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Interference of Light: Coherence, Polarization and Beats

Two plane waves Ej = Aj êj cos(kjx − ωjt + φj) in a non-dispersive medium. Set the phase, amplitudes and wavelengths, the complex degree of coherence γ₁₂ and the angle between the polarizations. Then read what a detector with a finite integration time records at a probe point.

All views share one probe position and one clock • Units: λ₁, T₁, A₁, I₁ = A₁²/2
Δφ 0° Δφ at Probe
I 4.000 I₁ Detector Reading
V 1.000 Visibility
μ 1.000 |μ| = |γ₁₂ × overlap|
Δf 0 (steady) Intensity Beat |Δf|
📊 Synchronized views t = 0.00 T₁

Spatial field E(x, t) at the current instant click to move probe

Field components in units of A₁ along ê₁ (solid) and perpendicular to it (dotted, only when θ ≠ 0). Dashed envelope: |E₁ + E₂| of the co-polarized part.

Time trace at the probe, E(xₚ, t′)

Current instant at the right edge. Fields before the analyser.

Detector intensity at the probe I₁ units

Includes γ₁₂, the polarization overlap, the analyser and the integration window. Hover to read values.

Rotating phasors at xₚ

One representative realization with relative phase Δφ + arg γ₁₂. Only E₂ cos θ adds head-to-tail to E₁; E₂ sin θ (dotted) is orthogonal.

Polarization overlap (transverse y–z plane) schematic

Unit vectors ê₁ and ê₂ in the transverse plane, the projection cos θ that interferes and, if enabled, the analyser axis.

💡 How to Use

Learn with this tool

Learning objectives

  • Predict the two-beam intensity and visibility from I₁, I₂, the complex degree of coherence γ₁₂ and the polarization overlap.
  • Distinguish spatial interference, temporal beats at |Δf| and the averaging done by a detector with integration time T.
  • Explain with the Fresnel–Arago laws why crossed polarizations give no fringes and why an analyser brings them back.

Prerequisites

  • Phasors and complex exponentials, eiθ = cos θ + i sin θ.
  • Plane waves, k = 2π/λ and ω = 2πf = ck.
  • Linear polarization and Malus's law (polarization tool).
  • Time averages of cos² over a period.

Model

Two transverse plane waves travel in +x through a linear, lossless, non-dispersive medium. Beam 1 is polarized along ê₁ = ŷ and beam 2 along ê₂ = cos θ ŷ + sin θ ẑ, at angle θ to it in the transverse plane. The detector sees the cycle-averaged intensity (arbitrary units, I = ⟨|E|²⟩):

Two-field interference

I=I1+I2+2I1I2Re⁡ ⁣[γ12(e^1⋅e^2)eiΔφ]I=I_1+I_2+2\sqrt{I_1I_2}\operatorname{Re}\!\left[\gamma_{12}(\hat e_1\cdot\hat e_2)e^{i\Delta\varphi}\right]
I=I1+I2+2I1I2∣μ∣cos⁡(Δφ+arg⁡μ)I=I_1+I_2+2\sqrt{I_1I_2}|\mu|\cos(\Delta\varphi+\arg\mu)

Coherence and phase

μ=γ12cos⁡θ\mu=\gamma_{12}\cos\theta
Δφ(x,t)=(k2−k1)x−(ω2−ω1)t+(φ2−φ1)\Delta\varphi(x,t)=(k_2-k_1)x-(\omega_2-\omega_1)t+(\varphi_2-\varphi_1)

Finite detector exposure

Idet(t)=I1+I2+2I1I2∣μ∣sinc⁡ ⁣(ΔωT2)cos⁡ ⁣[Δφ(t−T/2)+arg⁡μ]I_{\mathrm{det}}(t)=I_1+I_2+2\sqrt{I_1I_2}|\mu|\operatorname{sinc}\!\left(\frac{\Delta\omega T}{2}\right)\cos\!\left[\Delta\varphi(t-T/2)+\arg\mu\right]

Fringe visibility

V=Imax⁡−Imin⁡Imax⁡+Imin⁡V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}
V=2I1I2I1+I2∣γ12∣ ∣cos⁡θ∣∣sinc⁡ ⁣(ΔωT2)∣V=\frac{2\sqrt{I_1I_2}}{I_1+I_2}|\gamma_{12}|\,|\cos\theta|\left|\operatorname{sinc}\!\left(\frac{\Delta\omega T}{2}\right)\right|
Aj, Ij
peak amplitude and intensity Ij = Aj²/2 of beam j (units A₁ and I₁)
kj, ωj
wavenumber and angular frequency, ω = ck (non-dispersive)
φj
fixed phase offset of beam j; the slider sets φ₂ − φ₁
γ₁₂
complex degree of coherence ⟨eiδ⟩, the average over the random extra relative phase δ(t); |γ₁₂| ≤ 1
θ
angle between the linear polarization directions ê₁ and ê₂
α
transmission axis of the optional analyser, measured from ê₁
T
detector boxcar integration time (on top of optical-cycle averaging)
V
fringe visibility (contrast) as Δφ is scanned

Assumptions: scalar quasi-monochromatic beams (each with one frequency), stationary and ergodic phase noise, uniform plane waves (no spatial coherence structure), ideal detector that averages over many optical cycles, ideal lossless analyser. With the analyser on, beam amplitudes become A₁|cos α| and A₂|cos(α − θ)| and both leave along the same axis, so |ê₁·ê₂| = 1 after it.

Derivation

Write Ej = Re[Aj êj eiψj] with ψj = kjx − ωjt + φj, and allow a random extra phase δ(t) on beam 2. Then |E|² = |E₁|² + |E₂|² + 2E₁·E₂. Averaging over an optical cycle removes terms at ω₁ + ω₂ and leaves ⟨|E|²⟩ = A₁²/2 + A₂²/2 + A₁A₂ (ê₁·ê₂) cos(Δφ + δ).

Averaging over the noise: ⟨cos(Δφ + δ)⟩ = Re[eiΔφ⟨eiδ⟩] = Re[γ₁₂ eiΔφ]. With Ij = Aj²/2, A₁A₂ = 2√(I₁I₂), which gives the two-beam law. Only E₂'s projection onto ê₁ can interfere with E₁: the orthogonal part E₂ sin θ adds I₂ sin²θ to the background.

A boxcar detector over [t − T, t] averages cos(Δφ(t′) + arg μ) with Δφ linear in t′ at rate −Δω. The mean of a cosine over a window equals its value at the window centre times sin(ΔωT/2)/(ΔωT/2), which gives the sinc factor. The extremes as Δφ is scanned then give V.

Exercise 1 — Measure |γ₁₂| from fringe contrast

  1. With A₂ = 0.5 A₁, θ = 0, T = 0 and |γ₁₂| = 0.6, what visibility do you expect?
  2. Load Beating (it scans Δφ in time), then set A₂/A₁ = 0.5 and |γ₁₂| = 0.6.
  3. Hover the detector trace and read Imax and Imin, or read them in the readouts. Compute V.
  4. Why is V smaller than |γ₁₂| here, and when would V equal |γ₁₂|?
Show answer

I₁ = 1, I₂ = 0.25 (in I₁). V = 2√0.25 · 0.6 / 1.25 = 0.48. The readouts give Imax = 1.85 and Imin = 0.65, so (1.85 − 0.65)/(1.85 + 0.65) = 0.48. Unequal intensities reduce the amplitude factor 2√(I₁I₂)/(I₁ + I₂) to 0.8. V = |γ₁₂| only when I₁ = I₂ (and θ = 0, T = 0).

Exercise 2 — Limiting case: crossed polarizations

  1. Two equal, fully coherent beams have θ = 90°. What does the detector read as φ₂ − φ₁ goes from 0° to 180°? What changes with a 45° analyser?
  2. Load Crossed pol. and drag the phase slider. Then load + Analyser 45° and drag it again.
  3. Record the detector reading at φ = 0° and 180° in both cases.
  4. Where did the fringes go and why can a polarizer bring them back? Try α = −45°.
Show answer

Without the analyser: I = I₁ + I₂ = 2 I₁ for every φ, so V = 0 (Fresnel–Arago: orthogonal states do not interfere). The phase still changes the polarization state of the sum (linear → elliptical → linear), which the polarization tool shows. With α = 45° each beam passes with amplitude A/√2. They are now parallel, so I swings between 2 I₁ (φ = 0) and 0 (φ = 180°) and V = 1. At α = −45° the projections have opposite signs and the fringes shift by 180°.

Exercise 3 — Beats and a slow detector

  1. For λ₂/λ₁ = 1.1, predict the beat frequency and the smallest integration time that removes the beat completely.
  2. Load Beating, press Start, then increase T.
  3. Read V and the averaging factor sinc(ΔωT/2) at T = 5.5, 11 and 22 T₁.
  4. Why is the residual visibility not monotonic in T?
Show answer

f₂ = f₁/1.1, so |Δf| = 1 − 1/1.1 = 0.0909 f₁ and the beat period is 11 T₁. At T = 11 T₁ (one beat period), sinc(π) = 0 and V = 0. At T = 5.5 T₁, V = sinc(π/2) = 0.637. At 22 T₁ V is 0 again. Between zeros the sinc has side lobes (|sinc(3π/2)| = 0.212 at T = 16.5 T₁), so V rises again before it decays like 1/T.

Worked example — Heterodyne beat of two He–Ne lines through a polarizer

Two He–Ne laser modes at λ₁ = 632.8 nm have a frequency spacing Δf = 438 MHz, with orthogonal linear polarizations (θ = 90°). This is common for two-mode lasers. Each mode delivers 0.5 mW onto a photodiode with a 1 GHz bandwidth. The mutual coherence of the modes over the measurement is |γ₁₂| ≈ 0.95. (a) What does the photodiode see without a polarizer? (b) With a polarizer at α = 45°? (c) With a 10 µs integrating power meter?

(a) cos θ = 0, so μ = 0 and P = P₁ + P₂ = 1.0 mW with no 438 MHz signal. (b) Malus: each mode passes cos²45° = 0.5, so P₁′ = P₂′ = 0.25 mW, and after the polarizer |ê₁·ê₂| = 1. The photocurrent then beats at 438 MHz: P(t) = 0.5 mW + 2√(0.25·0.25) · 0.95 cos(2πΔf t + ϕ) mW = 0.5 ± 0.475 mW, so V = 0.95. The beat period is 2.28 ns, which the 1 GHz photodiode resolves. (c) T = 10 µs ≈ 4380 beat periods: |sinc(πΔf T)| ≤ 1/(πΔf T) ≈ 7×10⁻⁵, so the meter reads 0.5 mW. Set λ₂/λ₁ close to 1, θ = 90°, and switch the analyser on at 45° to reproduce the same behaviour in normalized units.

When the model fails

  • Broadband light: γ₁₂ depends on the path delay τ through the source spectrum (Wiener–Khinchin). Here γ₁₂ is a fixed input, so no coherence length is computed. The interferometers tool derives γ(τ).
  • Dispersion: with ω(k) nonlinear the envelope moves at the group velocity and the spatial and temporal beat periods are no longer related by c.
  • Non-plane waves: real beams cross at an angle and produce spatial fringes whose visibility depends on spatial coherence (van Cittert–Zernike). See the double-slit tool.
  • Elliptical or partially polarized light: the overlap becomes a complex Jones inner product, and partially polarized beams need coherency matrices. The model has jonesOverlap, but the page only sets linear angles.
  • Detectors: shot noise, finite bandwidth other than a boxcar, and saturation are not modelled. A classical field animation says nothing about single-photon statistics.

References

  • E. Hecht, Optics, 5th ed., §7.1 (addition of waves, beats) and §9.1–9.2 (conditions for interference, Fresnel–Arago laws).
  • M. Born and E. Wolf, Principles of Optics, 7th ed., §10.3–10.4 (mutual coherence, complex degree of coherence, visibility).
  • B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 2nd ed., ch. 12 (statistical optics, interference of partially coherent light).
  • J. W. Goodman, Statistical Optics, 2nd ed., ch. 5 (coherence of optical waves).

📚 Physics Background

🎯 Reading the result quantitatively

Equal amplitudes, same frequency, θ = 0, γ₁₂ = 1: Δφ = 0 gives I = 4I₁, twice the sum of the separate intensities. Δφ = π gives exactly zero, and Δφ = π/2 gives 2I₁, the incoherent sum.

Unequal amplitudes: Imax = (A₁ + A₂)²/2 and Imin = (A₁ − A₂)²/2. With A₂ = 0.5A₁, Imin = 0.25I₁ and V = 0.8.

Energy bookkeeping: the cross term moves energy around and does not create or destroy it. Averaged over all Δφ, I = I₁ + I₂ for any γ₁₂ and θ.

🔗 What γ₁₂ means

If the relative phase of the two beams wanders by a random δ(t), each instant still gives perfect fringes, but they move. The detector sees their average, which is weighted by γ₁₂ = ⟨eiδ⟩. A phase that jitters uniformly over a range w gives |γ₁₂| = sinc(w/2), and a phase that wanders over the full 2π gives γ₁₂ = 0 (incoherent addition). A nonzero mean of δ appears as arg γ₁₂, which moves the fringe maxima to Δφ = −arg γ₁₂ without changing V.

🧭 Polarization overlap and the polarization tool

Fields are vectors, so the interference term is E₁·E₂ ∝ ê₁·ê₂ = cos θ for linear states. This is the Fresnel–Arago result: orthogonally polarized beams from the same source do not form intensity fringes. Their relative phase still sets the polarization of the sum (linear, elliptical or circular), which is exactly the Jones-vector superposition in the polarization tool. A linear analyser at α projects both beams onto one axis (Malus's law: amplitude factors cos α and cos(α − θ)), and the fringes come back. For general Jones vectors J₁, J₂ the factor is the normalized inner product J₁†J₂, which can be complex. The model's jonesOverlap implements it.

🔄 Phasors when the frequencies differ

At the probe, each wave is the real part of a vector Ajeiθj with θj = kjxp − ωjt + φj, which rotates clockwise at ωj. If ω₁ = ω₂, both rotate together and one stationary phasor sum describes the interference. If ω₁ ≠ ω₂, the angle between them changes at Δω and no stationary sum exists, so the tool draws the instantaneous rotating vectors. The resultant length pulses at |Δf|.

🌟 Spatial interference versus temporal beats

  • Spatial beat: a snapshot at fixed t shows an envelope with period Λ = 1/|1/λ₂ − 1/λ₁|.
  • Temporal beat: a detector at fixed x sees I oscillate at fbeat = |f₁ − f₂|.

In a non-dispersive medium the envelope moves at the group velocity Δω/Δk = c, so fbeat = c/Λ. A detector with T ≫ 1/|Δf| reports I₁ + I₂.

🔬 Where this appears

  • Heterodyne detection: the beat between a signal and a local oscillator carries optical information to radio frequencies. Polarization matching (cos θ) sets the mixing efficiency.
  • Coherence measurement: fringe visibility gives |γ₁₂| directly when I₁ = I₂ and the polarizations are parallel.
  • Polarization-sensitive interferometry (OCT, fibre sensors): polarization fading happens when cos θ → 0.