BrainFlux

Interferometers and coherence

Michelson interferometer

seen from above

Drag the mirror M2 along its beam (its knob: nanometres; its yellow top: tilt). The screen and the trace follow.

What you see

Spectrum S(σ)

draw on it

Interferogram I(Δ)

Experiments

Start here

A Michelson interferometer splits light in two, sends the halves to two mirrors and brings them back together; the screen shows the fringes they make, in true colour. The page opens with white light and the motor sweeping the mirror M2 back and forth through equal paths.

Drag the mirror M2 on the drawing (or stop the motor, ■, and turn M2's knob for nanometre steps; drag its yellow top to tilt it): the screen and the detector's trace follow. Press a light source on the right, or draw on the spectrum to make your own light.

Look for how far from equal paths the fringes last: a few micrometres for white light, the whole travel for a laser. That reach is the light's coherence, and the trace and the spectrum are each other's Fourier transform. Stellar, at the top, switches to Michelson's stellar interferometer, which measures a star's size the same way.

1. White light: fringes only near equal paths

What the page opens with: white light, M2 tilted a little so that the fringes are straight bands (experiment 3), and the motor sweeping it back and forth through equal paths. Stop the motor (■) and drag M2 yourself, or click on the trace to put it where you like. The fringes appear only within two or three micrometres of equal paths (the visibility is 1/2 at $\Delta=0.44$ μm and below 0.1 beyond 2.5 μm): every wavelength makes fringes of its own spacing, and they are all in step only at $\Delta=0$. The detector's record is the same on both sides of the central bright fringe, which is white; next to it the fringes are coloured, and a few fringes away the colours have mixed back into white. These are the interference colours that soap films show too (a soap film's start from black, because one of its two reflections gains half a wave; so can a beam splitter's, and then the colours are the complementary ones). On a real instrument nothing tells you where equal paths are: you find them by hunting for these fringes. The compensating plate gives both beams the same thickness of glass; without it, glass's dispersion would put each colour's zero path at a different place, and the pattern would be spread out and lopsided. Try the hot filament too, and change its temperature.

2. Count the fringes: the laser's wavelength

A helium-neon laser. The motor moves M2 at 0.5 μm/s; each time it moves $\lambda/2$, the path changes by $\lambda$ and one more ring is born at the centre of the screen. The counter under the drawing counts the bright fringes passing the detector (the cross). When the mirror has moved 10 μm the count is 31 or 32, so $\lambda=2d/N\approx 2\times10\ \text{μm}/31.6=633$ nm. Counting is good to one fringe, so the longer you count, the better you know the wavelength. (Press Reset to start again.) Rings are born at the centre while M2 moves away from equal paths, on either side; while it moves towards them the rings shrink into the centre and vanish. Run the motor backwards (◀) and watch them go, and the count go down.

3. Straight fringes and rings

Tilting M2 by $\alpha$ makes a thin wedge of air between M1 and the image of M2: the path difference grows by $2\alpha$ per unit length across the mirror and the fringes are straight, $\lambda/2\alpha$ apart. Here $\alpha=150$ μrad and the screen shows 12 mm of mirror: about six fringes. Drag the yellow dot at the top of M2 to tilt it more or less. Then move M2 away from equal paths with the slider $d$ and watch the fringes curve into rings: light crossing the gap at an angle $\theta$ travels $2d\cos\theta$, so each ring is a cone of light at one angle. They curve one way for $d>0$ and the other way for $d<0$ (at $d=\pm60$ μm the centre of the rings is off to the right, or to the left), so the straightest fringes mark equal paths. (The screen adds the two effects in a simple model; in a real instrument they form in different planes.)

4. The sodium doublet: fringes that fade and return

A sodium lamp's yellow light is two lines, 588.995 and 589.592 nm. Each makes its own fringes; where the bright fringes of one fall on the dark ones of the other the fringes fade, and they come back in step every $$\Delta=\frac{1}{\sigma_2-\sigma_1}=\frac{\lambda^2}{\Delta\lambda}=0.582\ \text{mm}$$ of path (0.291 mm of mirror travel). The motor sweeps M2 through equal paths, out to $\Delta=\pm1.2$ mm, and the detector's trace records the fading, the same on both sides: the fringes are clearest at zero path and fade at $\Delta=\pm0.29$ and $\pm0.87$ mm. (The screen blurs while the mirror moves fast: stop the motor to see the rings.) Measure the period and work out $\Delta\lambda$, as Michelson did. The D₂ line is twice as strong as D₁, so the fringes fade only to a visibility $(2-1)/(2+1)=1/3$: drag the dot on the weaker line (the spectrum is zoomed in on the pair) up to the height of the other and the fringes vanish completely at the minima.

5. Coherence length

A line of width $\Delta\lambda$ (a Gaussian), with the motor sweeping through equal paths. Its fringes fade as the path difference grows, on either side of zero, because the wavelengths in the line get out of step. The envelope of the interferogram is the Fourier transform of the line's shape, here a Gaussian whose visibility falls to one half at $$\Delta=\pm\frac{2\ln 2}{\pi}\frac{\lambda^2}{\Delta\lambda}\approx\pm0.44\frac{\lambda^2}{\Delta\lambda},$$ ±13.3 μm for 550 nm and 10 nm; $\lambda^2/\Delta\lambda=30$ μm is called the coherence length. Change $\Delta\lambda$ (under Light source) and watch the envelope stretch and shrink: a ten times narrower line has fringes ten times further.

6. A white LED

A white LED is a blue LED chip (here 450 nm, 20 nm wide) under a yellow phosphor (565 nm, 110 nm wide). At zero path its fringes are white-light fringes. The broad phosphor band loses its fringes within a micrometre or two, but the narrower blue band keeps them for several (half visibility at 4.5 μm, against 1.3 μm for the phosphor). So a few micrometres from zero path, on either side, only the blue light still interferes, and the fringes alternate between lavender (blue added to the phosphor's yellow) and yellow-green (blue taken away).

7. Draw a spectrum

Draw any spectrum on the spectrum plot (and drag a line's dot to change its strength); the interferogram, recorded here from −12 to +12 μm, follows as you draw, always the same on both sides of zero. Two separate bands give fringes that beat; a narrow band gives fringes that last; a broad one, fringes only near zero path. The width of a feature in one plot sets the extent of the other. Switch the spectrum's axis between $\sigma$ and $\lambda$: the same light, but its density per unit wavenumber and per unit wavelength differ by the factor $|d\sigma/d\lambda|=1/\lambda^2$.

8. Fourier-transform spectroscopy

A Fourier-transform spectrometer steps the mirror through the interferogram, from $-\Delta_\text{max}$ through zero path to $+\Delta_\text{max}$, takes a sample every $\delta$, and transforms the samples back: $$S(\sigma)\approx 2\delta\sum_{k=-K}^{K}\left[I(k\delta)-\tfrac12 I(0)\right]\cos 2\pi\sigma k\delta ,\qquad K=\Delta_\text{max}/\delta .$$ ($\tfrac12 I(0)$ is the level with no interference.) Here a mercury lamp: press Scan and watch. The samples on the two sides of zero path are mirror images (scroll over the interferogram to zoom in on zero and see). The spectrum (blue) is computed from the part of the scan on both sides of zero, so it appears once the scan passes zero path, and sharpens as the scan reaches further out. The ripples beside each line come from the scan's abrupt ends; Apodize weights the samples down to zero at the ends, which removes them at the cost of wider lines. The yellow pair, 577.0 and 579.1 nm, is $\Delta\sigma=0.0063$ μm⁻¹ apart: with $\Delta_\text{max}=60$ μm it is one bump. Raise $\Delta_\text{max}$ to about 120 μm to see two (140 μm with Apodize).

9. Both sides, or one?

The two sides of zero path carry the same information, so one side would do: scan from zero path out to $\Delta_\text{max}$ and take the other side as its mirror image, in half the time. But that needs the first sample exactly at zero path, and nothing marks it to a fraction of a wavelength. Here the spectrometer's zero is off by $\varepsilon=0.137$ μm (the zero error, under the spectrometer), a quarter of the green line's wavelength: the samples are at $k\delta+\varepsilon$, not $k\delta$. With both sides, the spectrum is right. Now choose One side and press Scan: each line comes out multiplied by $\cos(2\pi\sigma\varepsilon)$, plus a wiggle, so the green one is all wiggle, and the blue and violet ones point partly down. With both sides, the shift only gives the transform a phase, $2\pi\sigma\varepsilon$ at each line, which the spectrometer measures and takes out (phase correction). That is why spectrometers scan both sides of zero path, or at least a little of the second side, to measure the phase. (A real instrument can be lopsided in other ways too: without a compensating plate, the glass of the beam splitter delays each colour differently, and the phase is no longer in proportion to $\sigma$. A two-sided scan measures that phase too.) Set $\varepsilon$ to 0 and the two kinds of scan agree.

10. Resolution comes from the scan's reach

Two lines $\Delta\sigma$ apart only show as two when the scan covers a good part of their beat, whose period is $1/\Delta\sigma$: the resolution is about $1/(2\Delta_\text{max})$. For sodium, $\Delta\sigma=0.0017$ μm⁻¹ and $1/\Delta\sigma=0.58$ mm. Press Scan: the scan runs from −1.2 mm to +1.2 mm, and once it is past zero path the single line splits in two when the scan passes 0.41 mm: a little more than $1/(2\Delta\sigma)=0.29$ mm, because each line is smeared into a sinc and the weaker one hides in the shoulder of the stronger. (With Apodize the lines are wider and need 0.57 mm.) The resolving power $\sigma/\delta\sigma=2\sigma\Delta_\text{max}$ is twice the number of wavelengths in $\Delta_\text{max}$: about 4000 at the end, where sodium needs 1000.

11. Aliasing comes from the step

The samples must catch each fringe at least twice: $\delta<\lambda/2$, or $\sigma$ below the Nyquist wavenumber $\sigma_N=1/(2\delta)$. Here $\delta=0.4$ μm, $\sigma_N=1.25$ μm⁻¹, and the laser line at $\sigma=1.580$ μm⁻¹ shows up at $2\sigma_N-\sigma=0.920$ μm⁻¹ (1087 nm, in the infrared): the samples (dots) cannot tell its fringes (grey) from a slower wave through the same points. Lower the step below $\lambda/2=0.316$ μm and the line goes back where it belongs.

12. Draw an interferogram

Draw on the spectrometer's samples, and the spectrum becomes that of the light which would make your drawing. What you draw on one side of zero path appears on the other side too: the interferogram of any light is even. Flatten the outer half of the samples onto the dashed line (half the middle sample: no interference) and the line widens, since a shorter interferogram means a wider line. Draw a lone spike at zero path, flat elsewhere: white light. Draw a slow wave: infrared. Most drawings need negative power at some wavenumbers (drawn in red): no light could make them, because an interferogram must be the transform of a spectrum that is nowhere negative.

The stellar interferometer

Press Stellar at the top of the page. Two mirrors on a beam across a telescope collect a star's light, and the fringes at the focus fade as the mirrors are moved apart: how soon they fade tells the star's size. Drag an outer mirror, or let the baseline sweep.

13. Measure a star

A star 20 mas (0.020″) across, in light of 550 nm; the baseline sweeps back and forth. The stripes over the star are the fringes the interferometer projects onto the sky, $\lambda/B$ apart: when 1.22 of them fit across the disk, the bright and dark stripes cover equal light and the fringes vanish. For a uniform disk the visibility is $$|V|=\left|\frac{2J_1(x)}{x}\right|,\qquad x=\frac{\pi B\theta}{\lambda},$$ which first vanishes at $B=1.22\lambda/\theta=6.92$ m. Past it the fringes come back, weakly and shifted by half a fringe (the middle of the pattern turns dark). Real stars are darker towards the edge: set the limb darkening to 1 and the first zero moves out to $1.43\lambda/\theta$, so a uniform-disk fit would make the star too small. The Sun seen from 10 parsecs is 0.93 mas across and would need a 149 m baseline (the CHARA array on Mount Wilson today has 330 m).

14. Betelgeuse, December 1920

Michelson and Pease mounted the mirrors on a 20-foot (6.1 m) beam across the 100-inch telescope at Mount Wilson and moved them apart until Betelgeuse's fringes disappeared, at about 3.07 m (121 inches). For light of 575 nm that gives $\theta=1.22\lambda/B=0.047″$: the first star other than the Sun to be measured. Turn on Atmosphere: the air above the telescope delays the two beams differently and the fringes dance, as they did for Michelson's observers, but their contrast can still be judged. They judged it by eye, in a broad band of light: set $\Delta\lambda$ to 100 nm. The middle fringe still vanishes at almost the same baseline, 3.09 m instead of 3.08 m (the visibility changes sign there, and the wavelengths on either side cancel), but further out the dashed curve, the contrast of the middle fringe over the band, is smoothed.

15. A binary star

Two stars about 50 mas apart, of nearly equal brightness, like Capella. Each makes its own fringes, shifted by the stars' separation $\rho$ along the baseline: the fringes fade when the shift is half a fringe and come back when it is a whole one. The visibility beats with period $\lambda/(\rho\cos\varphi)=2.27$ m, dipping to $(1-r)/(1+r)$ for a brightness ratio $r$. Only the separation along the baseline counts: turn the angle $\varphi$ and the beats slow down; at 90° they are gone. In 1920 John Anderson measured Capella's orbit this way at Mount Wilson: two stars too close together for any telescope to show apart.

How it works

Moving the mirror M2 a distance $d$ from where the two paths are equal changes the path difference by $\Delta=2d$ (the light goes there and back); $d$ and $\Delta$ are negative when M2 is nearer the beam splitter than that. Light of wavenumber $\sigma=1/\lambda$ comes back with a phase difference $2\pi\sigma\Delta$, and light with the spectrum $S(\sigma)$ gives the detector the signal $$I(\Delta)=\int S(\sigma)\,[1+\cos 2\pi\sigma\Delta]\,d\sigma .$$ The cosine is even, so $I(-\Delta)=I(\Delta)$: the interferogram is the same on both sides of zero path. Apart from the constant, it is the Fourier (cosine) transform of the spectrum: they are a Fourier pair, and you can change either one here. The dashed line on the interferogram is the level with no interference, half the signal at zero path. The screen shows what you would see looking into the instrument, in true colour; over it, the visibility is the contrast $(I_\text{max}-I_\text{min})/(I_\text{max}+I_\text{min})$ of the fringes around the present $\Delta$: 1 for a laser, smaller where the wavelengths in the light have got out of step. Scroll over the spectrum or the interferogram to zoom; the buttons above them set their ranges.

The stellar interferometer. Light from a single point far away arrives as plane waves, and two mirrors anywhere in it give perfect fringes. A star of angular size $\theta$ is many points, each making its own fringes, shifted according to its direction; with the mirrors a distance $B$ apart these get out of step once $B\theta$ is about $\lambda$. The fringe visibility against baseline is the Fourier transform of the star's brightness, at spatial frequency $B/\lambda$ (the van Cittert–Zernike theorem). In Michelson's design the inner mirrors are fixed, so the fringes keep their spacing while the outer mirrors set $B$; with two slits instead, the fringes would also get finer as $B$ grows.

What is approximate

Colours come from the CIE 1931 colour matching functions (the analytic fit of Wyman, Sloan and Shirley, 2013) and are shown in sRGB; pure spectral colours lie outside what a screen can show and are drawn with the negative channel cut to zero. The mercury lamp's line strengths are approximate; sodium's 2 : 1 is the value for a thin source. Drawn spectra live between 357 and 833 nm. The fringe counter counts a fringe each time the signal rises through 1.3 times its mean after falling below 0.7 times it, so it misses fringes of visibility below 0.3. The spectrometer is perfect but for its zero error: no noise, and its steps exact. Its phase correction takes the phase of the two-sided sum weighted by a triangle (Mertz's method, here with all of the scan): where there is light, that phase is the light's own. The stellar fringes are drawn inside the diffraction pattern of the inner mirrors (at the middle wavelength of the band), without the blurring of the image by the air.

New in 2026 for BrainFlux. Source: github.com/gallicchio/brainflux, MIT licence.

Interferometers and coherence

Michelson interferometer

What you see

Spectrum S(σ)

Interferogram I(Δ)

Michelson's stellar interferometer

The star

The fringes at the focus

Visibility |V| against baseline