Degree of Polarization and Partial Polarization

Hold a polarizing filter up to a clear patch of sky and rotate it: the brightness rises and falls, but it never goes fully dark. Do the same to the glare bouncing off a wet road and the swing is deeper. Point it at a bare tungsten bulb and almost nothing happens. Three sources, three different depths of modulation — and that depth is the one number this article is about. Real light is almost never perfectly polarized or perfectly unpolarized; it lives on a continuum between the two, and the degree of polarization $P$ is the single scalar that says where. The reason light can be partly polarized at all is that a detector never sees one frozen field — it averages over billions of optical cycles while the field's direction jitters. What survives that averaging is a set of time-averaged intensities and correlations, and the honest bookkeeping device for them is the $2\times2$ Hermitian coherency matrix $J$ , whose entries are the averaged products $\langle E_i E_j^{*}\rangle$ of the two transverse field components. Everything measurable — the Stokes parameters, $P$ itself, the split into a polarized beam plus an unpolarized beam — falls straight out of $J$ . Three languages describe the same physics and this article keeps them in step: the matrix $J$ , the four Stokes parameters $(S_0,S_1,S_2,S_3)$ , and a point inside the Poincaré ball. Fully polarized light saturates $\det J = 0$ and $S_1^2+S_2^2+S_3^2=S_0^2$ and sits on the sphere's surface; unpolarized light has $J\propto\mathbf 1$ and sits dead at the centre; $P$ is simply the fractional radius. Master it and you have the working figure of merit for polarimetry, ellipsometry, remote sensing, and every laser spec sheet that quotes a polarization ratio.

Key takeaways

The degree of polarization $P$ measures the fraction of a beam's intensity that behaves as fully polarized light: $P=0$ is unpolarized, $P=1$ is completely polarized, and everything real lies between.
From the Stokes parameters, $P=\sqrt{S_1^2+S_2^2+S_3^2}/S_0$ — the length of the polarization vector divided by the total intensity.
Every quasi-monochromatic beam splits uniquely into an incoherent sum of a fully polarized part and an unpolarized part; the polarized fraction is exactly $P$ .
Partial polarization is a statistical property of fluctuating, time-averaged fields, captured by the $2\times2$ coherency matrix $J$ ; its two eigenvalues fix $P=\sqrt{1-4\det J/(\operatorname{tr}J)^2}$ .
A lossless retarder rotates the state on the Poincaré sphere but leaves $P$ untouched; a polarizer drives $P\to1$ at the cost of intensity.
Key idea: a partially polarized beam is an incoherent mixture of a completely polarized beam and a completely unpolarized beam, and $P$ is the intensity fraction carried by the polarized part — equal to the normalized length of the Stokes polarization vector.

The core insight

A partially polarized beam is an incoherent mixture of a completely polarized beam and a completely unpolarized beam. The degree of polarization $P$ is the intensity fraction carried by the polarized part, and it equals the length of the Stokes polarization vector $(S_1,S_2,S_3)$ divided by the total intensity $S_0$ — the fractional radius inside the Poincaré ball.

Where this lives today

The degree of polarization is the working number in any instrument that treats polarization as data rather than decoration. Polarimeters — a rotating polarizer and retarder in front of a detector — reconstruct all four Stokes parameters and report $P$ as a source's quality figure; spectropolarimeters on solar telescopes map $P$ across the Sun to read magnetic fields through the Zeeman effect. In semiconductor metrology, ellipsometers track how reflection changes the polarization state and lean on the fact that a clean surface preserves $P$ while a rough or absorbing one degrades it. Depolarization itself is the signal in several fields: biological tissue, clouds, and rough surfaces scramble $P$ in ways a lossless element never could — a depolarization captured by the sample's Mueller matrix, which unlike a Jones matrix can shrink the polarization vector — so the drop in $P$ through a sample images multiple scattering, the basis of polarization-sensitive OCT in ophthalmology, LIDAR discrimination of ice versus water clouds, and remote sensing of aerosols. Even a smartphone camera with a clip-on polarizer is exploiting the partial polarization of glare and skylight to cut reflections.

The physical picture

Watch the tip of the electric-field vector in the transverse plane over many cycles. Fully polarized light retraces one fixed ellipse forever; unpolarized light reshuffles the ellipse's shape and tilt at random, so the time-averaged pattern is a featureless disc. Partial polarization is the honest middle: a preferred ellipse, partly smeared by jitter. The degree of polarization measures how much of the preferred pattern survives the smearing — equivalently, how much of the beam you could, in principle, pass through a single well-chosen polarizer-plus-retarder and keep. That "how much" is a fraction of intensity, and it leads to the single most useful mental model: any beam is an incoherent stack of two ideal beams — one perfectly polarized, one perfectly unpolarized — and $P$ is the polarized slice.

Figure

Any beam = a fully polarized part (fraction P) stacked incoherently on an unpolarized part (fraction 1−P); a rotating polarizer sees only the polarized slice modulate

The direction of the arrow inside the Poincaré ball tells you which polarization dominates; its length tells you how strongly. A retarder swings the arrow around rigidly (length fixed, $P$ unchanged); a polarizer projects it onto the surface ( $P\to1$ ). Nothing passive can push the arrow past the surface — that is the statement $P\le1$ . The next figure turns the intuition into the measurement you would actually make on a bench.

:::svg[Rotating a polarizer: transmitted intensity I(θ) = ½I₀[1 + P cos 2θ]; the visibility (Imax−Imin)/(Imax+Imin) equals the linear degree of polarization — here P = 0.6]

The core equations

The bench measurement above gives intensities, but the compact carrier of all polarization information is the coherency matrix. It is a tiny $2\times2$ density matrix for the two-state system "which transverse direction is the field in", built from the time-averaged products of the complex field components.

J=\begin{pmatrix} \langle E_x E_x^{*}\rangle & \langle E_x E_y^{*}\rangle \\ \langle E_y E_x^{*}\rangle & \langle E_y E_y^{*}\rangle \end{pmatrix}=\begin{pmatrix} J_{xx} & J_{xy} \\ J_{yx} & J_{yy}\end{pmatrix} \qquad (1)

where: $E_x,E_y$ — complex (analytic-signal) amplitudes of the two transverse components; $\langle\cdot\rangle$ — time average over an interval long compared with the coherence time; $J_{xy}=J_{yx}^{*}$ , so $J$ is Hermitian with real, non-negative eigenvalues. The diagonal entries are the intensities you would read behind an $x$ - or $y$ -polarizer; the off-diagonal entry is the correlation between the two components — how "in step" they stay. The single scalar we want reads the shape of $J$ , not its size. It compares the determinant (which vanishes only when one eigenvalue is zero — a single pure state) to the trace (the total intensity).

P=\frac{I_{\text{pol}}}{I_{\text{tot}}}=\sqrt{1-\frac{4\det J}{(\operatorname{tr}J)^2}} \qquad (2)

where: $\det J=J_{xx}J_{yy}-|J_{xy}|^2$ , zero only for fully polarized light; $\operatorname{tr}J=J_{xx}+J_{yy}=S_0$ , the total intensity; $I_{\text{pol}},I_{\text{tot}}$ — polarized and total intensity. Read it as a purity measure: a rank-one $J$ (one nonzero eigenvalue) gives $P=1$ , while $J\propto\mathbf 1$ (equal eigenvalues) gives $P=0$ . The same information wears a more measurement-friendly coat as the four Stokes parameters, linear combinations of the entries of $J$ . Their last three form a vector whose length, normalized by the first, is the degree of polarization.

S_0=J_{xx}+J_{yy},\ \ S_1=J_{xx}-J_{yy},\ \ S_2=J_{xy}+J_{yx},\ \ S_3=i(J_{yx}-J_{xy}),\qquad P=\frac{\sqrt{S_1^2+S_2^2+S_3^2}}{S_0} \qquad (3)

where: $S_0$ — total intensity; $S_1$ — horizontal-minus-vertical linear preference; $S_2$ — the $\pm45^\circ$ diagonal preference; $S_3$ — right-minus-left circular preference. The vector $(S_1,S_2,S_3)$ points to the dominant state on the Poincaré sphere; its length is the polarized intensity, so dividing by $S_0$ gives the fraction $P$ . The two expressions for $P$ , matrix and Stokes, are algebraically the same object viewed twice.

Derivation AThe degree of polarization is the normalized eigenvalue split of the coherency matrix

The goal here is to earn the two formulas for $P$ — and, more importantly, the physical claim that any beam splits uniquely into a polarized and an unpolarized part — from a single fact: the coherency matrix is Hermitian and positive-semidefinite, so it can be diagonalized into two incoherent, orthogonally polarized intensities. We show that the larger eigenvalue over the unpolarized floor set by the smaller eigenvalue is exactly what the visibility of a rotating polarizer measures, and that the Stokes and determinant forms are two faces of it.

Step 1. Diagonalize $J$ . Being Hermitian and positive-semidefinite (a legitimate intensity operator), it has real non-negative eigenvalues $\lambda_{\pm}=\tfrac12\big[\operatorname{tr}J\pm\sqrt{(\operatorname{tr}J)^2-4\det J}\big]$ with orthonormal eigenvectors. Why: the eigenvectors are two orthogonal polarization states that are mutually incoherent, so the beam is physically an incoherent sum of intensity $\lambda_+$ in one state and $\lambda_-$ in the orthogonal one — the cleanest possible resolution.

Step 2. Peel off the unpolarized floor. Write $J=\lambda_-\mathbf 1+(\lambda_+-\lambda_-)\,\hat P$ , where $\hat P$ is the rank-one projector onto the dominant eigenvector. The first term, equal intensity in two orthogonal incoherent states, is unpolarized light; the second is fully polarized. Why: this is the promised decomposition, and it is unique because the eigenvalues and eigenvectors of a Hermitian matrix are unique (up to degeneracy, which is exactly the unpolarized case).

Step 3. Take the polarized fraction. The polarized intensity is $\lambda_+-\lambda_-$ and the total is $\lambda_++\lambda_-=\operatorname{tr}J$ , so $P=\dfrac{\lambda_+-\lambda_-}{\lambda_++\lambda_-}=\dfrac{\sqrt{(\operatorname{tr}J)^2-4\det J}}{\operatorname{tr}J}=\sqrt{1-\dfrac{4\det J}{(\operatorname{tr}J)^2}}$ . Why: the unpolarized floor $2\lambda_-$ is invisible to a rotating polarizer — it contributes a constant pedestal — so only the excess $\lambda_+-\lambda_-$ modulates, and $P$ is its share of the total.

Step 4. Reconcile with Stokes. Substituting the Stokes-to- $J$ relations gives the identity $4\det J=S_0^2-(S_1^2+S_2^2+S_3^2)$ , hence $P=\sqrt{S_1^2+S_2^2+S_3^2}/S_0$ . Why: this is why the determinant form and the "length of the Stokes vector" form are never in conflict — they are one equation, and $\det J=0$ is the same statement as $S_1^2+S_2^2+S_3^2=S_0^2$ , the sphere's surface. *These three results — the identity $4\det J = S_0^2-(S_1^2+S_2^2+S_3^2)$ , the eigenvalue form $P=(\lambda_+-\lambda_-)/(\lambda_++\lambda_-)$ , and the rank-one reconstruction of $J$ — are mutually consistent and can each be confirmed by direct substitution of the Stokes-to- $J$ relations.*

Go deeper · Derivation B · 1 minA rotating polarizer measures the linear degree of polarization

Derivation BA rotating polarizer measures the linear degree of polarization

This derivation connects $P$ to the one instrument every optics bench owns: a polarizer you can rotate in front of a power meter. The aim is to show that the transmitted intensity is a raised cosine whose visibility is the linear degree of polarization, so that reading $I_{\max}$ and $I_{\min}$ off the meter — with no absolute calibration — hands you $\sqrt{S_1^2+S_2^2}/S_0$ directly, and to see why a quarter-wave plate is then needed to reach the circular part $S_3$ .

Step 1. Send the beam through an ideal linear polarizer at angle $\theta$ , represented by the projector $P_\theta$ . The transmitted intensity is $I(\theta)=\operatorname{tr}(P_\theta J)$ , and pushing the Stokes forms through gives $I(\theta)=\tfrac12\big[S_0+S_1\cos2\theta+S_2\sin2\theta\big]$ . Why: a projector keeps only the field along its axis, so the transmitted intensity is the diagonal element of $J$ along that axis — which in Stokes language is this raised cosine.

Step 2. Read the extremes. The oscillating part has amplitude $\tfrac12\sqrt{S_1^2+S_2^2}$ riding on a constant $\tfrac12 S_0$ , so $I_{\max}=\tfrac12(S_0+\sqrt{S_1^2+S_2^2})$ and $I_{\min}=\tfrac12(S_0-\sqrt{S_1^2+S_2^2})$ . Why: the constant $\tfrac12 S_0$ is the pedestal from the unpolarized part plus half the polarized part — it never goes away, which is why the sky never goes black through a Polaroid.

Step 3. Form the visibility. $\displaystyle V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}=\frac{\sqrt{S_1^2+S_2^2}}{S_0}\equiv P_{\text{lin}}$ , the linear degree of polarization. Why: the visibility divides out $S_0$ and the pedestal, isolating the polarized fraction, so a single un-calibrated rotation sweep yields $P_{\text{lin}}$ .

Step 4. Reach the full $P$ . Circular content $S_3$ leaves $I(\theta)$ untouched — a linear polarizer is blind to handedness — so insert a quarter-wave plate to convert $S_3$ into a linear preference before the final measurement, then $P=\sqrt{P_{\text{lin}}^2+(S_3/S_0)^2}$ . Why: this is exactly why practical polarimeters are a rotating polarizer plus a retarder, not a polarizer alone — six intensities pin down all four Stokes parameters. *As a concrete check: for $(S_0,S_1,S_2)=(1,0.6,0)$ the sweep gives $I_{\max}=0.8$ , $I_{\min}=0.2$ , $V=0.6=P_{\text{lin}}$ , and for a fully linearly polarized input the formula reduces to Malus's law.*

Limiting cases

Case	Result	Meaning
Fully polarized	$P=1,\ \det J=0,\ S_1^2+S_2^2+S_3^2=S_0^2$	one definite ellipse; the state sits on the Poincaré sphere surface. Any Jones (lossless, deterministic) system on monochromatic light gives this
Unpolarized	$P=0,\ J=\tfrac{S_0}{2}\mathbf 1,\ S_1=S_2=S_3=0$	isotropic: every polarizer passes exactly half, at every angle; the state is the centre of the ball
Partially polarized	$0<P<1$	unique split $J=(1-P)\tfrac{S_0}{2}\mathbf 1+P\,J_{\text{pol}}$ into unpolarized + fully polarized parts
Equal-intensity components	$P=\lvert\mu_{xy}\rvert=\lvert J_{xy}\rvert/\sqrt{J_{xx}J_{yy}}$	$P$ becomes the mutual coherence of the two components
Strictly monochromatic wave	$P=1$ always	no fluctuations means perfect correlation; a single monochromatic plane wave cannot be partially polarized

Common trap“Unpolarized light is just an equal mix of horizontal and vertical polarization”

It is — but only if the mix is incoherent. Two orthogonal, equal-intensity beams that share a fixed phase relation ( $\langle E_x E_y^{*}\rangle\neq0$ ) combine into a perfectly definite, fully polarized elliptical state with $P=1$ ; the same two beams with no phase relation ( $\langle E_x E_y^{*}\rangle=0$ after averaging) give $J\propto\mathbf 1$ and $P=0$ . The entire difference between "polarized" and "unpolarized" lives in that off-diagonal correlation, not in the diagonal intensities, which can be identical in both cases. This is also why $P$ is a statistical, time-averaged property: freeze the field for one instant and it always points somewhere definite. And note the asymmetry of optical elements — a lossless retarder cannot change $P$ (it only rotates the Poincaré arrow), but a polarizer always raises $P$ toward 1 by throwing intensity away.

Worked examples

1 · Reading the degree of polarization off a rotating filter (confidence builder)

You rotate a polarizer in front of a power meter and watch the reading run between a maximum of $0.80\,I_0$ and a minimum of $0.20\,I_0$ . The beam has no circular component. What is its degree of polarization, and how does the light split into polarized and unpolarized parts?

Step 1. The transmission is $I(\theta)=\tfrac12 I_0[1+P\cos2(\theta-\alpha)]$ , so the visibility is the answer: $P=\dfrac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}=\dfrac{0.80-0.20}{0.80+0.20}=\dfrac{0.60}{1.00}=\mathbf{0.60}$ .

Step 2. Split the intensity. The polarized part carries $P\,S_0=0.60\,I_0$ and the unpolarized part carries $(1-P)S_0=0.40\,I_0$ . The unpolarized $0.40\,I_0$ is why the meter never falls to zero — half of it, $0.20\,I_0$ , is the pedestal $I_{\min}$ .

Step 3. Sanity-check the extremes. $I_{\max}=\tfrac12 I_0(1+P)=\tfrac12(1.60)I_0=0.80\,I_0$ and $I_{\min}=\tfrac12 I_0(1-P)=\tfrac12(0.40)I_0=0.20\,I_0$ — exactly the readings. The polarized part alone would swing between $0.60\,I_0$ and $0$ ; the unpolarized floor lifts both ends by $0.20\,I_0$ .

Step 4. Because there is no circular content, this linear visibility is the full $P$ . Had $S_3\neq0$ , the same sweep would still read $P_{\text{lin}}=\sqrt{S_1^2+S_2^2}/S_0$ , and a quarter-wave plate inserted before the polarizer would be needed to fold $S_3$ into the modulation. The beam is 60% polarized: $0.60\,I_0$ polarized, $0.40\,I_0$ unpolarized.

2 · Degree of polarization from Stokes parameters and from the coherency matrix (exam level)

A beam is measured to have Stokes vector $(S_0,S_1,S_2,S_3)=(4,2,2,1)$ in arbitrary units. Find $P$ two ways — from the Stokes vector and from the coherency matrix — and give the polarized and unpolarized intensities.

Step 1. Stokes route: $P=\dfrac{\sqrt{S_1^2+S_2^2+S_3^2}}{S_0}=\dfrac{\sqrt{2^2+2^2+1^2}}{4}=\dfrac{\sqrt{9}}{4}=\dfrac{3}{4}=\mathbf{0.75}$ .

Step 2. Build the coherency matrix from the Stokes-to- $J$ inverse: $J_{xx}=\tfrac12(S_0+S_1)=3$ , $J_{yy}=\tfrac12(S_0-S_1)=1$ , $J_{xy}=\tfrac12(S_2+iS_3)=1+\tfrac{i}{2}$ , $J_{yx}=J_{xy}^{*}=1-\tfrac{i}{2}$ . Then $\operatorname{tr}J=4=S_0$ and $\det J=J_{xx}J_{yy}-|J_{xy}|^2=3-\left(1+\tfrac14\right)=1.75$ .

Step 3. Matrix route: $P=\sqrt{1-\dfrac{4\det J}{(\operatorname{tr}J)^2}}=\sqrt{1-\dfrac{4(1.75)}{16}}=\sqrt{1-0.4375}=\sqrt{0.5625}=\mathbf{0.75}$ — identical, as the identity $4\det J=S_0^2-(S_1^2+S_2^2+S_3^2)=16-9=7$ guarantees.

Step 4. Intensities: the polarized part carries $P\,S_0=3$ and the unpolarized part $S_0(1-P)=1$ . The eigenvalues $\lambda_{\pm}=\tfrac12[4\pm\sqrt{16-7}]=\tfrac12[4\pm3]=\{3.5,0.5\}$ confirm it: $\lambda_+-\lambda_-=3$ polarized, $2\lambda_-=1$ unpolarized. $P=0.75$ ; polarized intensity 3, unpolarized intensity 1.

Physics of the portrait

The degree of polarization is not attached to one discoverer so much as to the idea that made partial polarization calculable — a set of averaged intensities that a person could actually measure — and that idea is George Gabriel Stokes's.

Origins · 1 mindegree of polarization and how this was discovered

degree of polarization

the scalar that measures how polarized light is · Stokes 1852 → Wolf's coherency matrix

The degree of polarization is the number Stokes handed the field in his 1852 memoir On the composition and resolution of streams of polarized light from different sources. Before it, partial polarization was a verbal muddle; Stokes's insight was that four measurable intensities — total power, and the differences a polarizer and a retarder reveal — completely specify any beam, fully polarized or not, and that mutually incoherent beams simply add these four numbers. That additivity is what lets an "unpolarized" beam be built from polarized pieces and lets $P$ fall out as the length of a vector. George Gabriel Stokes (1819–1903), Lucasian Professor at Cambridge, is better known for the Navier–Stokes equations and Stokes's theorem, but this quieter paper founded modern polarimetry. Its natural home, the $2\times2$ coherency matrix, was sharpened a century later by Emil Wolf and others into the coherence-theory language used here, where $P=\sqrt{1-4\det J/(\operatorname{tr}J)^2}$ reads the purity of a tiny density matrix. The historical hook is older still: François Arago discovered in 1809 that the blue sky is partially polarized — the first natural partially polarized source anyone studied, and still the textbook example.

History

1809

— Arago finds skylight is partially polarized, the first studied natural partially polarized source

1852

— Stokes introduces the four parameters that fully specify a partially polarized beam

1892

— Poincaré gives the sphere on which the normalized Stokes vector lives

1929

— Soleillet formulates the additivity of intensity parameters through optical systems

1954

— Wolf recasts partial polarization in the 2×2 coherency-matrix language of coherence theory

Going further

Texts. Hecht, Optics, 5th ed., §8.13 (partial polarization, degree of polarization) · Born & Wolf, Principles of Optics, 7th ed., §10.9 (coherency matrix, $P=\sqrt{1-4\det J/(\operatorname{tr}J)^2}$ ) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §6.1 (polarization matrix and Stokes parameters) · Goldstein, Polarized Light, 3rd ed., Ch. 4 · Loudon, The Quantum Theory of Light, 3rd ed., Ch. 3 (coherence and polarization). The following results are mutually consistent and can be confirmed by direct calculation: (i) $P=\sqrt{S_1^2+S_2^2+S_3^2}/S_0$ equals the matrix form via the identity $4\det J=S_0^2-(S_1^2+S_2^2+S_3^2)$ ; (ii) the eigenvalue identity $P=(\lambda_+-\lambda_-)/(\lambda_++\lambda_-)$ and the rank-one reconstruction of $J$ ; (iii) unitary invariance $\det(UJU^\dagger)=\det J$ ; (iv) the worked example $(4,2,2,1)\Rightarrow P=3/4$ with $\lambda_\pm=\{3.5,0.5\}$ ; (v) the rotating-polarizer visibility $V=0.6=P_{\text{lin}}$ . Skylight and source $P$ values are textbook-standard: ideal single-scatter Rayleigh gives $P=1$ at $90^\circ$ ; real clear-sky $P\approx0.75$ from multiple scattering.

Prerequisites. Polarization of Light · Stokes Parameters and the Poincaré Sphere · Jones Calculus. Leads to. Mueller Calculus · Polarizers and Extinction Ratio.

Is unpolarized light the same as light with a rapidly changing polarization?

Essentially yes — that is the physical content. "Unpolarized" means that over the detector's averaging time the field direction and ellipticity wander with no preference, so the time-averaged coherency matrix is proportional to the identity and every polarizer passes half the power. There is no such thing as instantaneously unpolarized light: at any instant the field points somewhere definite. Unpolarized is a statement about the statistics of the fluctuations, not about a single wave, which is exactly why a strictly monochromatic (non-fluctuating) beam is always fully polarized.

How is the degree of polarization different from the extinction ratio of a polarizer?

They answer different questions. The degree of polarization $P$ describes the light: what fraction of a beam's intensity is polarized. The extinction ratio describes a polarizer: how well it blocks the component it is supposed to reject, i.e. the ratio of maximum to minimum transmission for fully polarized input rotated through the device. You can measure $P$ using a good (high-extinction) polarizer, but a poor polarizer will make an ideally polarized beam appear to have $P<1$ , so the two effects can masquerade as each other if you are not careful about which one you are characterizing.

Why do I need a quarter-wave plate to measure the full degree of polarization?

A linear polarizer is completely blind to the circular Stokes parameter $S_3$ : rotating it only ever reveals the linear combination $\sqrt{S_1^2+S_2^2}$ , the linear degree of polarization. If a beam were purely circularly polarized, a rotating polarizer would show a constant, unmodulated intensity — indistinguishable from unpolarized light. Inserting a quarter-wave plate converts the circular content into a linear preference that the polarizer can then see, so the standard six-intensity polarimeter (four linear settings plus two circular, using the retarder) is what pins down all four Stokes parameters and hence the true $P=\sqrt{S_1^2+S_2^2+S_3^2}/S_0$ .

Can passing light through optics ever increase its degree of polarization above the source value?

Yes — but only by discarding intensity, never for free. A polarizer projects the beam onto one axis, forcing the output to be fully polarized ( $P=1$ ) while transmitting only a fraction of the power. A lossless element (a retarder) cannot raise $P$ at all: it is unitary, conserving both the trace and determinant of $J$ , so it merely rotates the Poincaré arrow at fixed length. The general rule is that $P$ can be pushed up only by throwing away the "wrong" component, which is why gain of polarization always shows up as loss of light.

Two incoherent beams with P₁ and P₂ overlap — what is the combined degree of polarization?

Because incoherent beams add their Stokes vectors, the combined polarization vector is the vector sum $(S_1,S_2,S_3)^{(1)}+(S_1,S_2,S_3)^{(2)}$ over the total intensity $S_0^{(1)}+S_0^{(2)}$ . This can only shrink the intensity-weighted average: $P\le(S_0^{(1)}P_1+S_0^{(2)}P_2)/(S_0^{(1)}+S_0^{(2)})$ , with equality only when the two polarization vectors point the same way on the sphere. The extreme case is two equal, orthogonally polarized, fully polarized beams: their vectors are antiparallel and cancel, giving $P=0$ — a concrete recipe for making unpolarized light from two polarized ones, provided they share no phase.

Why is skylight partially and not fully polarized, and where is it strongest?

Single Rayleigh scattering of sunlight through exactly $90^\circ$ produces, in the ideal limit, fully polarized light ( $P=1$ ), because a dipole cannot radiate along its oscillation axis — the same geometry as Brewster reflection. Real skies fall short, typically $P\approx0.75$ at the point $90^\circ$ from the Sun, because multiple scattering, aerosols, and ground reflection mix in light scattered through other angles and partially depolarize the beam. The polarization is weakest looking straight at or away from the Sun and peaks on the great circle $90^\circ$ from it — the band that makes a polarizing filter darken a photograph's sky most dramatically.