In short

Coherent sources emit waves of the same frequency and maintain a constant phase difference over the time interval in which you observe them. Only coherent sources produce a stable interference pattern.

Why two independent bulbs do not interfere. An ordinary light source is a crowd of atoms that each emit a short wave train (about 10^{-8} s long, or a few metres in space). The phase of each emission is random. Two independent sources therefore have a phase difference that shuffles billions of times per second — bright and dark fringes form, move, and wipe each other out faster than any detector (or your eye) can resolve. The time-averaged intensity becomes the sum I_1 + I_2, with the cross-term averaging to zero.

Temporal coherence. A single wave train has finite length. The coherence time \tau_c is the duration over which the wave stays in-phase with itself; the coherence length is l_c = c\,\tau_c. Both are related to the spectral width \Delta\nu of the source:

\tau_c \approx \frac{1}{\Delta\nu}, \qquad l_c \approx \frac{\lambda^2}{\Delta\lambda}

Typical coherence lengths. White light: \sim 1 μm. Sodium lamp: \sim 1 mm. He-Ne laser: \sim 30 cm (multi-mode) or several metres (single-mode). Stabilised narrow lasers: kilometres.

Spatial coherence refers to how well the phase correlates across the wavefront (between two points at the same time). A point source is perfectly spatially coherent. An extended source (the Sun, a filament bulb) is spatially coherent only within a small patch on the wavefront, of size

d_c \approx \frac{\lambda \,R}{a}

where a is the source size and R the distance to it.

Two strategies for coherent sources.

  • Division of wavefront (Young's slits, Fresnel biprism, Lloyd's mirror): split one wavefront into two spatially separated portions.
  • Division of amplitude (thin films, Newton's rings, Michelson interferometer): split one ray into two via partial reflection at a surface, so both secondaries come from the same atom's emission.

Both strategies share the same trick: derive both beams from a single atomic emission, so they are guaranteed to share a phase.

Point two torches at the same patch of wall in a darkened hostel room at IIT Kanpur. The wall gets brighter — you have added two amounts of light — but nothing else happens. No stripes. No dark lines. Now replace the two torches with one torch shining through two pinholes punched close together in a piece of aluminium foil. The wall lights up with a pattern of parallel bright and dark bands, and the bands stay put long enough for you to photograph them with a phone.

Both setups have two sources of light. Both sources emit waves of roughly the same wavelength. And in both cases, the waves arrive at every point on the wall after travelling slightly different paths, so there is a path difference — the ingredient for interference. Yet one setup interferes and the other does not. The difference has a name: coherence. The two pinholes, fed from the same torch, share a history — every wavelet passing through them was created by the same atomic emission a moment earlier, so they leave the pinholes in lock-step. Two separate torches have no such history, and their phases drift apart faster than anything can keep up with.

This chapter is the manual for when you can get an interference pattern and when you cannot. You will learn what "constant phase difference" really means, how long light from a real source stays in-phase with itself (the coherence time), how far across a wavefront the phase stays correlated (the spatial coherence), and the two architectural strategies — division of wavefront and division of amplitude — that every interference experiment uses to cheat the randomness of real light and produce a stable pattern.

The interference condition, written plainly

Two waves of equal frequency arriving at the same point superpose. If their electric fields are E_1 = E_0 \cos(\omega t + \phi_1) and E_2 = E_0 \cos(\omega t + \phi_2), the total field is

E = E_0 [\cos(\omega t + \phi_1) + \cos(\omega t + \phi_2)]

The intensity (what a detector or your eye measures) is the time average of E^2:

I \;=\; I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos(\phi_1 - \phi_2) \tag{1}

The first two terms are the background — what you would get if you just added the two light intensities without interference. The third term is the interference term. It oscillates between +2\sqrt{I_1 I_2} (constructive) and -2\sqrt{I_1 I_2} (destructive) depending on the phase difference \Delta\phi = \phi_1 - \phi_2.

Interference is visible only if the third term is there — not averaged to zero. That requires \Delta\phi to stay fixed long enough for the detector (your eye, a camera, a photodiode) to register the pattern. This is the whole of coherence compressed into one sentence: coherence means the phase difference stays constant on the timescale of the observation.

Coherent vs incoherent superpositionLeft panel: two sinusoidal wave trains with a fixed phase offset of 90 degrees; their sum is a shifted sinusoid with a steady envelope. Right panel: two wave trains that each periodically jump to a new random phase at different times; their sum fluctuates chaotically and averages out. coherent — fixed Δφ E₁ E₂ sum steady sinusoid → visible fringes incoherent — Δφ shuffles E₁ E₂ sum chaotic average → uniform brightness
Left: two waves with a fixed phase difference add to give a steady wave of definite amplitude. Right: two waves whose phase differences reset randomly every few cycles (schematic — the actual resets happen on sub-nanosecond timescales for ordinary light) add to give a chaotic sum that, averaged over time, produces only the background $I_1 + I_2$ with the interference term washed out.

For equal intensities I_1 = I_2 = I_0, equation (1) becomes I = 2I_0 (1 + \cos\Delta\phi) = 4I_0 \cos^2(\Delta\phi/2), so the fringes run from 0 (perfect dark) to 4I_0 (four times one source alone, not two — the hallmark of constructive interference). When \Delta\phi randomises, \langle\cos\Delta\phi\rangle = 0 and I = 2I_0 uniformly — twice one source, no fringes.

Why two bulbs on the ceiling do not interfere

A light bulb is a crowd of roughly 10^{23} atoms, each of which from time to time makes a quantum jump and emits a short burst of electromagnetic wave — a wave train — lasting about \tau \sim 10^{-8} s. The bursts are independent: each atom has no knowledge of what any other atom is doing. The resulting light field is a random, jumbled sum of \sim 10^{23} short wave packets, each with its own random phase, and the net phase of the field at any instant is the phase of whichever wave trains happen to be present at that instant.

Two bulbs give two such jumbles, neither of which has any memory of the other. The phase of bulb A and the phase of bulb B bear no relation. Even if by coincidence they happen to be in phase at t = 0, within 10^{-8} s — the duration of a single wave train — one of them will have reset to a new random phase. In one full second, the phase difference has reshuffled about 10^8 times. Your eye, your phone camera, even a high-speed scientific detector averages over this shuffle — you see only the DC background I_1 + I_2. No stripes.

The numbers drive this home. A typical detector response time is of order a microsecond (10^{-6} s); light's wave train is of order a hundredth of a microsecond. During one detector frame, the phase difference visits essentially every value in [0, 2\pi) uniformly, so the cosine term in equation (1) averages to zero.

This is not an accident of poorly-behaved torches. It is the fundamental character of ordinary thermal light. No amount of careful alignment will make two independent bulbs produce stable fringes. To get interference you must derive both beams from the same emission event — and that is what every interference experiment in physics, from Young's slits to Michelson's interferometer to LIGO's gravitational-wave detector, does.

Temporal coherence — how long a wave train lasts

Temporal coherence asks: for how long does a wave field stay in phase with itself? Formally, if the field at time t is E(t) and at time t + \tau is E(t+\tau), how large can \tau be before E(t) and E(t+\tau) lose their phase correlation?

For an ideal monochromatic wave of a single frequency \omega_0, E(t) = E_0 \cos(\omega_0 t + \phi_0), and the correlation persists for ever. But no real source is perfectly monochromatic. Real sources emit a band of frequencies around \omega_0, spanning some width \Delta\omega (or equivalently a wavelength spread \Delta\lambda). The band shows up in the emission spectrum: a sodium lamp's yellow line is not a single frequency but a narrow line of width about \Delta\nu \approx 5 \times 10^{11} Hz. A white-light bulb's spectrum is broad — \Delta\nu \approx 3 \times 10^{14} Hz, roughly its entire visible range.

The duration of a wave train emitted by a single atom, \tau_c, is inversely related to this spectral width. This is a consequence of the Fourier relationship between time and frequency: a pulse confined to a time interval \tau_c contains Fourier components within a frequency range \Delta\nu \sim 1/\tau_c. Equivalently,

\boxed{\;\tau_c \;\approx\; \frac{1}{\Delta\nu}\;} \tag{2}

The coherence length l_c is the distance light travels in one coherence time:

l_c \;=\; c\,\tau_c \;\approx\; \frac{c}{\Delta\nu} \;\approx\; \frac{\lambda^2}{\Delta\lambda} \tag{3}

Why the \lambda^2/\Delta\lambda form: \nu = c/\lambda, so \Delta\nu = c\,\Delta\lambda/\lambda^2 (differentiating). Substituting into l_c = c/\Delta\nu gives l_c = \lambda^2/\Delta\lambda. Either form is correct; physicists tend to quote whichever is more directly measurable.

This one number — the coherence length — tells you how far apart two paths in an interferometer can be before their fringes disappear. If the path difference exceeds l_c, the two wave trains no longer overlap, and there is nothing to interfere. If the path difference is much smaller than l_c, you see crisp high-contrast fringes. In between, the contrast gradually fades.

Typical coherence lengths, worth memorising:

Source \Delta\lambda (approx) l_c (approx)
Incandescent bulb / white light 300 nm \approx 1 μm
LED (single colour) 30 nm \approx 10 μm
Sodium lamp (D line) 0.002 nm \approx 2 cm (modern) / \sim 1 mm (classroom)
Mercury lamp (green line) 0.001 nm \approx 30 cm
He-Ne laser (multi-mode) 0.002 nm \approx 20–30 cm
He-Ne laser (single longitudinal mode) 10^{-6} nm \gtrsim 100 m
Stabilised Nd:YAG laser <10^{-9} nm \gtrsim 100 km

The IIT-Kanpur or IUCAA Pune laser-physics lab's red He-Ne laser is why you can set up Michelson interferometers with arms differing by tens of centimetres and still see fringes. A classroom sodium lamp limits you to a few millimetres of path difference — enough for thin-film fringes, not enough for deep interferometry. And the reason stellar spectroscopy works at all is that even from a distant star, the spectral line has a narrow enough \Delta\lambda that the coherence length is centimetres, enough to interfere with itself on a spectrograph.

Explore — phase drift between two sources

Animated: phase drift between two sources A coherent sinusoid E1 equals sin of 2 pi times x, and a second sinusoid E2 equals sin of 2 pi times x plus a time-dependent phase shift. At small t the two are locked; as t grows the second source drifts, and their sum alternately adds and cancels, visualising the loss of coherence. source A (locked phase) source B (phase drifts with time t) horizontal axis: position along beam | time progresses as bodies move right
A simple visual for the phase drift that destroys coherence. Source A's peak (red) advances at a steady rate. Source B's peak (dark) advances at the same rate plus an extra drift that grows linearly with time — its phase relative to A is not fixed. For small $t$ the two stay close (in-phase); for large $t$ they separate and reunite chaotically. In real thermal light the drift is random, not linear, but the consequence is the same: a phase difference that is only fixed for a time $\tau_c$, after which coherence is lost. Click replay to watch again.

Spatial coherence — how wide across the wavefront the phase holds

Temporal coherence is about two moments. Spatial coherence is about two places: at the same instant, how similar are the phases at two separated points on the wavefront?

For a point source, the wavefront is a sphere of equal phase — two points anywhere on the sphere are in phase with each other, so spatial coherence extends across the whole wavefront. But for an extended source (the Sun, a filament bulb, an LED chip), the light arriving at an observer is a superposition of many spherical waves from many different points on the source. Each source point contributes a slightly different wavefront, tilted by a slightly different angle. Two observer points receive contributions from the same source points at slightly different path differences, and those path differences scramble the phase correlation if the observer points are far enough apart.

The quantitative result is the Van Cittert–Zernike theorem, which gives the coherence diameter d_c at distance R from a source of angular size a/R:

d_c \;\approx\; \frac{\lambda R}{a} \tag{4}

Why: you can think of the source as an incoherent collection of point emitters spread over size a. Each emitter's wavefront reaches the observer at a slightly different angle, spanning a tilt of \approx a/R. At observer spacing d, two points receive contributions whose path difference from one source edge to the other differs by roughly d \cdot a/R. When this is less than \lambda/2, the two observer points are still in phase; when it exceeds \lambda, coherence is lost. The crossover is equation (4), which turns out to be a fairly universal result even for other source shapes.

For the Sun (angular diameter \approx 1/2° = 0.0087 rad at Earth) and \lambda = 500 nm, the coherence diameter at ground level is

d_c \approx \frac{500 \times 10^{-9}}{0.0087} \approx 6 \times 10^{-5} \text{ m} = 60\ \mu\text{m}

So even sunlight is spatially coherent over patches about the thickness of a human hair. If you design a Young's double slit with slits 60 μm apart illuminated by direct sunlight, you will get faint fringes. Push the slits apart to 1 mm and the fringes disappear — sunlight is not spatially coherent over that distance.

This is why Young's original 1801 experiment used a single slit placed in front of the double slit: the single slit is a narrow effective source, and by equation (4) it dramatically increases d_c at the double slit's location, enough for the double slit to lie within the coherence patch. Without the single slit, Young would have seen nothing.

Spatial coherence from an extended sourceAn extended source of width a on the left emits wavefronts at multiple angles. At distance R, two observer points P1 and P2 separated by d receive superposed waves from each point of the source. If d times a over R is less than lambda, the observer points are within a coherence patch; otherwise they are not. source (size a) wavefront from top wavefront from bottom P₁ P₂ d distance R
An extended source of width $a$ sends wavefronts from every point on itself. At distance $R$, two observer points $P_1, P_2$ separated by $d$ each see the superposition of these wavefronts, with slightly different path differences. If $d < \lambda R/a$, the path differences are small compared to $\lambda$ and the observer points share nearly identical phase — they lie inside the coherence patch. Outside this patch, the phases scramble.

How to get coherent sources — the two strategies

If independent sources will not work, every interference experiment needs a way to manufacture two coherent beams from one. There are exactly two strategies.

Strategy 1: division of wavefront

Take one wavefront from a single source, and cut it geometrically into two separated portions. The two portions, because they came from the same wavefront, carry identical phases — they are inherently coherent.

Examples:

The common theme: divide the wavefront spatially. One source, two secondary positions, both phase-locked.

Strategy 2: division of amplitude

Take one ray and split it at a semi-reflecting surface: part reflects, part transmits. Now you have two beams, each carrying some fraction of the original amplitude. Because they came from the same ray, they share the same atomic emission — they are coherent — no matter how far apart they travel before recombining.

Examples:

Division of wavefront and division of amplitudeTop panel: a single source sends a wavefront through two pinholes, each a new coherent source (division of wavefront). Bottom panel: a single ray strikes a semi-transparent surface, splitting into a reflected and a transmitted beam that later recombine (division of amplitude). division of wavefront source slit 1 slit 2 screen division of amplitude ray semi-reflective transmitted reflected (recombines later)
Two architectural strategies for getting coherent beams from a real source. Top: division of wavefront — one wavefront is sampled at two spatially separated apertures. Bottom: division of amplitude — one ray splits into two at a partially reflecting surface, then the two reunite after traversing different paths. Both strategies ensure the two interfering beams share a common atomic emission, guaranteeing a fixed phase difference.

Why lasers changed everything

A laser is a source of light where many atoms are forced into phase by stimulated emission. Every photon emitted by a lasing atom arrives in-phase with the stimulating photon, and the cascade of such emissions yields a beam in which essentially all the photons share one phase and one direction. The coherence length jumps from micrometres (ordinary light) to metres or kilometres (lasers). Interferometers that were hopeless with classical sources became easy.

In the Indian context: the He-Ne laser (\lambda = 632.8 nm, red) in every JEE-level optics lab is coherent over about 30 cm — long enough to do Michelson interferometry with arm differences up to 15 cm without fringe loss. The diode lasers in laser pointers (\lambda = 650 nm, red or \lambda = 532 nm, green) have coherence lengths of centimetres — enough for Young's slits but not for heroic interferometry. A stabilised laser at Bangalore's Raman Research Institute or the Saha Institute in Kolkata can achieve l_c > 1 km, supporting experiments that classical optics never dreamed of.

Worked examples

Example 1: Coherence length of a sodium lamp

A sodium vapour lamp emits its yellow D line at \lambda_0 = 589.0 nm with a spectral width \Delta\lambda = 0.01 nm (dominated by Doppler broadening from the hot sodium atoms). Find the coherence time and coherence length. A Michelson interferometer using this lamp has one arm of length L_1 = 20 cm and a second arm of length L_2 variable by a micrometer. Estimate the maximum |L_2 - L_1| at which visible fringes still form.

Sodium D-line spectrumA narrow Gaussian peak centred at 589 nm with full width at half maximum 0.01 nm. The narrow width corresponds to a long coherence length, approximately 3 centimetres. intensity Δλ = 0.01 nm λ₀ = 589.0 nm wavelength
The sodium D line: a narrow Gaussian-shaped peak. Full width at half maximum $\Delta\lambda = 0.01$ nm. The narrowness of this line is what makes a sodium lamp a useful classroom source for interference.

Step 1. Convert the wavelength width to a frequency width.

\Delta\nu = \frac{c\,\Delta\lambda}{\lambda_0^2} = \frac{(3.00 \times 10^8)(0.01 \times 10^{-9})}{(589 \times 10^{-9})^2}
= \frac{3.00 \times 10^{-3}}{3.47 \times 10^{-13}} \approx 8.65 \times 10^{9} \text{ Hz} \approx 8.65 \text{ GHz}

Why: differentiate \nu = c/\lambda to get |d\nu| = c\,|d\lambda|/\lambda^2. The negative sign from the derivative is absorbed into taking absolute values.

Step 2. Coherence time from equation (2).

\tau_c \approx \frac{1}{\Delta\nu} = \frac{1}{8.65 \times 10^9} \approx 1.2 \times 10^{-10} \text{ s} = 0.12 \text{ ns}

Why: equation (2) is the Fourier-inverse relation: a wave packet of duration \tau_c has frequency spread \approx 1/\tau_c. The sodium lamp's light thus emits in bursts of about 0.1 nanoseconds each.

Step 3. Coherence length from equation (3).

l_c = c\,\tau_c = (3.00 \times 10^8)(1.2 \times 10^{-10}) \approx 3.5 \times 10^{-2} \text{ m} \approx 3.5 \text{ cm}

Why: light moves at c, so the spatial extent of a wave train is c\,\tau_c. Or, use the direct form l_c = \lambda^2/\Delta\lambda = (589 \times 10^{-9})^2 / (0.01 \times 10^{-9}) = 3.5 \times 10^{-2} m — same answer, different parametrisation.

Step 4. Apply to the Michelson setup. In a Michelson interferometer, the path difference between the two arms is 2|L_2 - L_1| (light goes to each mirror and back). Fringes are visible if this path difference is smaller than the coherence length:

2|L_2 - L_1| \lesssim l_c
|L_2 - L_1| \lesssim \frac{l_c}{2} \approx 1.8 \text{ cm}

Why: the factor of 2 appears because the light traverses each arm twice. If the path difference exceeds l_c, the two returning wave trains no longer overlap in time — there is no interference because there is nothing to interfere with.

Result: \tau_c \approx 0.12 ns, l_c \approx 3.5 cm, and arm mismatches up to \approx 1.8 cm preserve visible fringes.

What this shows. Coherence length sets a hard ceiling on the path difference in any interferometer. For a sodium lamp, a few centimetres; for a stabilised laser, kilometres. Your instrument's precision is set by the source's coherence, not the mechanical quality of the mirrors.

Example 2: Can sunlight illuminate a Young's slit directly?

A Young's double slit has two slits separated by d = 0.3 mm, placed at R = 1 m from a direct sunlight source (no primary slit). The Sun subtends an angle of \approx 0.5° as seen from Earth. Using \lambda = 550 nm (middle of visible), decide whether the slit separation lies within the spatial-coherence patch of the incident sunlight. Then find the slit separation at which direct-sun fringes would just fail.

Spatial coherence patch from the SunA schematic showing the Sun as an extended source on the left, Earth observer position 1 AU away, and the coherence patch of diameter about 60 micrometres at the observer's location. Two slits separated by 0.3 mm lie well outside this patch. Sun angular size 0.5° Earth d_c ≈ 60 μm (shown enlarged) slits d = 0.3 mm
The Sun's coherence patch at Earth is $\approx 60$ μm wide. A slit pair separated by $0.3$ mm (300 μm) spans *five* coherence patches — too wide for stable fringes with direct sunlight. A primary slit narrows the effective source, enlarging the patch and enabling interference.

Step 1. Compute the sunlight coherence diameter from equation (4). Angular diameter a/R \approx 0.5° = 0.5 \cdot \pi/180 \approx 8.73 \times 10^{-3} rad.

d_c \approx \frac{\lambda}{a/R} = \frac{550 \times 10^{-9}}{8.73 \times 10^{-3}} \approx 6.3 \times 10^{-5} \text{ m} = 63\ \mu\text{m}

Why: equation (4) with R and a combined into the angular size a/R (which is what you measure from Earth). Any two observer points within this radius share nearly the same phase; points farther apart do not.

Step 2. Compare to the slit separation d = 0.3 mm = 300 μm.

\frac{d}{d_c} = \frac{300}{63} \approx 4.8

The slit separation is about five times larger than the coherence diameter. The two slits lie in different coherence patches. The phase difference between them is not fixed — it fluctuates with the contribution from different parts of the Sun. No stable fringes.

Why: when the slit spacing exceeds d_c, each slit samples a different random phase. Equation (1) still holds for each slit pair, but the phase difference \Delta\phi averages to nothing.

Step 3. Find the maximum slit separation at which fringes would just form.

d_{\max} \approx d_c \approx 63\ \mu\text{m}

A Young's double slit with slits < 60 μm apart, illuminated by direct sunlight, would give faint visible fringes. This is not a theoretical construct — it has been done, and the fringes are faint but real.

Step 4. How Young actually got his pattern. He placed a primary slit in front of the Sun, narrowing the effective source to about 100 μm. This primary slit reduced the effective source size a by a factor of \sim 100 or more, multiplying d_c by the same factor. The coherence patch at his double slit was now of order millimetres, comfortably wider than the \sim 0.5 mm slit separation.

Why: the coherence diameter scales as 1/a, so narrowing the primary source widens the coherence patch proportionally. Young's primary slit is the one non-negotiable element of his setup — take it out, and the fringes vanish.

Result: Direct sunlight illumination of a 0.3-mm double slit produces no visible fringes because the slit separation lies outside the spatial coherence patch. A 60-μm separation would just barely work; a primary slit in the setup widens the coherence patch enough to allow any reasonable slit pair.

What this shows. Spatial coherence is a geometric gatekeeper. The pattern-forming condition d \lesssim d_c must be satisfied before anything else in an interference experiment matters. This is why classroom demonstrations always include the primary slit — Young's original, 225-year-old insight, still required.

Common confusions

If you understand why two independent sources don't interfere, can estimate coherence lengths from spectral widths, and know the two strategies, you have what JEE and a first course in optics demand. What follows is the formal theory of partial coherence, the visibility function, and a glance at Hanbury Brown–Twiss intensity interferometry.

Visibility and the degree of coherence

Define the fringe visibility V of an interference pattern as

V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}

For perfectly coherent light, V = 1; for perfectly incoherent light, V = 0. Real sources give 0 \leq V \leq 1. In terms of the phase difference and the statistical properties of the light, V is directly the magnitude of the complex degree of coherence \gamma_{12} between the two interfering beams:

V = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2} |\gamma_{12}|

For equal intensities I_1 = I_2, this simplifies to V = |\gamma_{12}|. The complex degree of coherence is defined as the normalised cross-correlation of the two field amplitudes:

\gamma_{12}(\tau) = \frac{\langle E_1^*(t)\,E_2(t + \tau)\rangle}{\sqrt{\langle|E_1|^2\rangle\langle|E_2|^2\rangle}}

When the beams are temporally delayed by \tau, |\gamma_{12}(\tau)| decays from 1 at \tau = 0 to 0 at \tau \gg \tau_c. Its exact shape depends on the spectral profile: for a Gaussian-shaped spectrum, |\gamma_{12}(\tau)| is a Gaussian in \tau; for a Lorentzian spectrum (natural-linewidth emission), it decays exponentially.

The Wiener–Khinchin theorem

A central result of coherence theory says: the complex degree of coherence of a stationary random source is the Fourier transform of its normalised power spectrum.

\gamma_{12}(\tau) = \int_0^\infty \tilde S(\nu) e^{-i2\pi\nu\tau}\,d\nu

where \tilde S(\nu) is the normalised power spectral density. This is the Wiener–Khinchin theorem. It quantifies what equation (2) only estimates: the exact relationship between spectral width and coherence time. For a Gaussian spectrum of FWHM \Delta\nu, the precise coherence time is

\tau_c = \sqrt{\frac{2\ln 2}{\pi}} \cdot \frac{1}{\Delta\nu} \approx \frac{0.66}{\Delta\nu}

The \approx 1/\Delta\nu estimate is good to within a small numerical factor for any reasonable spectrum.

Spatial coherence — the Van Cittert–Zernike theorem

The spatial analogue of Wiener–Khinchin is the Van Cittert–Zernike theorem: the complex degree of spatial coherence between two points on a wavefront is the Fourier transform of the intensity distribution of the source. For a uniform disc source of angular diameter \theta_a = a/R, the spatial coherence at observer separation d is

|\gamma_{12}(d)| = \left|\frac{2 J_1(\pi d \theta_a / \lambda)}{\pi d \theta_a / \lambda}\right|

where J_1 is a Bessel function. This reaches its first zero at d\theta_a/\lambda = 1.22, giving

d_c^{\text{first zero}} = \frac{1.22\,\lambda}{\theta_a}

— the "resolution limit" familiar from diffraction theory. The rough estimate d_c \approx \lambda/\theta_a used in equation (4) is the same formula with the 1.22 absorbed into the approximation.

Hanbury Brown–Twiss intensity interferometry

In the 1950s, Robert Hanbury Brown and Richard Twiss measured the angular sizes of stars by correlating not the electric fields (impossible at optical frequencies over long baselines) but the intensities at two detectors. Intensity fluctuations at two points separated by d remain correlated as long as d < d_c, even though the fields themselves cannot be directly combined. This works because the photon arrival statistics at a thermal source are super-Poissonian: photons cluster (bunching), and the bunching correlation extends across the coherence patch. HBT interferometers measured the angular diameters of stars like Sirius with modest equipment, inaugurating the field of quantum optics and leading — by way of a slight detour through the question "why do thermal photons bunch?" — to the first clean demonstration of photon antibunching from a single atom, a signature of non-classical light.

The coherence of gravitational-wave detection

LIGO's twin 4-kilometre interferometer arms need a coherence length of at least 4 km to function — a stabilised Nd:YAG laser at 1064 nm with linewidth \lesssim 75 Hz provides l_c \gtrsim 4000 km, so the mirrors at each end reflect wave trains that are still mutually coherent when they recombine at the beam-splitter. The signal LIGO looks for — gravitational waves from colliding black holes — modulates the arm-length difference by about 10^{-18} m. This is a fraction of the coherence length approximately 10^{-21}. That the signal can be pulled out at all is a testament to how well coherence engineering has advanced since Huygens' two-century-old wavelets.

Where this leads next