advanced28 min studyLesson 222 in Physics

Spatial coherence encodes angular source size

Derive a far-field visibility function from an incoherent source brightness distribution.

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What you’ll understand

  • State the incoherent-source and far-field assumptions.
  • Transform a uniform angular strip into a sinc visibility.
  • Separate baseline, angular width, and wavelength.

Two apertures illuminated by a distant extended source can show fringes even when different source points radiate independently. The same source point illuminates both apertures, creating a correlation; summing over source points can then cancel that correlation. The van Cittert–Zernike theorem turns this geometry into a Fourier transform. We will use a one-dimensional strip so that its visibility and first zero can be calculated directly.

Independent emitters can illuminate a correlated pair

Place two receiving apertures a transverse distance b apart. Radiation from a source direction θ reaches them with an approximate path difference bθ for small angles. Each source element therefore contributes a definite relative phase, even though its absolute phase fluctuates. Distinct source elements have vanishing mutual correlation in the spatially incoherent source model, so their cross terms disappear after averaging. Their individual contributions to the receiver correlation remain and add with direction-dependent phase.

Assume quasi-monochromatic stationary scalar radiation of wavelength λ, a distant source, small angular extent, and receiver geometry in the far-field or equivalent Fraunhofer regime. Match temporal paths sufficiently well that bandwidth averaging is negligible. These conditions allow brightness to depend only on direction and the phase to be linear in baseline times angle. Nearby sources with important wavefront curvature need a more general propagation calculation; the simple angular transform cannot silently be reused there.

γ(b)=∫B(θ)e−i2πbθ/λ dθ∫B(θ) dθ\gamma(b)=\frac{\int B(\theta)e^{-i2\pi b\theta/\lambda}\,d\theta}{\int B(\theta)\,d\theta}
B is nonnegative angular brightness per radian, b and λ are in metres, and θ is a small angle in radians. A removable geometric phase is suppressed; the source elements are mutually incoherent.↔ Scroll wide equations horizontally.

The uniform strip has an exact first null

Let the brightness be constant across a full angular width α centered on the axis and zero outside it. Integrating the complex exponential from minus α over two to plus α over two gives a sine divided by its argument. The normalization makes the coherence equal to one at zero baseline. The first zero occurs when the two edge directions differ in phase by a complete cycle, so contributions across the strip cancel in the average.

The result explains the inverse size relation: a smaller angular source stays coherent across a larger receiver separation. It also shows why coherence need not decrease monotonically. A sharp-edged strip has sidelobes beyond its first null, with alternating signs of the real coherence. Visibility measures the magnitude and loses that sign unless fringe phase is also measured. A smooth Gaussian angular profile instead produces a smooth Gaussian baseline dependence, without these sharp-boundary zeros.

γ(b)=sin⁡(πbα/λ)πbα/λ,bfirst null=λα\gamma(b)=\frac{\sin(\pi b\alpha/\lambda)}{\pi b\alpha/\lambda},\qquad b_{\mathrm{first\ null}}=\frac{\lambda}{\alpha}
α is the full angular width of the uniform one-dimensional strip. The sinc value at zero is defined by its limit. Balanced, polarization-matched fringe visibility is |γ|.↔ Scroll wide equations horizontally.

Interpreting a baseline measurement

A single visibility value generally does not determine an arbitrary source shape. Many brightness distributions can share one Fourier sample, and visibility magnitude alone leaves further phase ambiguities. Measuring several baselines, orientations, and where possible complex correlations supplies more information. Atmospheric phase errors, finite aperture averaging, and spectral bandwidth can alter the observations. The theorem provides the ideal forward relation; calibrating an instrument and reconstructing a source are additional inference problems.

Use the two-wave lab to examine the cancellation underlying the integral. Reset and set both controls to phase zero degrees and ratio one; predict the largest sum. For another fresh comparison Reset and set phase 180 degrees and ratio one; predict zero. These are two coherent contributions at one fixed frequency, not an extended incoherent source. The lab can test phasor cancellation between two terms but cannot vary a baseline, integrate angular brightness, or demonstrate the theorem by itself.

Infer the width of a uniform strip

A distant uniform strip is observed at wavelength 600 nm. Its first visibility null occurs at receiver separation 0.30 m. Find the full angular width, then the ideal visibility at half that baseline.

  1. Use the first-null relation with both lengths in metres: α=(600×10⁻⁹)/0.30=2.00×10⁻⁶ rad. This is the full strip width, not its half width.

  2. At baseline 0.15 m, the dimensionless ratio bα/λ is one half. The sinc argument is therefore π/2, with sine equal to one.

  3. Divide one by π/2 to obtain coherence 2/π≈0.637. With equal aperture intensities and matched polarizations, visibility has the same value.

The result

The strip has full angular width 2.00 microradians and balanced visibility 0.637 at 0.15 m.

Explore the model

Interactive lab

Interactive lab · waves

Cancellation among angular contributions

Reset and set BOTH controls to phase 0° and ratio 1; predict amplitude 2. Reset and set phase 180° and ratio 1 for a separate cancelling pair. There is no source-size or baseline control.

Cancellation among angular contributionsTwo 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 forThe spatial-coherence integral sums direction-dependent phases; this pair illustrates only its cancellation mechanism.

Check your understanding

Which assumption allows the basic theorem to transform angular brightness alone into receiver mutual coherence?

Try it yourself

At wavelength 500 nm, a uniform strip has full angular width 5 microradians. Find its first-null baseline. If its width doubles while its distance regime remains far-field, what happens to that baseline?

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 the full width in radians and keep wavelength fixed.

Compare with the solution

The first baseline is (500×10⁻⁹)/(5×10⁻⁶)=0.100 m. Doubling angular width makes the first null 0.050 m. The larger apparent source is less spatially coherent across a given separation; this does not imply a change in its temporal linewidth.

Take it with you

  • Mutually incoherent emitters can produce receiver correlations.
  • Baseline samples the Fourier transform of angular brightness in the stated regime.
  • Source shape and measurement of phase matter beyond the first null.

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