
Optional unit · institution-dependent · Interactive lesson
Polarization
Interference told you where a beam is bright. Polarization tells you which way its electric field points — a second label every beam carries. Learn to project a field onto an axis, predict what survives each sheet, and read the angle at which a plain reflection hands you polarized light for free.
UPII-14.07 · Lesson 7 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
Light carries a direction across its travel, and every polarizer projects onto one.
The key idea
Polarization names the direction along which a light wave's electric field oscillates, always at right angles to the direction the wave travels.
Example. Light polarized at 60° to a sheet's transmission axis keeps cos²60° = 0.25 of its intensity: 80 W m⁻² in, 20 W m⁻² out, and what leaves is polarized along the sheet's axis, not at 60°.
The complete physical model and assumptions
An electromagnetic wave travelling along z keeps its electric field in the x–y plane, so beyond a wavelength and an intensity a beam carries a direction: the line along which E oscillates. Light is linearly polarized when that line holds still, and unpolarized when the atoms of an ordinary lamp reset it at random every few nanoseconds. A polarizing sheet does not sieve the beam into lucky and unlucky rays. It transmits the component of E along its transmission axis and absorbs the perpendicular one, so the surviving amplitude is E₀ cos θ and the intensity, which goes as the square of the field, is I₀ cos²θ — Malus's law. Averaging cos²θ over every orientation gives ½, which is why unpolarized light always loses exactly half at the first sheet and leaves polarized along that sheet's axis, with every later angle measured from there rather than from the lab vertical. Nature supplies polarizers of its own: reflection from a dielectric at tan θB = n₂/n₁ returns light polarized parallel to the surface, and a birefringent crystal, having two indices, retards one component against the other. The model assumes an ideal sheet, one fixed axis, and fields that add as amplitudes. It stops short of partial polarization, and it reaches circular light only qualitatively through a wave plate, since a full account carries both components and their phase difference together; a real Polaroid also transmits less than the ideal law promises.
1A transverse wave carries a direction as well as an energy
In an electromagnetic wave E and B are both perpendicular to the direction of travel, and by convention the polarization is named after E. That leaves a whole plane of possible directions for a beam going one way, and which one it uses is a genuine physical property. A laser or a radio transmitter emits with E along a fixed line: linear polarization. A filament lamp or the Sun does not. Its light comes from enormous numbers of atoms radiating independently, each wave train lasting of order 10⁻⁸ s before the direction resets at random, so over any measurable time every orientation appears equally often. That is unpolarized light — not light with no direction, but light whose direction is reshuffled far faster than any instrument can track. A longitudinal wave has no such freedom: sound displaces air along its own direction of travel, so there is no second axis to distinguish and no rotation of any filter could change what gets through.
2A polarizer projects a vector; it does not sieve a beam
An ideal polarizing sheet has one transmission axis. Resolve the incoming field into a component along that axis and a component across it; the sheet absorbs the second and passes the first. Since the field is a vector, the surviving amplitude is E₀ cos θ, where θ is the angle between the incident polarization and the axis. Intensity carries the square of the field, so I = I₀ cos²θ. That squaring is where most of the arithmetic errors live. At θ = 45° the field keeps cos 45° = 0.707 of its size, but the intensity keeps only 0.500. At θ = 30° the field keeps 0.866 and the intensity 0.750; at θ = 60° the field keeps 0.500 and the intensity just 0.250. Notice also what leaves the sheet: light polarized along the sheet's own axis, whatever direction it arrived from. Nothing about the incident direction survives except the single factor cos θ in the amplitude.
3Unpolarized light loses half and gains an axis
Malus's law needs a definite θ, and unpolarized light has none. Take instead the average over every orientation. Because cos²θ swings between 0 and 1 symmetrically about ½, its mean over a full turn is exactly ½, so the first sheet always transmits I₁ = ½ I₀ no matter how it is oriented — turning a single polarizer in the beam of a filament lamp dims nothing, while skylight, already partly polarized by scattering, does dim as the sheet turns. What the sheet does change is the state: the light emerging is linearly polarized along its axis. From that point on the beam has a definite direction, and every later angle must be measured from that axis rather than from the lab vertical. Put 200 W m⁻² of unpolarized light into a vertical sheet and 100 W m⁻² leaves, polarized vertically; send it through a second sheet 35° away and 100 cos²35° = 67.1 W m⁻² survives, polarized at 35°. Rotating the first sheet alone changes nothing, because θ is a relative angle.
4Slip a third sheet between two crossed ones
Two sheets at 90° to each other pass nothing: the first prepares the light along its axis, and cos²90° = 0. Now insert a third sheet between them, its axis at θ to the first. Sheet 1 gives ½I₀. Sheet 2 is θ away, so it transmits ½I₀ cos²θ and re-prepares the light at θ. Sheet 3 is now 90° − θ from that, giving ½I₀ cos²θ cos²(90° − θ) = ½I₀ cos²θ sin²θ, which the identity sin θ cos θ = ½ sin 2θ tidies to I₃ = (I₀/8) sin²(2θ). At θ = 45° the output peaks at I₀/8: a beam of 400 W m⁻² delivers 50 W m⁻² through a pair of sheets that on their own delivered nothing. Adding an absorber increased the transmission. That is only paradoxical if you think the sheets remove rays; it is inevitable if each one projects the field onto a new axis.
5Reflection polarizes, and Brewster's angle says where
Shine unpolarized light onto glass or water and the reflected beam comes back partially polarized parallel to the surface, and at one special incidence completely so. That is Brewster's angle, tan θB = n₂/n₁. Air onto water, n₂/n₁ = 1.33, gives θB = 53.1°; air onto crown glass at 1.52 gives 56.7°. The mechanism is worth having. At θB the reflected and refracted rays are exactly 90° apart — for water, 53.1° + 36.9° = 90° — and the reflected light is radiated by charges in the surface driven along the refracted beam's own field direction. A dipole radiates nothing along its own axis, so the component that would have been polarized in the plane of incidence has no way to leave in the reflected direction, and only the component parallel to the surface survives. Glare from a road, a lake or a car roof is therefore horizontally polarized, which is why polarizing sunglasses are cut with a vertical transmission axis.
6Birefringence: one crystal, two speeds
Calcite, quartz and stretched plastic are birefringent: the refractive index depends on the polarization direction. A uniaxial crystal has an ordinary index nₒ and an extraordinary index nₑ, so the two components of a field resolved onto the fast and slow axes travel at different speeds and arrive with a phase difference Δφ = 2π (nₑ − nₒ) d / λ₀. Nothing is absorbed — the components are simply re-timed. Choose d so that Δφ = π and you have a half-wave plate, which flips the sign of the slow component and so reflects the polarization direction in the fast axis: light at α to the fast axis leaves at −α, rotated by 2α. Choose Δφ = π/2 and the two components leave a quarter cycle apart, turning linear polarization into circular. Quartz at 589 nm has nₒ = 1.5443 and nₑ = 1.5534, so Δn = 0.0091 and a quarter-wave plate needs only d = 16.2 µm. Because λ₀ sits in the denominator, a wave plate is exact at one wavelength alone.
Key equations
Read the units, assumptions and limits before substituting.
Field projection onto the axis
Eₜ = E₀ cos θ
The sheet keeps one component of a vector, so the amplitude falls as cos θ — the squaring happens only when you move to intensity.
E in V m⁻¹; θ is the angle between the incident polarization and the transmission axisMalus's law
I = I₀ cos²θ
Turns the amplitude projection into the quantity a photodiode actually reads.
I in W m⁻²; valid only when the light reaching the sheet is already linearly polarizedUnpolarized light at the first sheet
I₁ = ½ I₀
Supplies the one step Malus's law cannot, and fixes the axis every later angle is measured from.
⟨cos²θ⟩ = ½ averaged over every incident orientation; I in W m⁻²Brewster's angle
tan θB = n₂/n₁
Gives fully polarized light with no filter at all, and explains why glare from a horizontal surface is horizontal.
n₁ the incident medium, n₂ the transmitting medium; both indices are dimensionlessCrossed sheets with one between
I₃ = (I₀/8) sin²(2θ)
Shows that a polarizer re-prepares the light along its own axis instead of filtering photons away.
unpolarized I₀ into sheets at 0°, θ and 90°; I in W m⁻², peak at θ = 45°Retardance of a birefringent plate
Δφ = 2π (nₑ − nₒ) d / λ₀
Set Δφ = π for a half-wave plate that rotates linear polarization, π/2 for circular light.
d the plate thickness in m, λ₀ the vacuum wavelength in m, Δφ in radInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set θ = 45°: the projection arrow still keeps 0.71 of the field, but the curve has already fallen to half, because intensity squares the projection. Push on to 90° and the arrow vanishes. Raising I₀ lifts the whole curve and its ceiling without moving the zeros.
Live results
- AFTER SHEET 1 I₀/2
- 60.0 W m⁻²
- AFTER ANALYZER
- 45.0 W m⁻²
- INTENSITY FACTOR cos²θ
- 0.750
- FIELD FACTOR |cos θ|
- 0.866
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Analyzer angle θ from sheet 1: 30 °; Unpolarized source I₀: 120 W m⁻²
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The picture to lose is the picket fence: a sheet that lets through the light already aligned with it, stops the rest, and leaves the survivors untouched. Two things go wrong. Applied to unpolarized light it invites I = I₀ cos²θ at the first sheet, when the arriving field has no single θ to substitute and the honest answer is the average, I₀/2. And it cannot survive the three-sheet experiment: two crossed sheets pass nothing, yet slipping a third between them at 45° restores I₀/8, and a fence can only ever remove more. A polarizer absorbs one component and re-prepares what is left along its own axis, so the light reaching sheet 3 arrives polarized at 45° to it, not at 90°. Past each sheet the beam has forgotten where it came from except through one factor of cos θ in the amplitude.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Unpolarized light of intensity 240 W m⁻² falls on a polarizing sheet whose transmission axis is vertical, and then on a second sheet whose axis is 25° from the vertical. Find the intensity after each sheet, and state the polarization direction of the light that finally emerges.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The source is unpolarized, so no single angle can be put into Malus's law at the first sheet. Averaging cos²θ over every incident orientation gives ½: I₁ = ½ × 240 W m⁻² = 120 W m⁻², and that light leaves polarized vertically.
- The first sheet's axis is now the reference. The second is θ = 25° from it, so the surviving field is E₁ cos 25° = 0.9063 E₁.
- Intensity goes as the square of the field: I₂ = I₁ cos²25° = 120 × 0.8214 = 98.6 W m⁻².
- The emerging light is linearly polarized along the second sheet's axis, 25° from the vertical. It is not 'mostly vertical with a tilt' — the sheet has re-prepared it entirely.
Answer. I₁ = 120 W m⁻², I₂ = 98.6 W m⁻², polarized at 25° to the vertical. The second sheet costs only 18%, because cos²25° is still 0.82.
MediumExample 2
An unpolarized beam of intensity 60 W m⁻² passes through three sheets whose transmission axes lie at 0°, θ and 90°. Derive the transmitted intensity as a function of θ, evaluate it at θ = 20°, and find the θ that lets through the most light.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Sheet 1 halves the unpolarized beam: I₁ = ½ × 60 = 30 W m⁻², polarized at 0°.
- Sheet 2 is θ from that axis, so I₂ = 30 cos²θ, and the light now leaves polarized at θ.
- Sheet 3 is at 90°, which is 90° − θ away from sheet 2's axis, so I₃ = 30 cos²θ cos²(90° − θ) = 30 cos²θ sin²θ.
- Use sin θ cos θ = ½ sin 2θ, so cos²θ sin²θ = ¼ sin²(2θ): I₃ = 7.5 sin²(2θ) W m⁻², which is the general result (I₀/8) sin²(2θ).
- At θ = 20°: sin 40° = 0.6428, so I₃ = 7.5 × 0.4132 = 3.10 W m⁻².
- sin²(2θ) peaks when 2θ = 90°, so θ = 45° and I₃ = 7.5 W m⁻² — one eighth of the source, out of a crossed pair that alone passes nothing.
Answer. I₃ = 7.5 sin²(2θ) W m⁻²; 3.10 W m⁻² at θ = 20°, rising to a maximum of 7.5 W m⁻² at θ = 45°.
HardExample 3
A 589 nm laser beam is linearly polarized at 0° and meets an analyzer set at 90°, so nothing passes. A quartz plate with its optic axis in the face (nₒ = 1.5443, nₑ = 1.5534) is then inserted, cut as a half-wave plate with its fast axis at 30° to the incident polarization. Find the thinnest such plate, the direction of the light leaving it, and how much of the 4.0 W m⁻² beam the analyzer now passes.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Half-wave means the two components leave exactly π out of phase: Δφ = 2π Δn d / λ₀ = π, so the thinnest plate has d = λ₀ / (2Δn).
- Δn = nₑ − nₒ = 1.5534 − 1.5443 = 0.0091. Quartz is positive uniaxial, so the extraordinary axis is the slow one and the ordinary axis is fast. Then d = (589 × 10⁻⁹ m) / (2 × 0.0091) = 3.24 × 10⁻⁵ m = 32.4 µm.
- A half-wave plate reverses the sign of the slow component, which reflects the polarization direction in the fast axis. The incident light sits at −30° from the fast axis, so it leaves at +30° from it: at 60° in the lab frame, rotated by 2 × 30° = 60°.
- The analyzer is at 90°, so the angle between it and the arriving polarization is 90° − 60° = 30°. Malus's law gives I = 4.0 cos²30° = 3.0 W m⁻².
- The plate absorbed nothing; it only re-timed one component against the other, and only at this colour. At 450 nm the same 32.4 µm gives Δn d / λ₀ = 0.654, a retardance of 1.31π rather than π, so the trick fails.
Answer. d = 32.4 µm; the polarization emerges rotated 60° from its original direction, and the analyzer passes 3.0 W m⁻² — 75% of a beam it was blocking completely a moment earlier.
Quick check
Select an answer and explain it before checking.
Review & continue
Polarization names the direction along which a light wave's electric field oscillates, always at right angles to the direction the wave travels.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.