Polarization of Light

Beginner

Step 1. Tie a skipping rope to a door handle and flick your end. You can make the rope wiggle up and down, from side to side, or along any slanted line in between, while the wave itself always runs towards the door. Light is a wave of this sideways kind, and the direction in which it wiggles is called its polarization.

Step 2. A lamp or the Sun sends out light whose wiggle direction changes constantly and at random, so every direction is mixed in. Such light is called unpolarized. In the animation, the yellow wiggles leaving the lamp point every which way.

Step 3. A polarizing filter behaves like a row of narrow slots: it lets through only the part of the wiggle that lines up with the slots and swallows the rest. Behind filter 1 the blue wave wiggles only up and down, and about half of the light is gone.

Step 4. Now watch filter 2 turn. While its slots match those of filter 1, the light passes and the spot on the screen is bright. As it turns, the red wave behind it shrinks; after a quarter turn the slots are crossed and the screen goes dark. Keep turning and the light returns.

Step 5. Sunlight bouncing off a lake or a wet road comes back wiggling mostly sideways. Polarized sunglasses have upright slots, so they block much of this glare.

Check yourself: wearing polarized sunglasses, you look at the shine on a puddle and then tilt your head onto your shoulder. What happens? Answer: the glare comes back, because your slots now lie sideways, lined up with the reflected wiggle.

Intermediate

Light is a transverse wave

Polarized sunglasses and LCD screens depend on a property that light has and sound in air lacks. Light is an electromagnetic wave: an electric field $\vec{E}$ and a magnetic field oscillate together, each perpendicular to the other and to the beam. Its polarization is the direction along which $\vec{E}$ oscillates: vertical, horizontal or anything in between. Sound in air vibrates along its direction of travel, so it has no such direction and cannot be polarized.

Light whose field keeps one direction is linearly polarized. Sunlight and lamplight come from countless atoms emitting independently, so their direction of vibration changes at random, far faster than any instrument can follow: they are unpolarized. The glare from a wet road and the light of a blue sky are partially polarized, a mixture of the two.

An eye or a light meter registers the intensity $I$, the power per unit area, in watts per square metre (W/m²). It is proportional to the square of the field amplitude $E_0$, measured in volts per metre:
$I=\frac{1}{2}c{\epsilon }_0{E}_0^2$
where $c=3.00\times 10^{8}m/s$ is the speed of light and ${\epsilon }_0=8.85\times 10^{-12}F/m$ the permittivity of free space. Halving the amplitude divides the intensity by four.

What a polarizer does

A polarizer has a transmission axis: it passes the component of $\vec{E}$ along that axis and absorbs the component across it, so the light that leaves is linearly polarized along the axis. A common polarizing sheet is a plastic film of long polymer chains, stretched into line and soaked in iodine so that electrons can move along them. A field along the chains drives currents and is absorbed, so the transmission axis is perpendicular to the chains, not along them as a picket fence would suggest; the slots in the animation picture the axis, not the molecules.

Unpolarized light contains every direction equally. A direction at an angle $\phi$ to the axis keeps the fraction ${\cos }^2\phi$ of its intensity, which averages to one half over all directions, so an ideal polarizer passes half of the light, however it is turned. In the animation, $I_0=100W/{\mathrm{m}}^2$ from the lamp becomes $I_1=50W/{\mathrm{m}}^2$ behind filter 1, and the blue wave vibrates only up and down.

Malus's law

A second polarizer, the analyzer, has its axis at an angle $\theta$ to the polarization of the light reaching it. The front view in the animation splits the blue field $E_1$ in two: the part along the axis, $E_1\cos \theta$, passes as the red field $E_2$, and the part across it, $E_1\sin \theta$, is absorbed. Squaring the amplitude gives the intensity, so the working equations are:

• $I_1=\frac{I_0}{2}\text{(unpolarized light, ideal polarizer)}$
• $E_2=E_1\cos \theta$
• $I_2=I_1{\cos }^2\theta \text{(Malus’s law)}$

The last one is named after Étienne-Louis Malus, who discovered in 1808 that light reflected from a window is polarized. The absorbed part, $I_1{\sin }^2\theta$, makes up the difference, since ${\cos }^2\theta +{\sin }^2\theta =1$. At $90^{\circ }$ the filters are crossed and nothing passes, and because an axis turned through $180^{\circ }$ is the same axis, the pattern repeats every half turn.

Worked example: the analyzer in the animation

Filter 2 turns at 22.5° per second, so $\theta$ reaches $22.5^{\circ }$, $45^{\circ }$ and $67.5^{\circ }$ after 1, 2 and 3 s. With $I_1=50W/{\mathrm{m}}^2$, Malus's law gives
$I_2=50{\cos }^{2}22.5^{\circ }=50\cdot 0.854=42.7W/{\mathrm{m}}^2$
$I_2=50{\cos }^{2}45^{\circ }=50\cdot 0.500=25.0W/{\mathrm{m}}^2$
$I_2=50{\cos }^{2}67.5^{\circ }=50\cdot 0.146=7.3W/{\mathrm{m}}^2$
and zero at $90^{\circ }$, after 4 s. These are the values on the $I_2$ readout; in the second half of the loop they return in reverse order.

Compare the waves with the readout. At $45^{\circ }$ the red wave still has $\cos 45^{\circ }=0.71$ of the blue wave's amplitude, yet it carries only half the intensity. At $60^{\circ }$, reached after 2.7 s, the amplitude is halved and the intensity quartered, $I_2=50\cdot 0.25=12.5W/{\mathrm{m}}^2$: in field units, $E_1=\sqrt{2I_1/(c{\epsilon }_0)}\approx 194V/m$ becomes $E_2\approx 97V/m$. In the front view, the green arc marks $\theta$ while the red arrow $E_2$ shrinks.

Worked example: three polarizers

Two crossed polarizers, at $0^{\circ }$ and $90^{\circ }$, block everything. Slip a third one between them at $30^{\circ }$ to the first, send in unpolarized light with $I_0=100W/{\mathrm{m}}^2$, and always use the angle between neighbouring axes. The first filter passes $100/2=50W/{\mathrm{m}}^2$. The middle one passes $50{\cos }^{2}30^{\circ }=37.5W/{\mathrm{m}}^2$, now polarized at $30^{\circ }$. The last one, $60^{\circ }$ further on, passes $37.5{\cos }^{2}60^{\circ }=9.4W/{\mathrm{m}}^2$.

For a middle polarizer at any angle $\theta$, the last one is $90^{\circ }-\theta$ from it, and ${\cos }^2(90^{\circ }-\theta )={\sin }^2\theta$:
$I_3=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta \text{(polarizers at }{0}^{\circ },\theta ,90^{\circ }\text{)}$
This equals $(I_0/8){\sin }^{2}2\theta$, largest at $\theta =45^{\circ }$, where $I_0/8=12.5W/{\mathrm{m}}^2$ gets through.

Adding a filter has let light through where there was none. The middle filter opens no leak: it re-polarizes the light along its own axis, which is no longer at right angles to the last one. Each polarizer resets the direction of the field.

Polarization by reflection

Light reflected from water, glass or a road is partially polarized, with $\vec{E}$ mostly parallel to the surface. At Brewster's angle ${\theta }_B$ it is completely polarized:
$\tan {\theta }_B=\frac{n_2}{n_1}\text{(Brewster’s angle)}$
${\theta }_B+{\theta }_2=90^{\circ }\text{(reflected }\perp \text{ refracted)}$
Here $n_1$ is the refractive index of the medium the light comes from, $n_2$ that of the reflecting one and ${\theta }_2$ the angle of refraction: the reflected and refracted rays are then perpendicular.

From air into water ($n=1.33$), ${\theta }_B=\arctan 1.33=53.1^{\circ }$ and ${\theta }_2=36.9^{\circ }$; for glass ($n=1.50$), ${\theta }_B=56.3^{\circ }$. When the Sun stands $36.9^{\circ }$ above a calm lake, its light meets the water at exactly $53.1^{\circ }$, and the glare is horizontally polarized. Sunglasses with a vertical transmission axis block it, while the mostly unpolarized rest of the scene is only halved.

Common mistakes

• Using $\cos \theta$ instead of ${\cos }^2\theta$. At $60^{\circ }$ the amplitude is halved but the intensity is quartered: from $I_1=50W/{\mathrm{m}}^2$ you get 12.5 W/m², not 25 W/m².
• Applying Malus's law to unpolarized light, or forgetting the factor $\frac{1}{2}$ at the first filter: without it the three-polarizer answer doubles to 18.75 W/m².
• Measuring every angle from the first filter. Use the angle between neighbouring axes: the last of the three polarizers is $60^{\circ }$ from the middle one, and taking $90^{\circ }$ would wrongly give zero.
• Taking the picket fence literally. A polarizing sheet absorbs the field along its chains, and a wire grid reflects the field along its wires, so the transmission axis is perpendicular to them.
• Using the sine instead of the tangent for Brewster's angle: $\sin {\theta }_B=1.33$ has no solution, and $\arcsin (1/1.33)=48.8^{\circ }$ is the critical angle for light leaving water, a different effect.

Where you meet it

Polarizing filters on sunglasses and cameras cut reflections from water and shop windows and darken a blue sky, whose light is partly polarized by scattering, most strongly $90^{\circ }$ from the Sun; bees navigate by that pattern. An LCD screen holds liquid crystal between two polarizers, and each pixel sets its brightness by how far it turns the polarization, so a phone seen through polarized sunglasses goes dark at one particular tilt. 3D cinemas send the two eyes' images in different polarizations, and engineers see stress in plastic models between crossed polarizers as coloured fringes.

Reflection and Refraction gave the angles used here. Electromagnetic Waves explains why light is transverse and where $I=\frac{1}{2}c{\epsilon }_0{E}_0^2$ comes from, and Wave–Particle Duality introduces the photons, for which Malus's law becomes a probability ${\cos }^2\theta$ of getting through.

Advanced

The field as a two-dimensional oscillator

Polarization describes how the electric field of a light wave moves across the beam. Every polarizer, reflecting surface and crystal in this lesson acts on that motion, so we start with a clear picture of it.

Take a wave travelling along the $z$ axis. Its electric field $\vec{E}$ lies across the beam, in the $xy$ plane, so it has two components, $E_x$ and $E_y$. Both oscillate at the light's angular frequency $\omega$, but each has its own amplitude, $a$ for $E_x$ and $b$ for $E_y$, and $E_y$ may run ahead of $E_x$ by a phase difference $\delta$. At a fixed point in the beam,
$E_x=a\cos \omega t,E_y=b\cos (\omega t+\delta )$
The two components are independent of each other, so the field behaves like a two-dimensional oscillator, much as a pendulum bob can swing to and fro in any direction, or round in an ellipse.

Legend

$a$	Amplitude of $E_x$: half the width of the box.	$V/m$
$b$	Amplitude of $E_y$: half the height of the box.	$V/m$
$\vec{E}$	Electric field at the instant drawn; its tip traces the ellipse.	$V/m$
$E_x$	Component of $\vec{E}$ along $x$.	$V/m$
$E_y$	Component of $\vec{E}$ along $y$.	$V/m$
$\omega t$	Phase of the wave at the point: angular frequency times time. It is the horizontal axis of the graph.	$\mathrm{rad}$
$\delta$	Phase difference between the components: $E_y$ reaches its peak a phase $\delta$ before $E_x$.	$\mathrm{rad}$

Figure 1 shows this motion in two ways. On the left is the plane across the beam. At each instant the tip of $\vec{E}$ sits at the point $(E_x,E_y)$, so the two components are the shadows of the arrow on the axes. Because $E_x$ never exceeds $a$ and $E_y$ never exceeds $b$, the tip stays inside the box of width $2a$ and height $2b$; and because each component reaches its peak once per cycle, the curve touches all four sides of that box. On the right the same two components are plotted against the phase $\omega t$. Here $\delta =60^{\circ }$, so each peak of $E_y$ comes a phase $\delta$ before the matching peak of $E_x$. The dots on the two curves mark the instant drawn on the left.

To find the shape of the curve, eliminate the time. Expanding the cosine gives $E_y/b=\cos \omega t\cos \delta -\sin \omega t\sin \delta$. The first equation says $\cos \omega t=E_x/a$, so $\sin \omega t\sin \delta =(E_x/a)\cos \delta -E_y/b$. Squaring both sides and replacing ${\sin }^2\omega t$ by $1-E_x^2/a^2$ leaves
$\frac{E_x^2}{a^2}+\frac{E_y^2}{b^2}-\frac{2E_x{E}_y}{ab}\cos \delta ={\sin }^2\delta$

This is the equation of an ellipse. In general, then, the tip of $\vec{E}$ traces an ellipse inside the $2a$ by $2b$ box, and the three numbers $a$, $b$ and $\delta$ fix its size, shape and tilt.

Two special cases have names. For $\delta =0$ the right side is zero and the left side is a perfect square, $(E_x/a-E_y/b{)}^2=0$, so $E_y=(b/a)E_x$: the ellipse collapses to a straight line at $\arctan (b/a)$ to the $x$ axis. This is linear polarization. For $\delta =\pi$ the sign of the cross term flips and the line lies at $-\arctan (b/a)$. On the right of Figure 1 the two curves would then rise and fall together ($\delta =0$) or in opposition ($\delta =\pi$).

For $a=b$ and $\delta =\pm \pi /2$ the cross term vanishes and the equation becomes $E_x^2+E_y^2=a^2$: the tip goes round a circle, and the light is circularly polarized. The two curves are then a quarter of a cycle apart, and the two signs of $\delta$ make the tip turn in opposite senses. Every other case is elliptical polarization. Two opposite conventions exist for calling a sense left or right, so always say whether you are looking along the beam or towards the source.

The polarization does not change how much energy the wave carries. The intensity is proportional to the average of $E_x^2+E_y^2$, and each ${\cos }^2$ averages to one half over a cycle, whatever its phase. So $I=\frac{1}{2}c{\epsilon }_0(a^2+b^2)$ for every $\delta$: the polarization only decides how the field is shared between the two components. The magnetic field, of size $E/c$, is locked at right angles to $\vec{E}$, so it carries no extra information. It is $\vec{E}$ that pushes on electrons, which is why polarizers and surfaces respond to its direction, and why "the polarization" always means the direction of the electric field.

A polarizer as a projection

A polarizer is a filter that lets through only one direction of the field. Seeing it as a projection tells us exactly how much light passes, and why two crossed polarizers go dark.

Legend

$\hat{u}$	Pass axis of the polarizer, across the wires: a unit vector.
${\vec{E}}_1$	Field arriving at the polarizer, at the angle $\theta$ to $\hat{u}$.	$V/m$
${\vec{E}}_2$	Field that passes: the part of ${\vec{E}}_1$ along $\hat{u}$, of size $E_1\cos \theta$.	$V/m$
$E_1\sin \theta$	Part of ${\vec{E}}_1$ along the wires; it does not get through.	$V/m$
$\theta$	Angle between ${\vec{E}}_1$ and $\hat{u}$; marked in degrees on the graph.	$\mathrm{rad}$
$I_1$	Intensity of the light arriving at the polarizer.	$W/{\mathrm{m}}^2$
$I_2$	Intensity that passes, $I_1{\cos }^2\theta$.	$W/{\mathrm{m}}^2$

The left of Figure 2 shows a wire-grid polarizer face-on: parallel metal wires, spaced much closer than one wavelength. Its pass axis, the unit vector $\hat{u}$, runs across the wires. The incoming field ${\vec{E}}_1$ makes the angle $\theta$ with $\hat{u}$, so it splits into two parts. The part along the wires, of size $E_1\sin \theta$, drives electrons up and down the wires. The currents it sets up radiate a field that cancels this part beyond the grid and sends it back as a reflected wave. The part along $\hat{u}$, of size $E_1\cos \theta$, would have to drive electrons across the wires, where they cannot travel far, so it passes almost untouched as the field ${\vec{E}}_2$, the horizontal arrow. A sheet polarizer works in the same way with long iodine-doped polymer chains in place of the wires; the chains absorb the blocked part instead of reflecting it.

Either way, the polarizer keeps the projection of the field on its pass axis:
${\vec{E}}_2=({\vec{E}}_1\cdot \hat{u})\hat{u},E_2=E_1\cos \theta$
Intensity goes as the square of the field amplitude, so the passed intensity is
$I_2=I_1{\cos }^2\theta =\frac{1}{2}{I}_1(1+\cos 2\theta )$
This is Malus's law. The second form, from ${\cos }^2\theta =\frac{1}{2}(1+\cos 2\theta )$, shows that the curve repeats every $180^{\circ }$: an axis turned half a turn is the same axis. The rest of the light, $I_1{\sin }^2\theta$, is absorbed or reflected.

The right of Figure 2 plots $I_2$ against $\theta$. The dashed line at the top marks $I_1$, and the dot marks the angle drawn on the left. In the animation, filter 1 delivers $I_1=50W/{\mathrm{m}}^2$ and the analyzer turns steadily, so the graph of this level traces the same curve: 50 W/m² at $\theta =0$, 25 W/m² at $45^{\circ }$, zero at $90^{\circ }$ and 50 W/m² again at $180^{\circ }$.

The slope says how strongly the reading reacts to a small turn of the polarization: $d{I}_2/d\theta =-I_1\sin 2\theta$. It is steepest at $45^{\circ }$ and $135^{\circ }$, where it is 50 W/m² per radian, or 0.87 W/m² per degree. Those are the best settings for detecting a small rotation of a polarization.

At $90^{\circ }$, with the polarizers crossed, the curve touches zero with zero slope. An analyzer a small angle $ϵ$ (in radians) away from crossing passes $I_1{\sin }^2ϵ\approx I_1{ϵ}^2$. One degree lets through only $3\times 10^{-4}$ of $I_1$, and anywhere within $5.7^{\circ }$ of crossing less than 1 % gets through; that is why the animation's screen stays dark from 3.74 s to 4.26 s. Such a flat minimum is hard to pin down by eye, so the null is best found as the midpoint between two settings, one on each side, that look equally bright.

Unpolarized light

Light from a lamp or the Sun shows no preferred direction: a single polarizer passes the same amount however it is turned. Why is that, and why is the amount exactly one half?

Such a source is a crowd of independent emitters, each starting and stopping at random. Visible sunlight spans 400 to 700 nm, a spread of about $3\times 10^{14}\mathrm{Hz}$ in frequency. A wave with that spread keeps one ellipse only for a time of about one over the spread, roughly 3 fs, before the direction and phase of its field drift to new values.

No detector is that fast, so every measurement averages over all directions $\phi$ of the field. A polarizer passes the fraction ${\cos }^2\phi$ of each direction, and the average over a full turn is
$\frac{1}{2\pi }{\int }_0^{2\pi }{\cos }^2\phi d\phi =\frac{1}{2}$
So a polarizer passes half of unpolarized light, $I_1=I_0/2$, whichever way it points. In the animation that is the 50 W/m² behind filter 1, from a lamp giving $I_0=100W/{\mathrm{m}}^2$. The animation models natural light in just this way, each half-wave vibrating in a random direction.

"Unpolarized" therefore describes statistics, not a special kind of wave. An equal random mixture of horizontal and vertical light gives the same result in every experiment as an equal random mixture of left- and right-circular light, so the two cannot be told apart. Four measurable intensities, the Stokes parameters, describe the polarization of any beam completely, whether it is polarized, unpolarized or in between.

Partially polarized light

Most real light is neither fully polarized nor fully unpolarized. A turning analyzer shows how much of each kind a beam contains.

Legend

$I$	Intensity passed by the analyzer.	$W/{\mathrm{m}}^2$
$\theta$	Angle between the analyzer's axis and the polarized part; marked in degrees on the axis.	$\mathrm{rad}$
$I_{\max }$	Largest intensity, $\frac{1}{2}{I}_u+I_p$, with the axis along the polarized part.	$W/{\mathrm{m}}^2$
$I_{\min }$	Smallest intensity, $\frac{1}{2}{I}_u$, with the axis across it.	$W/{\mathrm{m}}^2$
$I_p$	Polarized part of the intensity: the height of the swing.	$W/{\mathrm{m}}^2$
$\frac{1}{2}{I}_u$	Half of the unpolarized part $I_u$: what the analyzer always passes of it, and the floor of the curve.	$W/{\mathrm{m}}^2$

Model such a beam as the sum of an unpolarized part, of intensity $I_u$, and a polarized part, of intensity $I_p$. The two parts are independent, so their intensities simply add. Figure 3 shows what an analyzer passes as it turns through the angle $\theta$, measured from the direction of the polarized part; it is drawn for $I_u=60$ and $I_p=40$, in W/m².

Look first at the floor of the curve. The unpolarized part always passes half of itself, $\frac{1}{2}{I}_u$, whatever $\theta$ is, so the curve never drops below this level. On top of the floor sits the polarized part, which follows Malus's law and adds $I_p{\cos }^2\theta$, the same swing as in Figure 2. Together,
$I(\theta )=\frac{1}{2}{I}_u+I_p{\cos }^2\theta$
The curve swings between $I_{\max }=\frac{1}{2}{I}_u+I_p$, with the analyzer along the polarized part, and $I_{\min }=\frac{1}{2}{I}_u$, with it across, and never reaches zero.

The depth of the swing measures how polarized the beam is. The degree of polarization is
$P=\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{I_p}{I_u+I_p}$
because the difference of the two readings is $I_p$ and their sum is $I_u+I_p$. It is the polarized fraction of the total intensity: 0 for unpolarized light, 1 for fully polarized light. For the beam of Figure 3 the analyzer reads between 30 and 70 W/m², so $P=(70-30)/(70+30)=0.4$.

Why reflection polarizes

Glare from water or glass is strongly polarized, and at one angle of incidence completely so. The reason lies in how the electrons of the surface radiate.

Legend

$n_1$	Refractive index of the medium the light comes from.
$n_2$	Refractive index of the reflecting medium.
${\theta }_B$	Brewster's angle: the angle of incidence, equal to the angle of reflection and measured from the normal, at which no p light is reflected; $\tan {\theta }_B=n_2/n_1$.	$\mathrm{rad}$
${\theta }_2$	Angle of refraction, measured from the normal.	$\mathrm{rad}$
$p$	Light polarized in the plane of incidence (the page): its field drives the electrons along the double arrow, across the refracted ray.
$s$	Light polarized across the plane of incidence: its field points out of the page, drawn as a ring with a dot.

In Figure 4 light comes from a medium of refractive index $n_1$ (air) and meets a second medium of index $n_2$ (water). The angle of incidence is Brewster's angle ${\theta }_B$, the angle of reflection is the same, and the refracted ray leaves at the angle ${\theta }_2$; all three are measured from the normal, the dashed vertical line. The reflected light is not bounced off the surface like a ball: it is radiated by the electrons of the second medium. Those electrons are driven by the refracted wave, so they oscillate along its field, which is always across the refracted ray.

Now split the light into its two polarizations. For s light, polarized across the plane of incidence (the page), the field points out of the page, drawn as a ring with a dot labelled $s$. The electrons then oscillate parallel to the surface and across the reflected ray, so they radiate towards it freely. For p light, polarized in the plane of incidence, the electrons oscillate in the page, across the refracted ray, along the double arrow labelled $p$.

One fact about an oscillating dipole settles the matter: it radiates in proportion to ${\sin }^2\alpha$, where $\alpha$ is measured from its axis, so it sends nothing along that axis. At Brewster's angle the reflected ray is at right angles to the refracted ray, as the $90^{\circ }$ mark in the figure shows. The double arrow $p$, being across the refracted ray, then points straight along the reflected direction: it is parallel to the short dashed line, which runs parallel to the reflected ray. The p electrons would have to radiate along their own axis, which they cannot, so no p light is reflected.

The right angle means ${\theta }_B+{\theta }_2=90^{\circ }$, so $\sin {\theta }_2=\cos {\theta }_B$. Putting this into Snell's law, $n_1\sin {\theta }_B=n_2\sin {\theta }_2$, gives $n_1\sin {\theta }_B=n_2\cos {\theta }_B$, that is
$\tan {\theta }_B=\frac{n_2}{n_1}$
Figure 4 is drawn for $n_2/n_1=4/3$, water in air, which gives ${\theta }_B=53.1^{\circ }$ and ${\theta }_2=36.9^{\circ }$.

The dipole picture explains the zero; Maxwell's equations give the whole story. For a transparent, non-magnetic surface, the fractions of the intensity reflected for the two polarizations, the reflectances, are
$R_s=\frac{{\sin }^2({\theta }_1-{\theta }_2)}{{\sin }^2({\theta }_1+{\theta }_2)},R_p=\frac{{\tan }^2({\theta }_1-{\theta }_2)}{{\tan }^2({\theta }_1+{\theta }_2)}$
with ${\theta }_1$ the angle of incidence. $R_p$ vanishes when ${\theta }_1+{\theta }_2=90^{\circ }$, because the tangent in its denominator becomes infinite: the same condition as the dipole argument gave. $R_s$ never vanishes, so the light reflected at Brewster's angle is pure s light.

For water at Brewster's angle, ${\theta }_1-{\theta }_2=16.3^{\circ }$ and ${\theta }_1+{\theta }_2=90^{\circ }$, so $R_s={\sin }^{2}16.3^{\circ }=0.078$. Unpolarized sunlight is half s light, so it loses $\frac{1}{2}{R}_s$, or 3.9 % of its intensity, to a glare that is completely s-polarized; on a lake, that means horizontal. For glass ($n_2/n_1=1.5$) the same steps give $R_s=0.148$ and a loss of 7.4 %. Away from Brewster's angle the glare is only partly polarized: at $30^{\circ }$ incidence on water, $R_s=0.031$ and $R_p=0.012$, so the reflected light has $P=(R_s-R_p)/(R_s+R_p)=0.44$.

Gas lasers make use of this zero. The windows of their tubes are tilted at Brewster's angle, so p light crosses them back and forth with no reflection loss, while s light loses a little at every pass; only p light builds up, and the laser emits linearly polarized light. Skylight is polarized for the same reason: the air molecules that scatter sunlight are dipoles driven across the Sun's rays, and a dipole sends nothing along its axis. The polarization is strongest at $90^{\circ }$ from the Sun, though light scattered more than once keeps $P$ below about 0.7 to 0.8 even in a clear sky.

Birefringence and wave plates

Some crystals do more than filter a polarization: they slow one polarization more than the other. That lets us change the phase difference $\delta$ of Figure 1, and so reshape the ellipse at will.

In a crystal whose structure differs from one direction to another, the refractive index depends on the polarization. Calcite, for example, has the two indices 1.658 and 1.486. A beam entering a calcite crystal generally splits into two beams polarized at right angles, so print seen through the crystal looks doubled. This splitting is called birefringence, or double refraction.

A wave plate is a slice of such a crystal, of thickness $d$, cut so that its optic axis (the crystal's special direction) lies in the face of the plate. At normal incidence the light polarized along the axis and the light polarized across it then follow the same path, but at different speeds. Over the thickness $d$ one falls behind the other by the optical path $\Delta nd$, where $\Delta n$ is the difference of the two indices. In terms of phase, with $\lambda$ the wavelength in vacuum, that is
$\delta =\frac{2\pi \Delta nd}{\lambda }$
so the plate adds this amount to the $\delta$ of the ellipse.

A quarter-wave plate gives $\delta =\pi /2$. With its axis at $45^{\circ }$ to a linear polarization, it splits the field into two equal components, $a=b$, and puts them a quarter of a cycle apart: by the cases of Figure 1, that is circular light. Its thickness is $d=\lambda /(4\Delta n)$; in quartz, with $\Delta n\approx 0.009$, it is only about 15 μm for 550 nm light.

A half-wave plate ($\delta =\pi$) reverses the component across its axis, so a linear polarization at the angle $\alpha$ to the axis leaves at $-\alpha$. It has been turned through $2\alpha$, and nothing has been absorbed. 3D-cinema glasses pair a quarter-wave plate with a linear polarizer to pass only one sense of circular light, so the left and right images stay apart even when you tilt your head.

Worked problem: rotating the polarization

Polarizers can turn a polarization as well as block it, but at a cost in intensity. This problem works out that cost and how to reduce it.

The light leaving filter 1 in the animation is vertical, with $I_1=50W/{\mathrm{m}}^2$. How much of it can be made horizontal using polarizers alone?

A single horizontal sheet passes nothing, since $\theta =90^{\circ }$. Put one extra sheet at $45^{\circ }$ in front of it. Each sheet is then $45^{\circ }$ from the polarization reaching it and passes the fraction ${\cos }^{2}45^{\circ }=1/2$, so $50{\cos }^{4}45^{\circ }=12.5W/{\mathrm{m}}^2$ comes out horizontal. Counted from the lamp's $I_0=100W/{\mathrm{m}}^2$, this is the $I_0/8$ of the familiar three-polarizer case.

Now use $N$ sheets, each turned a further $90^{\circ }/N$ from the one before, the last one horizontal. Each sheet receives light polarized along the previous sheet, at the angle $\pi /(2N)$ to its own axis, so it passes the fraction ${\cos }^2(\pi /2N)$, and
$I_N=I_1{\cos }^{2N}\left(\frac{\pi }{2N}\right)$
Three sheets pass 21.1 W/m², ten pass 39.0 W/m².

Why does the loss shrink as sheets are added? For a small angle $x$, $\ln \cos x\approx -x^2/2$. With $x=\pi /(2N)$, the logarithm of ${\cos }^{2N}x$ is about $-N{x}^2=-{\pi }^2/(4N)$, so
$I_N\approx I_1{e}^{-{\pi }^2/(4N)}\approx I_1\left(1-\frac{{\pi }^2}{4N}\right)$
Each sheet removes only a sliver, of order $(\pi /2N{)}^2$, which shrinks like $1/N^2$; the $N$ slivers together make a loss that falls like $1/N$.

Keeping 90 %, or 45 W/m², takes 24 sheets, which pass 45.1 W/m²; 23 sheets give only 44.9 W/m². Infinitely many sheets would drag the polarization round with no loss at all, a classical analogue of the quantum Zeno effect. A half-wave plate with its axis at $45^{\circ }$ does the whole job in one step and passes all 50 W/m², apart from small reflections at its surfaces, because it shifts a phase instead of projecting the field.

Photons and limits of validity

Malus's law was derived for waves, but light arrives as photons, and real polarizers and surfaces fall short of the ideal ones above. This section says where the simple picture holds.

In quantum terms, a photon's polarization is a two-state system: two complex amplitudes play the part of $a$, $b$ and $\delta$. A polarizer at the angle $\theta$ to a photon's polarization passes it with probability ${\cos }^2\theta$, and a photon that passes leaves polarized along the polarizer's axis. Malus's law is the average of an enormous number of such yes-or-no events: a beam of 50 W/m² of 550 nm light carries about $1.4\times 10^{20}$ photons per second through each square metre.

Real polarizing sheets pass only about 38 to 42 % of unpolarized light, not the ideal 50 %, and a crossed pair leaks $10^{-3}$ to $10^{-5}$ of it (crystal polarizers $10^{-5}$ to $10^{-6}$). Each type also works only over a limited band of wavelengths. Brewster's zero needs a smooth, transparent, isotropic surface: ripples spread the angles of incidence, and metals show only a shallow minimum of $R_p$, not a zero. Finally, the field is purely transverse only in wide beams: light focused to a spot about a wavelength across gains a field component along its direction of travel.

Where this leads

Electromagnetic Waves shows why light is transverse: in empty space Maxwell's equations allow no field along the direction of travel. Interference and Superposition adds waves with a phase difference, as Figure 1 adds two perpendicular components; but two beams polarized at right angles cannot interfere, so crossed polarizers over the two slits of Diffraction and the Double Slit wipe out the fringes. Wave–Particle Duality introduces the photons themselves, each of which passes a polarizer with the probability ${\cos }^2\theta$ described above.

Reflection and Refraction supplies the Fresnel amplitudes behind Brewster's angle, and Prisms, Dispersion and Colour the scattering that makes the sky blue. The gravitational waves of Gravity as Curved Spacetime also come in two polarizations, but $45^{\circ }$ apart rather than $90^{\circ }$. Such a wave stretches space along one axis while squeezing it along the perpendicular one. Turning that pattern by $90^{\circ }$ only swaps stretch and squeeze, which gives the same polarization reversed, so the second, independent polarization is the one turned by $45^{\circ }$.