative phase jitter randomly on the coherence time — nanoseconds or less — far faster than any detector integrates. So "unpolarized" is a statement about the average, not the instantaneous field. The tell is operational: rotate an analyzer and the reading never changes (V = 0). A second tell that trips students — circularly polarized light also gives a constant reading through a single rotating analyzer, yet it is fully polarized. The distinction (a quarter-wave plate unmasks the circular state but does nothing to unpolarized light) is the subject of circular-polarization (1461).

Worked examples

1 · Brewster angles for glass and water (confidence builder)

A camera points at a wet road and at a shop window; at what incidence angle is the glare from each fully linearly polarized, so a filter can extinguish it?

Step 1. Use $\tan\theta_B = n_2/n_1$ with air $n_1 = 1$ . For crown glass $n_2 = 1.52$ : $\theta_B = \arctan(1.52)$ . Why arctan: the condition solves directly for the angle once the indices are fixed.

Step 2. $\arctan(1.52) = 56.66^\circ \approx \mathbf{56.7^\circ}$ . For water $n_2 = 1.33$ : $\theta_B = \arctan(1.33) = 53.06^\circ \approx \mathbf{53.1^\circ}$ .

Step 3. At these angles the reflected glare is completely polarized perpendicular to the plane of incidence — i.e. horizontally for a horizontal road. A polarizer with a vertical transmission axis then blocks it entirely, which is exactly how polarized sunglasses work. Glass: 56.7°. Water: 53.1°.

2 · Reading the degree of polarization off an analyzer trace (exam level)

A beam through a rotating analyzer gives a maximum transmitted intensity of 8.2 units and a minimum of 1.8 units. What is its degree of polarization, and what fraction of the light is polarized?

Step 1. Apply $V = (I_{\max}-I_{\min})/(I_{\max}+I_{\min}) = (8.2-1.8)/(8.2+1.8) = 6.4/10$ . Why this ratio: the derivation above showed the contrast equals the polarized fraction.

Step 2. $V = \mathbf{0.64}$ . Because $V$ equals $I_p/I_0$ , 64% of the beam is polarized and 36% is an unpolarized background. A perfect polarizer would let this beam dip to $I_{\min} = 1.8$ but never to zero — the unpolarized half floors it.

Physics of the portrait

Origins · 1 minpolarization and how this was discovered

polarization

concept · named by Malus 1811 · the two-component fingerprint of a transverse wave, read from crystals and reflections a century before Maxwell explained it

Polarization was discovered as a puzzle in crystals and only later understood as light's transverse signature. In 1669 Erasmus Bartholin reported that a calcite (Iceland spar) crystal split a single image in two, and that rotating the crystal moved one image around the other — the first sighting of two states of light behaving differently. Huygens (1690) found that each of the two beams, passed through a second crystal, behaved as if it had already been "conditioned," but he had no word for the property. Étienne-Louis Malus supplied the physics in 1808: looking through a calcite crystal at sunlight reflected from the windows of the Luxembourg Palace, he saw one image extinguish as he rotated the crystal — reflection had polarized the light. He published the quantitative cos² law in 1809 and coined "polarization" (from the supposed "poles" of light rays) in 1811. The conceptual key came from Thomas Young and Augustin-Jean Fresnel: in 1817 Young proposed that light's vibrations are transverse, instantly explaining why two beams polarized at right angles refuse to interfere, and by 1821 Fresnel had built the full transverse-wave theory. Maxwell's equations (1865) later made transversality inevitable — but the fingerprint had been read off calcite and palace windows four decades earlier.

History

1669

— Erasmus Bartholin reports double refraction in Iceland spar (calcite)

1690

— Huygens shows the two calcite beams are differently conditioned

1808

— Malus sees reflected Luxembourg-Palace light extinguish through calcite: polarization by reflection

1809

— Malus publishes the cos² intensity law

1811

— Malus names the property "polarization"; Arago finds optical activity in quartz

1815

— Brewster states the tan θ_B = n law for the polarizing angle

1817

— Young proposes transverse vibrations to explain polarization

1821

— Fresnel builds the fully transverse wave theory; circular and elliptical states classified

1852

— Stokes introduces the four parameters that grade partial polarization

1929

— Land patents the first sheet polarizer (Polaroid)

Going further

Texts. Hecht, Optics, 5th ed., ch. 8 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 6 · Born & Wolf, Principles of Optics, ch. 1 & 10 · Loudon, The Quantum Theory of Light, 3rd ed. · Foot, Atomic Physics (photon polarization and selection rules).

Provenance. The polarization ellipse follows by eliminating the phase between $E_x = E_1\cos\tau$ and $E_y = E_2\cos(\tau+\delta)$ (Hecht ch. 8; Born & Wolf ch. 1) — its derivation is owned by elliptical-polarization (1462). Malus's law $I = I_0\cos^2\theta$ and the ½ unpolarized average follow from field projection (Hecht eq. 8.24) and are owned by malus-law (1463). Brewster's condition $\tan\theta_B = n_2/n_1$ is stated here and derived in full (Fresnel amplitudes) by brewster-angle (1473). Transversality $\mathbf k\cdot\mathbf E = 0$ is recalled by name from aq-em-polarization-em-waves. Numbers are exact: $\arctan 1.52 = 56.66^\circ$ , $\arctan 1.33 = 53.06^\circ$ , $\langle\cos^2\rangle = \tfrac12$ .

Builds on: aq-em-polarization-em-waves (transversality) · Plane Electromagnetic Waves. Leads to: Linear Polarization · Circular Polarization · Elliptical Polarization · Malus's Law · Polarizers and Extinction Ratio · Jones Calculus · Stokes Parameters and the Poincaré Sphere · Birefringence · Brewster's Angle.

Is polarization a property of a single photon or only of a beam?

Both, and they match. A single photon has a polarization state — a superposition of two basis states exactly analogous to the classical two-component field — and measuring it with an analyzer gives a click with probability cos²θ. A classical beam is an ensemble of such photons; its degree of polarization is the statistical alignment of the ensemble. Fully polarized classical light corresponds to every photon in the same pure state; unpolarized light corresponds to a maximally mixed ensemble. The classical Jones vector and the single-photon state vector are the same object, which is why polarization is a favourite bridge from classical optics to quantum information.

Why can't two beams polarized at right angles interfere?

Interference needs the fields to add and produce a cross term $\mathbf E_1\cdot\mathbf E_2$ . If the two beams are polarized perpendicular to each other that dot product is zero at every instant, so the intensities simply add with no fringes — this is one of the Fresnel–Arago laws of polarized-light interference. It is a statement about optics (the geometry of adding two vector fields), not something we attribute to the electromagnetic transversality article, which supplies only the two-component starting point. To recover fringes you must first project both beams onto a common axis with a polarizer, restoring a nonzero overlap.

What actually happens to the "missing" half when a polarizer halves unpolarized light?

An ideal absorptive polarizer converts the blocked component's energy to heat in the sheet; a reflective or birefringent polarizer sends it off in another direction instead. Either way, the surviving half emerges fully polarized along the transmission axis — the polarizer is not only an attenuator but a purifier, turning V = 0 light into V = 1 light at the cost of half the power (a 3 dB loss). This is why you never recover the lost half by adding a second aligned polarizer: it is gone as heat or deflected, not merely hidden.

How is the degree of polarization related to the Stokes parameters?

They are the same quantity written two ways. The Stokes parameters $S_1, S_2, S_3$ measure the horizontal/vertical, ±45°, and circular imbalances, and $S_0$ is the total intensity. The degree of polarization is $V = \sqrt{S_1^2+S_2^2+S_3^2}/S_0$ , which for a beam analyzed only in the linear basis reduces to the rotating-analyzer contrast $(I_{\max}-I_{\min})/(I_{\max}+I_{\min})$ . The full Stokes/Poincaré machinery — which also captures circular content a linear analyzer misses — is owned by stokes-parameters-poincare-sphere (1469); this article uses the linear-analyzer special case.

Why is the daytime sky polarized, and where is it strongest?

Sunlight scatters off air molecules (Rayleigh scattering). A molecule driven by the incoming field re-radiates like a tiny dipole antenna, and a dipole radiates nothing along its own oscillation axis. Looking at a patch of sky 90° away from the Sun, only the field component perpendicular to the scattering plane can reach you, so that patch is strongly — ideally fully — linearly polarized. The polarization weakens toward the Sun and the anti-solar point. This band of maximum polarization sweeps across the sky as the Sun moves, and animals from bees to ants read it as a compass; the full scattering treatment is Rayleigh scattering, owned by the electromagnetism track.

If I put on polarized sunglasses and tilt my head, why does an LCD screen go dark?

The screen's own output is already linearly polarized (its final layer is a polarizer). Your sunglasses are a second polarizer. When your head is upright the two axes are roughly aligned and light gets through; tilt 90° and the axes cross, so Malus's law drives the transmission to cos²90° = 0 and the screen goes black. At intermediate tilts you see the cos²θ dimming. The same trick reveals that many car windows and camera LCDs emit polarized light — a everyday demonstration that the final stage of these displays is a polarizer.