core34 min studyLesson 15 in Physics

Electromagnetic waves and polarization

Build a transverse plane wave, distinguish its polarization from propagation, and calculate its time-averaged intensity.

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What you’ll understand

  • Derive the vacuum plane-wave relation between E, B, and propagation direction.
  • Classify linear, elliptical, and circular polarization from component amplitudes and phases.
  • Calculate intensity using the correct peak-amplitude convention.

A vacuum electromagnetic wave is a self-consistent traveling configuration of electric and magnetic fields. Its propagation direction is different from the direction in which the electric field oscillates. Polarization records how that transverse electric vector changes during a cycle. Treating its two components carefully connects wave equations, interference, optical measurements, and the energy carried by light.

The curl equations determine the companion field

In a source-free vacuum, taking the curl of Faraday's law and using the other Maxwell equations yields a wave equation with speed c=1/μ0ϵ0c=1/\sqrt{\mu_0\epsilon_0}. For a single plane wave, the divergence equations require both fields to be perpendicular to its wave vector. The magnetic field is then fixed by the electric field and direction, rather than chosen as an independent oscillation with arbitrary phase.

An infinite plane wave is an idealization with infinite total transverse extent. It is useful locally for a broad beam far from its edges, but focusing and diffraction require more structure. Transversality here refers to each source-free plane-wave component. Fields near charges and antennas need not be purely transverse, so the plane-wave relation between magnitudes should not be applied to every electric or magnetic field in space.

∇2E−1c2∂t2E=0,B=1ck^×E,ω=c∣k∣\nabla^2\mathbf E-\frac{1}{c^2}\partial_t^2\mathbf E=0,\qquad \mathbf B=\frac{1}{c}\hat{\mathbf k}\times\mathbf E,\qquad \omega=c|\mathbf k|
These relations describe a traveling plane wave in source-free vacuum. Electric and magnetic peak amplitudes obey B₀ = E₀/c; their energy contributions are equal for that wave.↔ Scroll wide equations horizontally.

Two transverse components define a polarization ellipse

For propagation along z, write Ex=Acos⁡ψE_x=A\cos\psi and Ey=Bcos⁡(ψ+δ)E_y=B\cos(\psi+\delta), with ψ=kz−ωt\psi=kz-\omega t. At one fixed position, the tip of the electric vector generally traces an ellipse. If the relative phase is zero or π, the vector stays on a line. Equal amplitudes and a relative phase of a quarter-cycle produce a circle; unequal amplitudes at that phase produce an ellipse.

Handedness names depend on whether the observer looks along or against propagation and on the time-phase convention. Specify the actual component functions when a sign matters. Polarization does not mean that the beam moves sideways along the ellipse: energy propagates along the Poynting direction. A quarter-wave plate changes relative phase, whereas a linear polarizer projects the field onto an axis and generally discards part of its power.

E=Re⁡ ⁣[(Ax^+Beiδy^)ei(kz−ωt)],⟨I⟩=ϵ0c2(A2+B2)\mathbf E=\operatorname{Re}\!\left[(A\hat{\mathbf x}+Be^{i\delta}\hat{\mathbf y})e^{i(kz-\omega t)}\right],\qquad \langle I\rangle=\frac{\epsilon_0c}{2}(A^2+B^2)
A and B are real peak component amplitudes in V/m. Orthogonal components contribute additively to the cycle-averaged intensity regardless of their relative phase. They must not be treated as two scalar waves along the same axis.↔ Scroll wide equations horizontally.

Projection and interference answer different questions

Two coherent fields along the same direction can add constructively or destructively according to relative phase. Orthogonal components instead determine a vector polarization. After a polarizer projects them onto one common direction, their phases can again affect the measured intensity. This order of operations explains why adding component intensities before specifying the detector can hide observable interference.

For linearly polarized incident light and an ideal linear analyzer, the transmitted field amplitude is the projection onto the analyzer axis. Squaring that amplitude yields Malus's law. This statement concerns a fully polarized input; unpolarized light requires averaging over its polarization statistics. Real polarizers also have finite extinction and absorption, so the ideal cosine-squared dependence is a baseline rather than a complete device model.

Iout=Iincos⁡2α(ideal analyzer, linear input)I_{\mathrm{out}}=I_{\mathrm{in}}\cos^2\alpha\quad\text{(ideal analyzer, linear input)}
α is the angle between the incoming linear polarization and the transmission axis. The relation describes transmitted intensity, not electric-field amplitude, which is multiplied by cos α.↔ Scroll wide equations horizontally.

Polarization angle and carried power

A vacuum plane wave has Ex=3cos⁡ψ V/mE_x=3\cos\psi\,\mathrm{V/m} and Ey=4cos⁡ψ V/mE_y=4\cos\psi\,\mathrm{V/m}. Determine polarization and average intensity using c=3.0×108 m/sc=3.0\times10^8\,\mathrm{m/s} and ϵ0=8.85×10−12 F/m\epsilon_0=8.85\times10^{-12}\,\mathrm{F/m}.

  1. Both components share the same phase, so their ratio is always 4/3 when the common factor is nonzero. The polarization is linear at arctan⁡(4/3)=53.1∘\arctan(4/3)=53.1^\circ from the positive x axis, with its axis understood modulo 180°.

  2. The peak total electric magnitude is 32+42=5 V/m\sqrt{3^2+4^2}=5\,\mathrm{V/m}. Its magnetic peak magnitude is 5/c=1.67×10−8 T5/c=1.67\times10^{-8}\,\mathrm T.

  3. Use peak rather than RMS amplitudes: I=ϵ0c(32+42)/2=0.0332 W/m2I=\epsilon_0c(3^2+4^2)/2=0.0332\,\mathrm{W/m^2}. A 45° analyzer relative to this polarization would transmit half that intensity.

The result

The wave is linearly polarized at 53.1° and carries approximately 33.2 mW/m². The factor of one half comes from averaging the squared sinusoid over a cycle.

Explore the model

Interactive lab

Interactive lab · waves

Coherent scalar addition before a polarization calculation

Vary phase difference and amplitude ratio for sin(2x−t) + a sin(2x−t+phase). The lab adds two components on the same scalar axis and shows their sum. It teaches interference; it does not draw the two perpendicular electric components or a polarization ellipse.

Coherent scalar addition before a polarization calculationTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRelative phase has a different geometric meaning when components are parallel versus perpendicular.

Check your understanding

Two equal-amplitude perpendicular electric components have a relative phase of 90°. What is their polarization?

Try it yourself

A linearly polarized beam of intensity 80 W/m² passes an ideal analyzer at 60° to its polarization. Find the transmitted intensity. Then explain why circularly polarized input of the same intensity gives a different answer.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use cosine squared for the linear case; a circular wave has equal cycle-averaged power along every transverse axis.

Compare with the solution

For linear input, 80cos⁡260∘=20 W/m280\cos^2 60^\circ=20\,\mathrm{W/m^2}. An ideal analyzer transmits half the intensity of circular input, or 40 W/m240\,\mathrm{W/m^2}, independently of its transverse orientation. This does not make circular light unpolarized: its two components have a stable relative phase that a suitable phase plate and analyzer can reveal.

Take it with you

  • A vacuum plane wave has linked transverse electric and magnetic fields.
  • Polarization depends on both relative component amplitudes and phase.
  • Peak amplitudes require a cycle-average factor when calculating intensity.

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