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Optional extension · Interactive lesson

Transverse Waves & Polarization

Once a wave's oscillation has a direction of its own, a new question opens: which way does it point? This topic turns that question into something you can measure with a stack of plastic sheets, and into the argument that settled whether light is transverse at all.

UPII-PREFLIGHT-W.05 · Lesson 5 of 6 in this extension. Position is not a completion score.
Starting model: Transverse Waves & PolarizationThree ideal polarizers on unpolarized light. Bars, left to right: I₀, the half left after sheet 1, the 25.0 % after sheet 2 at θ = 45°, and the 12.50 % after sheet 3 at φ = 90°. The curve is that last value against θ at fixed φ; the dot marks the present setting and the dashed line is the ½ I₀ level.unpolarized in, axes at 0°, θ = 45°, φ = 90°I₃ = ½I₀ cos²θ cos²(φ − θ) = 12.50 % of I₀best θ at this φ is φ/2 = 45°, giving 12.50 %I₃ vs θ, 0° to 90°, at φ = 90°I₀½I₀I₂I₃
Start with a picture. Change the model in the lab below.

Optional extension. Use this unnumbered support sequence where your university’s syllabus or prerequisite needs require it. It is not an additional numbered unit. Optional work is not a required completion gate.

1

Build the model

A transverse wave has a spare direction, and a polarizer measures it.

The key idea

Polarization is the direction in which a transverse wave's oscillation actually points within the plane perpendicular to its travel, together with how that direction moves over one cycle.

Example. Light travelling along x with E oscillating only along y is linearly polarized at 0°. Put a polarizer axis at 60° in its path and the amplitude falls to E cos 60° = 0.500E, so the transmitted irradiance is 0.250 of what arrived.

The complete physical model and assumptions

A travelling wave carries two directions, and only one of them is forced. Propagation is fixed by the wave vector k; the oscillation must then sit somewhere. In a longitudinal wave it sits on the same axis, so once k is given nothing is left to choose. In a transverse wave it lives in the plane perpendicular to travel, and that plane holds a whole circle of directions — the leftover freedom we call polarization. Resolve the field into two perpendicular components and the entire state is fixed by two amplitudes and one relative phase δ: in step or exactly out of step gives a line, a quarter cycle apart with equal amplitudes gives a circle, and every δ in between an ellipse. Ordinary sources do not pick one, because each atom radiates an independent wave train lasting of order 10⁻⁸ s, so the direction rerandomizes far faster than any detector can follow and the beam averages to unpolarized. An ideal polarizer keeps the component of amplitude along its axis and absorbs the rest, and squaring that projection gives Malus's law for irradiance. The model stops where the idealization does: real sheets absorb some of the wanted component and leak a little of the unwanted one, the projection assumes the medium responds linearly, and for a single photon cos²θ is no longer a fraction transmitted but a probability.

1Two directions, and only one of them is forced

Every travelling wave has a propagation direction, set by the wave vector k. It also has an oscillation direction — whatever is actually displaced or changing. A longitudinal wave puts them on the same axis: the air in a sound wave moves back and forth along the direction the sound travels, so once k is given nothing is left to specify. A transverse wave puts the oscillation in the plane perpendicular to k, and a plane contains a whole circle of directions. That leftover choice is the polarization. For light the transverse quantities are E and B, perpendicular to each other and to k; by convention the polarization is the direction of E, because it is E that pushes the charges in a detector or an eye, and B follows from it.

2Two amplitudes and one phase name every state

Take a wave along z and resolve its field into components Eₓ cos(kz − ωt) and Ey cos(kz − ωt + δ). Only three numbers matter: the two amplitudes and the phase difference δ. If δ = 0 the components rise and fall together, the tip of E runs back and forth along a fixed line, and the wave is linearly polarized at arctan(Ey/Eₓ) from the x axis. If δ = π the y component is simply inverted, so the state is linear again, along the mirrored direction −arctan(Ey/Eₓ). If δ = ±π/2 with Eₓ = Ey, one component peaks while the other is zero, so the tip sweeps a circle of constant radius at angular frequency ω — circular polarization, right or left handed according to the sign of δ. Every δ other than 0 and π traces an ellipse, with the circle as the special case whose axes are equal. Notice what stays fixed: a circularly polarized beam has |E| constant, so its irradiance is steady even though its direction never is.

3Why natural light is unpolarized, and where the ½ comes from

A filament or the Sun is an enormous number of atoms radiating independently. Each emits a wave train of order 10⁻⁸ s with its own polarization direction, and the next train points somewhere else. A detector averaging over even a microsecond sees every direction equally represented, which is what unpolarized means — not that the light has no polarization, but that it has no persistent one. Send it through an ideal polarizer and each train contributes cos²θ of its own intensity, with θ spread uniformly round the circle. The average is (1/2π)∫₀(2π) cos²θ dθ = ½. So the first sheet passes exactly half whatever angle it is set to, and rotating that sheet while watching the transmitted power is precisely how you test whether a beam was unpolarized to begin with.

4The polarizer projects amplitude; irradiance squares it

Model an ideal sheet as a device that transmits the component of E along its transmission axis untouched and absorbs the perpendicular component completely. If the light arrives linearly polarized at angle θ to that axis, the surviving amplitude is E₀ cos θ. Irradiance goes as the square of amplitude, so I = I₀ cos²θ, which is Malus's law. The squaring is the whole content. At θ = 30° the amplitude keeps 0.866 of its value while the irradiance keeps 0.750; at 45° the amplitude keeps 0.707 and the irradiance exactly half; at 60° the amplitude is halved and only 0.250 of the irradiance survives. And whatever angle the light came in at, what emerges is linearly polarized along the new axis.

5Three sheets, and why an extra absorber can brighten a beam

Cross two ideal polarizers at 90° and nothing emerges: whatever leaves the first is perpendicular to the second, and cos 90° = 0. Now slide a third sheet between them, its axis at θ to the first. The cascade on unpolarized input I₀ runs I₀ → ½I₀ → ½I₀ cos²θ → ½I₀ cos²θ cos²(90° − θ), and with cos(90° − θ) = sin θ and sin θ cos θ = ½ sin 2θ that last expression tidies to (I₀/8) sin²2θ. It is zero at θ = 0° and 90° but rises to I₀/8 at 45°. The point is not the algebra but the re-preparation: the middle sheet does not select from a fixed population, it hands the last sheet a beam polarized at θ, which the last sheet no longer sees edge on.

6Polarization is the evidence that the wave is transverse

None of this works for sound. A longitudinal wave has no perpendicular plane in which to choose a direction, so no orientation of any filter can change what gets through, and no experiment on sound has ever found one that does. Light behaves the other way. Malus found in 1808 that light reflected from a window and viewed through a calcite crystal brightened and dimmed as he rotated the crystal, and within a decade Young and Fresnel had concluded that only a transverse wave can behave like that. That killed the longitudinal pressure wave in an elastic ether, and it was one of the strongest hints before Maxwell that light's oscillating quantity is a vector perpendicular to travel. Seismology repeats the argument: S waves in rock can be polarized, P waves cannot.

2

Key equations

Read the units, assumptions and limits before substituting.

Transverse condition

E ⋅ k̂ = 0

Two independent directions survive in the perpendicular plane, and that leftover freedom is the polarization.

E is the field in V m⁻¹; k̂ is the dimensionless unit vector along the direction of travel.

State of a transverse wave

δ = φy − φₓ

δ = 0 or π gives a line; δ = ±π/2 with equal amplitudes gives a circle; anything else an ellipse.

δ in rad, the phase of the y component minus that of the x component at a fixed position.

Amplitude through an ideal polarizer

Eₜ = E₀ cos θ

The sheet projects the amplitude onto its axis and absorbs the perpendicular component entirely.

E₀ and Eₜ in V m⁻¹; θ is the angle between the incident polarization and the transmission axis.

Malus's law

I = I₀ cos²θ

Because I ∝ E², the projection is squared: 45° halves the irradiance and 60° leaves a quarter.

I₀ and I in W m⁻², for light that is already linearly polarized before it reaches the sheet.

First sheet on unpolarized light

I₁ = ½ I₀

Transmitted power that ignores the sheet's orientation is the test for an unpolarized beam.

I₀ and I₁ in W m⁻². The ½ is the average of cos²θ over a full turn, so the sheet's angle is irrelevant.

Crossed pair with a sheet between

I₃ = (I₀/8) sin²(2θ)

Zero at 0° and 90° but I₀/8 at 45°: adding an absorber can raise the transmitted power.

I₀ unpolarized in W m⁻²; θ is the inserted sheet's angle from the first axis, the outer pair crossed.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

45 °
0 to 90 °; steps of 5.
90 °
0 to 90 °; steps of 10.

Try this. Set φ = 90° so the outer pair is crossed, then drag θ from 0°, where the last bar vanishes, to 45°, where the same crossed pair passes 12.50 % — the middle sheet re-prepares the light rather than filtering it.

Live results

AFTER SHEET 2
25.0 % of I₀
AFTER SHEET 3
12.50 % of I₀
BEST θ AT THIS φ
45.0 °
BEST I₃ AT THIS φ
12.50 % of I₀

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics II modelThree ideal polarizers on unpolarized light. Bars, left to right: I₀, the half left after sheet 1, the 25.0 % after sheet 2 at θ = 45°, and the 12.50 % after sheet 3 at φ = 90°. The curve is that last value against θ at fixed φ; the dot marks the present setting and the dashed line is the ½ I₀ level.unpolarized in, axes at 0°, θ = 45°, φ = 90°I₃ = ½I₀ cos²θ cos²(φ − θ) = 12.50 % of I₀best θ at this φ is φ/2 = 45°, giving 12.50 %I₃ vs θ, 0° to 90°, at φ = 90°I₀½I₀I₂I₃
Three ideal polarizers on unpolarized light. Bars, left to right: I₀, the half left after sheet 1, the 25.0 % after sheet 2 at θ = 45°, and the 12.50 % after sheet 3 at φ = 90°. The curve is that last value against θ at fixed φ; the dot marks the present setting and the dashed line is the ½ I₀ level.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Middle sheet angle θ: 45 °; Last sheet angle φ: 90 °

Baseline 45; choose 0 to 90, in steps of 5.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The error is treating a polarizer as a sieve: a grid that lets through whichever waves already happen to be aligned with it and blocks the rest. Two wrong answers follow. A sieve would pass the aligned fraction, giving I = I₀ cos θ, whereas the sheet projects the amplitude and irradiance goes as amplitude squared, so the cosine is squared. And a sieve can only ever remove, so inserting a third sheet between crossed ones could never help — yet it lifts the transmitted power from zero to I₀/8. What the sheet really does is re-prepare the light: everything leaving it is linearly polarized along its own axis, having forgotten the direction it arrived with. A polarizer is a projector, not a filter.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A laser beam is linearly polarized in the vertical direction and carries an irradiance of 24.0 W m⁻². It passes through one ideal polarizer whose transmission axis lies 35.0° from the vertical. Find the transmitted irradiance, and state what fraction of the electric field amplitude survives.

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.

MediumExample 2

An unpolarized beam of irradiance 45.0 W m⁻² passes through three ideal polarizers in succession. Measured from one fixed reference direction, their transmission axes lie at 0°, 25° and 70°. Find the irradiance leaving the third sheet, and express it as a percentage of the original.

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.

HardExample 3

Two ideal polarizers are crossed, and unpolarized light of irradiance 60 W m⁻² falls on the pair. A third ideal sheet is inserted between them with its axis at θ to the first. Derive the transmitted irradiance as a function of θ, locate the maximum using calculus, and find the two values of θ between 0° and 90° that give 3.0 W m⁻².

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.

6

Quick check

Select an answer and explain it before checking.

Unpolarized light of irradiance 8.0 W m⁻² falls on two ideal polarizers whose transmission axes are crossed at 90°. A third ideal polarizer is then slid between them with its axis at 30° to the first. What irradiance leaves the stack?
7

Review & continue

Polarization is the direction in which a transverse wave's oscillation actually points within the plane perpendicular to its travel, together with how that direction moves over one cycle.

  • 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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.