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Optics 041 · Interference, coherence, cavities, and metrology

Coherent Wave Composer

Why do a phasor diagram, a wavefront picture and a detector reading all describe the same interference? All three come from one sum of complex fields. The phasor diagram adds the fields as arrows, the wavefront picture shows the same sum spread out in space, and the detector reports the square of its size. A detector also averages over its exposure time, so beats between two slightly different frequencies can wash out.

A live landscape of the signed field, complex phasor polygons and two detector traces all come from one deterministic sum of plane waves. Combine two to six waves, move the probe, shift phase and optical path, and reveal beats through a finite detector window.

Interactive modelCoherent Wave Composer
Same-probe instantaneous and exposure intensity Iinst,I‾TI_{\mathrm{inst}},\overline I_T0.500.50
Time / peak / pair contrast or wave count t,Ipeak,VT,Nt,I_{\mathrm{peak}},V_T,N50%50\%
Beat cycles in exposure and phasor closure error ΔνT,εphasor\Delta\nu T,\varepsilon_{\mathrm{phasor}}0.00π0.00\pi
Model regimevalid model regime\text{valid model regime}

Physics tutorial

Audit the physics of Coherent Wave Composer

BackgroundLight waves add as fields, not as brightness. The detector here is scalar: it measures the squared electric-field amplitude, not the full vector energy flux through a tilted surface. The ideal plane waves share one polarization, and the phasor frame removes their optical carrier: S(r,t)=∑j=0N−1Ajei(kj⋅r+ϕj−2πjΔνt)S(\mathbf r,t)=\sum_{j=0}^{N-1}A_j e^{i(\mathbf k_j\cdot\mathbf r+\phi_j-2\pi j\Delta\nu t)}. The upper surface shows the electric field with its sign; it is not moving material or a rippling mirror. It magnifies micrometre distances. Only the shared optical carrier is replaced, by a slow rotation labelled as illustrative, and the true beat frequency offsets are kept.

Why it mattersWhy do a phasor diagram, a wavefront picture and a detector reading all describe the same interference?

Start with the essentials

Focus question
Why do a phasor diagram, a wavefront picture and a detector reading all describe the same interference?
One-sentence intuition
At the white probe, both phasor diagrams and the instantaneous detector intensity come from exactly the same complex sum: Iinst/I1=∣S∣2I_{\mathrm{inst}}/I_1=|S|^2. Exposure does something else: it averages the intensity over time, rather than averaging the fields before squaring. The cyan trace is instantaneous; the dashed gold trace averages over the exposure that just ended. When the frequencies differ, a short exposure follows the beat, while an exposure lasting a full beat period hides it.

Core mathematical model

Phase, optical path and crossing angle

θj=(j−N−12)α,ϕj=jΔϕ+j∥kj∥Δℓkj=2π(ν0+jΔν)c(sin⁡θj,cos⁡θj)\begin{aligned}\theta_j&=\left(j-\frac{N-1}{2}\right)\alpha,\quad \phi_j=j\Delta\phi+j\lVert\mathbf k_j\rVert\Delta\ell\\\mathbf k_j&=\frac{2\pi(\nu_0+j\Delta\nu)}{c}(\sin\theta_j,\cos\theta_j)\end{aligned}

Neighbouring waves differ by the same step in direction, phase and frequency. By definition, a positive extra optical path adds phase at the reference plane. Half a wavelength of path adds half a cycle; a full wavelength brings back the same pattern. All the directions lie in one plane, so the waves can physically share one linear polarization perpendicular to that plane.

Exact intensity for a finite exposure

I‾TI1=1T∫t−Tt∣S(r,τ)∣2 dτ=∑jAj2+2∑i<jAiAjsinc⁡(πΔνjiT)cos⁡ ⁣[ψji ⁣(t−T2)]\frac{\overline I_T}{I_1}=\frac1T\int_{t-T}^{t}|S(\mathbf r,\tau)|^2\,d\tau=\sum_j A_j^2+2\sum_{i<j}A_iA_j\operatorname{sinc}(\pi\Delta\nu_{ji}T)\cos\!\left[\psi_{ji}\!\left(t-\frac T2\right)\right]

The rectangular window ends at the displayed time. The sinc function is the sine divided by its argument and equals one at zero. Its sign must be kept, and so must the phase at the window's midpoint. At zero exposure the detector uses the continuous instantaneous limit. The fields are assumed steady before time zero, so the history can include negative times.

Two-wave maxima, minima and fringe contrast

Imax⁡,min⁡I1=(1±a)2,VT=2a1+a2∣sinc⁡(πΔνT)∣,fbeat=∣Δν∣\frac{I_{\max,\min}}{I_1}=(1\pm a)^2,\quad V_T=\frac{2a}{1+a^2}\left|\operatorname{sinc}(\pi\Delta\nu T)\right|,\quad f_{\mathrm{beat}}=|\Delta\nu|

These contrast and beat formulas hold for two waves only. Two equal waves can cancel completely; waves of unequal amplitude cannot. The visibility is the contrast of the phase scan or the beat envelope after exposure, not a parameter describing a randomly fluctuating source. With more waves, the detector includes every pair instead of reusing the two-wave formula.

Phasor closure for many waves, and the fixed plot scale

∑j=0N−1e2πij/N=0,Ipeak/I1=(∑jAj)2,0≤I‾T≤Ipeak\sum_{j=0}^{N-1}e^{2\pi i j/N}=0,\quad I_{\mathrm{peak}}/I_1=\left(\sum_j A_j\right)^2,\quad 0\le\overline I_T\le I_{\mathrm{peak}}

At the origin, six equal waves sixty degrees apart in phase form a closed polygon. Moving the probe changes the spatial phases, and the polygon can open elsewhere. The plot axes and the floor map are divided by the stated theoretical peak, not by the largest value observed. The physical readouts stay in units of the first wave's intensity on its own.

Common difficulties

A hidden beat does not mean the source lost coherence

Typical misconceptionA flat long-exposure trace means the waves have lost their phase relationship.

Better mental modelThe inputs here are deterministic and fully coherent; only the detector averages. With a one-second exposure and a one-hertz frequency difference, the gold trace stays at the sum of the two separate intensities, while the cyan beat and the phasor rotation carry on. The model has no random phase noise.

Run the experiment

  1. 01

    Scene 1: Two-wave fringes

    Pause the two-wave preset. Set the initial phase step to 180 degrees, then reduce the secondary amplitude or add half a wavelength of optical path.

    What to observe: The origin cancels completely only when the amplitudes match. Adding half a wavelength undoes the cancellation; the fringes move instead of vanishing everywhere.
  2. 02

    Scene 2: Six-wave phasor closure

    Choose the six-wave closure preset. Move the shared probe, then set the phase step to zero and watch the polygon stretch into a line.

    What to observe: At every probe position, the white endpoint matches the instantaneous intensity. Six unit waves in phase reach thirty-six times the intensity of the first wave alone at the origin.
  3. 03

    Scene 3: Beat hidden by long exposure

    Choose the hidden-beat preset and press play. Shorten the exposure from one second to zero. Then reverse the frequency difference and compare which way the pattern drifts.

    What to observe: When the exposure lasts one full beat period, the gold trace stays flat at twice the single-wave intensity. A short exposure brings the modulation back without any change to the source.

Related experiments and articles

Continue exploring

From sources to phasors

Which changes move the fringes, and which reduce their contrast?

  1. LabCoherent Wave StudioScan two spherical sources and vary their relative phase.
  2. You are hereCoherent Wave ComposerCombine plane waves, then test beating and exposure averaging.
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