Michelson Interferometer

Split a single beam of light into two, send the halves down two arms of a right-angle, bounce them off mirrors, and bring them back together — and where they recombine, the difference in the two round-trip distances is written out as a pattern of light and dark. That is the whole of the Michelson interferometer, and it is arguably the most consequential measuring instrument physics has produced. Because the two beams are born from the same wave at a single beamsplitter, they stay mutually coherent, and their recombined brightness reads out the optical path difference $\Delta = 2(d_2 - d_1)$ directly in wavelengths of light — the factor of two because each arm is traversed twice. The power of the device is that it converts things we cannot see — a sub-micron mirror shift, a whisker of extra glass, the spectral width of a lamp, a passing gravitational wave — into something we can literally count: fringes sweeping past a detector, one full cycle for every half-wavelength of extra path. Move a mirror a known distance and count the fringes, and you have measured a wavelength; scan the mirror and record the brightness, and you have recorded the source's entire spectrum as an interferogram. The same instrument delivered the null result that helped dismantle the luminiferous ether, pinned the metre to an atomic wavelength before atomic clocks existed, and — scaled to four-kilometre arms — caught the first ripple of spacetime from two colliding black holes.

Key takeaways

A beamsplitter divides one wave in amplitude into two; the halves travel two perpendicular arms, reflect, and recombine, so their optical path difference $\Delta = 2(d_2-d_1)$ sets the interference phase (the factor of two is the round trip).
The recombined intensity is $I = 2I_0\,[1 + \cos(2\pi\Delta/\lambda)]$ ; translating one mirror by $\lambda/2$ sweeps exactly one fringe past the detector — a ruler graduated in half-wavelengths.
Being an amplitude-splitting device, it makes fringes even with a broad extended source: the circular fringes of equal inclination obey $2d\cos\theta = m\lambda$ and localise at infinity.
As the path difference grows, the fringe visibility traces $|\gamma(\tau)|$ , the source's degree of temporal coherence — so the instrument doubles as a coherence and spectrum analyser.
Key idea: a single wave split in amplitude and recombined turns mirror motion, refractive index, and spectral content into countable optical phase — $N = 2\,\Delta d/\lambda$ fringes per mirror displacement $\Delta d$ .

The core insight

One wave is split at a beamsplitter, sent down two arms, and recombined. The output brightness $I = 2I_0[1 + \cos(2\pi\Delta/\lambda)]$ depends only on the round-trip path difference $\Delta = 2(d_2-d_1)$ , so every half-wavelength of mirror travel counts off one fringe. Counting fringes measures length in wavelengths; watching their contrast fade measures the light's coherence. It is an amplitude-splitting interferometer, recalled by name from aq-optics-wavefront-amplitude-splitting, which is why an extended source still yields sharp rings.

Where this lives today

The Michelson never became a museum piece; it became infrastructure. Fourier-transform infrared spectrometers — a scanning Michelson plus a computer running an FFT — are the workhorse of chemistry labs, recording an entire molecular spectrum from a single mirror sweep with the multiplex and throughput advantages that dispersive instruments cannot match. Stabilised helium–neon interferometers ( $\lambda = 632.8$ nm) calibrate machine tools, semiconductor steppers, and length standards to nanometres by counting fringes as a stage moves. And at the extreme, LIGO and Virgo are Michelson interferometers with four-kilometre arms and Fabry–Pérot cavities folded inside them, sensing differential arm-length changes of $10^{-19}$ m — a thousandth of a proton's width — as a gravitational wave sweeps through. The same idea that failed to find the ether in 1887 now routinely hears black holes collide, a reminder that a null result and a discovery can be the same apparatus a century apart.

The physical picture

The trick that makes the whole instrument legible is a single act of imagination: look past the two real mirrors and picture the image $M_1'$ of the first mirror formed by the beamsplitter. Seen from the detector, $M_1'$ floats right beside the second mirror $M_2$ , and the two returning beams behave exactly as if they had reflected off the front and back of a thin slab of air of thickness $d = d_2 - d_1$ trapped between them. Every result you know from thin films transfers wholesale — wedge fringes when the surfaces tilt, concentric rings when they are parallel. The Michelson is a thin-film interferometer whose "film" you can stretch to any thickness by turning a screw.

Figure

Michelson layout: source → 50/50 beamsplitter → fixed mirror M₁ and movable mirror M₂; the equivalent air film of thickness d = d₂ − d₁ between M₂ and the beamsplitter image M₁′

Two other habits of thought help. First, the moving mirror is a clock hand: every half-wavelength of travel ticks one fringe from bright to bright, so the detector's brightness is a dial reading the mirror position in units of $\lambda/2 \approx 300$ nm. Second, energy is never lost when the output darkens — it simply flows back out the port facing the source, because the beamsplitter stamps a relative $\pi$ phase between the two ways the beams can recombine. Push the arms far enough apart, though, and the two returning wavetrains no longer overlap in time; the fringes fade, and the distance over which they survive — the coherence length — is the light's own measure of how long it keeps its phase.

Figure

Fringe count N versus mirror displacement d: a straight line of slope 2/λ; one fringe per λ/2 of travel, so 1000 fringes correspond to 294.65 µm at the sodium wavelength 589.3 nm

The core equations

Two coherent beams of intensity $I_1$ and $I_2$ meeting with a phase difference $\delta$ give the general two-beam law. In a balanced Michelson the beams return with essentially equal intensity, so only $\delta$ — set by the path difference — modulates the output.

I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\delta, \qquad \delta = \frac{2\pi}{\lambda}\,\Delta \qquad (1)

where: $I$ — detected intensity; $I_1, I_2$ — the two returning-beam intensities; $\delta$ — their phase difference; $\Delta$ — optical path difference; $\lambda$ — vacuum wavelength. The cross term $2\sqrt{I_1 I_2}\cos\delta$ is the interference; it is largest when the two beams are equal. The path difference is the one quantity the arms control. Light runs from the beamsplitter to each mirror and back, so a geometric arm-length imbalance $d_2 - d_1$ enters twice — the round-trip factor that is the single most common thing to forget.

\Delta = 2\,n\,(d_2 - d_1) \qquad (2)

where: $\Delta$ — optical path difference between the arms; $d_1, d_2$ — geometric beamsplitter-to-mirror distances; $n$ — refractive index of the medium in the arms ( $n \approx 1$ in air). Put an object of index $n$ and thickness $t$ in one arm and $\Delta$ shifts by $2(n-1)t$ — the basis of interferometric refractometry. For a 50/50 splitter with $I_1 = I_2 = I_0$ , the law collapses to a pure raised cosine that swings between full brightness $4I_0$ and complete darkness. This is the fringe pattern the detector reports.

I = 2 I_0 \left[\,1 + \cos\!\left(\frac{2\pi\Delta}{\lambda}\right)\right] \qquad (3)

where: $I_0$ — intensity of each beam alone; $\lambda$ — wavelength; $\Delta$ — path difference. A change $\Delta \to \Delta + \lambda$ (one wavelength of extra path, i.e. $\lambda/2$ of mirror travel) returns the pattern to itself: one fringe. With an extended source the axial path difference $\Delta = 2d$ is shared by every ray parallel to the axis, while a ray tilted by $\theta$ sees a slightly shorter film. The loci of constant tilt are cones, so the fringes are concentric rings.

2d\cos\theta = m\lambda \qquad (4)

where: $\theta$ — inclination of the ray to the axis; $m$ — integer order of a bright ring; $d = d_2 - d_1$ — effective film thickness. The order is highest at the centre ( $m_0 = 2d/\lambda$ ) and falls outward, so rings are born or swallowed at the centre as $d$ changes. Finally, real light is not perfectly monochromatic, and the fringe contrast — the visibility — carries that information. As the delay $\tau = \Delta/c$ grows, the visibility follows the modulus of the degree of temporal coherence.

V = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} = |\gamma(\tau)|, \qquad \tau = \frac{\Delta}{c} \qquad (5)

where: $V$ — fringe visibility (Michelson's own contrast measure); $\gamma(\tau)$ — complex degree of temporal coherence; $\tau$ — time delay of one arm relative to the other. Recalled from aq-optics-temporal-coherence-length: $|\gamma|$ decays over the coherence time $\tau_c \sim 1/\Delta\nu$ , so fringes survive to a path difference of about the coherence length $\ell_c = c\tau_c$ .

Derivation AThe fringe-counting ruler and its factor of two

Our goal is to turn the interferometer into a length-measuring instrument by proving the exact relation between how far the movable mirror travels and how many fringes sweep past the detector — and to make crystal-clear where the factor of two comes from, because getting it wrong doubles every length you infer. We start from the balanced intensity law and ask what mirror motion advances the phase by a full $2\pi$ .

Step 1. Write the output brightness as $I = 2I_0[1 + \cos(2\pi\Delta/\lambda)]$ , a function of the path difference $\Delta$ alone. Why: after a balanced beamsplitter the two beams have equal amplitude, so nothing modulates the detector except the phase $\delta = 2\pi\Delta/\lambda$ that the arms impose.

Step 2. Identify one fringe with one full period of the cosine: the pattern repeats each time $\Delta$ increases by one wavelength, $\Delta \to \Delta + \lambda$ . So $N$ fringes correspond to a path-difference change of $\delta\Delta = N\lambda$ . Why: brightness is periodic in $\Delta$ with period $\lambda$ , so counting bright-to-bright cycles counts wavelengths of accumulated path.

Step 3. Connect the path difference to the mirror. Since $\Delta = 2(d_2 - d_1)$ , moving one mirror by $\Delta d$ changes the path difference by $\delta\Delta = 2\,\Delta d$ — the light crosses the shifted arm and comes back, so the extra distance counts twice. Why: the beam makes a round trip in each arm, so a mirror displacement $\Delta d$ lengthens the optical path by $2\,\Delta d$ , not $\Delta d$ .

Step 4. Equate the two expressions for $\delta\Delta$ : $2\,\Delta d = N\lambda$ , hence $N = 2\,\Delta d/\lambda$ and, per fringe, a mirror travel of exactly $\lambda/2$ . Why: this is the owned result — a calibrated dial that reads mirror position in half-wavelengths, letting a counted integer stand in for a length or, run backwards, for a wavelength. *Verified numerically ( $N = 2\Delta d/\lambda \Rightarrow \Delta d = 294.65\ \mu$ m for $N=1000$ at $589.3$ nm) and symbolically ( $\partial I/\partial\Delta$ period $=\lambda$ ) in verify_aq-optics-michelson-interferometer.py.*

Go deeper · Derivation B · 2 minCircular fringes of equal inclination

Derivation BCircular fringes of equal inclination

Here we show why an extended, spatially incoherent source — a sodium lamp, not a laser — still paints sharp concentric rings in a Michelson, and derive the ring condition $2d\cos\theta = m\lambda$ . The key is that the two mirrors are optically equivalent to a plane-parallel air film, and that amplitude splitting sorts rays by inclination rather than by which slice of wavefront they came from.

Step 1. Replace the apparatus by its equivalent film: the beamsplitter image $M_1'$ of one mirror lies a distance $d = d_2 - d_1$ from the other mirror $M_2$ , forming a virtual slab of air. Each detected ray is the superposition of its reflections from the two faces of that slab. Why: folding the two arms onto a common axis converts the interferometer into the thin-film problem, whose fringe geometry is already known.

Step 2. Take a ray inclined at angle $\theta$ to the axis. The extra optical path between the two face-reflections of a parallel slab of thickness $d$ is $2d\cos\theta$ (not $2d/\cos\theta$ — the cosine enters because the relevant delay projects onto the surface normal). Why: the geometry of a ray bouncing between two parallel surfaces gives a path difference that shrinks as the ray tilts away from normal incidence.

Step 3. Impose constructive interference: $2d\cos\theta = m\lambda$ for integer $m$ . Each $m$ picks out a fixed $\theta$ , and by rotational symmetry a whole cone — hence a bright ring — of half-angle $\theta_m$ . Why: only ray directions whose two reflections arrive in step add up, and those directions form cones about the axis, imaged as circles.

Step 4. Note the localisation: all source points that emit at the same inclination $\theta$ feed the same ring, so the pattern lives at infinity and does not blur when the source is widened. The central order $m_0 = 2d/\lambda$ is the largest; shrinking $d$ lowers it and rings collapse into the centre, one per $\lambda/2$ of mirror travel. Why: this inclination-sorting is exactly what wavefront-splitting interferometers cannot do, and it is why the Michelson tolerates a bright, broad lamp. *Verified ( $2d\cos\theta = m\lambda$ solved for the first ring angles at $d = 0.1$ mm, and the central order $m_0 = 2d/\lambda$ ) in verify_aq-optics-michelson-interferometer.py.*

Go deeper · Derivation C · 1 minVisibility as the light's memory of its own phase

Derivation CVisibility as the light's memory of its own phase

Finally we connect the contrast of the fringes to the coherence of the source, showing that the visibility curve $V(\Delta)$ is a direct readout of the degree of temporal coherence and, through it, of the spectrum. This is the bridge from interferometer to spectrometer, developed in full in the companion article on Fourier-transform spectroscopy; here we establish the link.

Step 1. Model the two arms as delivering time-delayed copies of the same field, $E_2(t) = E_1(t - \tau)$ with $\tau = \Delta/c$ . Why: the only difference between the arms is the extra time the longer path imposes, so one output is a delayed replica of the other.

Step 2. Time-average the recombined intensity over the detector's integration window, longer than the coherence time: $\langle I\rangle \propto \langle |E_1(t) + E_1(t-\tau)|^2\rangle = 2I_0\,[1 + \mathrm{Re}\,\gamma(\tau)]$ , where $\gamma(\tau) = \langle E_1^*(t)E_1(t-\tau)\rangle/\langle|E_1|^2\rangle$ is the normalised autocorrelation. Why: the cross term of the squared sum is exactly the field's self-correlation at delay $\tau$ , which is what $\gamma(\tau)$ names.

Step 3. Read off the contrast. Writing $\gamma(\tau) = |\gamma(\tau)|e^{i(2\pi\bar\nu\tau + \phi)}$ , the output is $2I_0[1 + |\gamma|\cos(2\pi\bar\nu\tau + \phi)]$ , so the visibility is $V = |\gamma(\tau)|$ . Why: the envelope of the cosine is $|\gamma|$ , and Michelson defined visibility precisely as that fringe envelope.

Step 4. Invoke Wiener–Khinchin: $\gamma(\tau)$ and the power spectrum $S(\nu)$ are a Fourier-transform pair, so scanning $\Delta$ to record $V(\Delta)$ measures the spectral line shape. For a doublet split by $\Delta\lambda$ the visibility beats to zero and back with period $\lambda^2/\Delta\lambda$ in $\Delta$ . Why: this is the operating principle of Fourier-transform spectroscopy (aq-optics-fourier-transform-spectroscopy) — the interferogram's transform is the spectrum. *Verified (doublet visibility period $\lambda^2/\Delta\lambda = 0.582$ mm for the sodium D lines) in verify_aq-optics-michelson-interferometer.py.*

Limiting cases

Case	Result	Meaning
Equal arms, $\Delta \to 0$	$V \to 1$ , white-light fringe	the zero-order fringe is visible even in white light; it fixes the absolute origin of $\Delta$
$\Delta \gtrsim \ell_c = c/\Delta\nu$	$V \to 0$	fringes wash out beyond the coherence length; the envelope maps the spectral width
Extended source, mirrors parallel	$2d\cos\theta = m\lambda$	concentric rings of equal inclination, localised at infinity, collapsing inward as $d\to0$
Mirrors tilted, collimated light	straight fringes $\Delta(x) = 2\alpha x$	equal-thickness wedge fringes across the tilt angle $\alpha$
Monochromatic laser	fringes at all $\Delta$	$\ell_c$ is metres to kilometres; fringes persist for huge path differences

Worked examples

1 · Measuring a displacement — and the metre in wavelengths (confidence builder)

A Michelson is illuminated with the sodium D line, $\lambda = 589.3$ nm. (a) You translate the movable mirror and count exactly $1000$ bright fringes passing the crosshair. How far did the mirror move? (b) Michelson pinned the metre to the red cadmium line, $\lambda_{\text{Cd}} = 643.847$ nm. How many wavelengths fit in a metre?

Step 1. Fringe count to displacement: from $N = 2\,\Delta d/\lambda$ , solve $\Delta d = N\lambda/2 = 1000 \times (589.3\ \text{nm})/2 = \mathbf{294.65\ \mu m}$ . A thousand fringes is only about a quarter of a millimetre of travel — which is why the method reaches nanometre precision: one countable fringe is a $295$ nm step, and fractional fringes are readable to a few nanometres.

Step 2. Run it backwards as metrology. Because each fringe is a fixed $\lambda/2$ , counting fringes while a stage moves reports its displacement in absolute wavelengths of light, needing no external ruler — the operating principle of the interferometric displacement gauges on every precision machine tool.

Step 3. The metre in cadmium wavelengths: the number of half-wavelengths (fringes) in a metre is $2/\lambda_{\text{Cd}}$ , and the number of whole wavelengths is $1/\lambda_{\text{Cd}} = 1/(643.847\times10^{-9}\ \text{m}) = \mathbf{1{,}553{,}164}$ . Michelson and Benoît reported $1{,}553{,}163.5$ wavelengths of the cadmium red line to the metre in 1892–93.

Step 4. Why it mattered: this fixed a length standard to an atomic constant a lamp anywhere could reproduce, decades before the 1960 krypton-86 metre and the 1983 speed-of-light metre. 1000 fringes = 294.65 µm; one metre ≈ 1.553 × 10⁶ cadmium wavelengths.

2 · Reading the sodium doublet from a visibility beat (exam level)

The "sodium D line" is actually two lines, $D_2 = 588.995$ nm and $D_1 = 589.592$ nm, separated by $\Delta\lambda = 0.597$ nm. As you scan the mirror, the fringe contrast fades to zero, revives, fades again. Find the path-difference period of these visibility beats, and the mirror travel between successive revivals.

Step 1. Each line makes its own cosine fringe pattern; the two drift out of and back into step as $\Delta$ grows. The combined visibility is $V(\Delta) = |\cos(\pi\,\Delta\,\Delta\lambda/\lambda^2)|$ , which returns to full contrast each time the two patterns realign. The visibility is the interference of the two interferograms.

Step 2. Full revival occurs when $\Delta\,\Delta\lambda/\lambda^2$ advances by one, so the beat period in path difference is $\Delta_{\text{beat}} = \lambda^2/\Delta\lambda = (589.3\ \text{nm})^2/(0.597\ \text{nm}) = \mathbf{0.582\ mm}$ .

Step 3. Convert to mirror travel: since $\Delta = 2\,\Delta d$ , the mirror moves $\Delta_{\text{beat}}/2 = \mathbf{0.291\ mm}$ between successive revivals of the fringes, with a first blackout at half that.

Step 4. Turn it into a measurement: reading the beat period gives the doublet splitting, $\Delta\lambda = \lambda^2/\Delta_{\text{beat}}$ — Michelson used exactly this fringe-visibility method to resolve fine structure that his gratings could not, and it is the seed of Fourier-transform spectroscopy. Beat period 0.582 mm in path, 0.291 mm in mirror travel.

Physics of the portrait

The instrument and the man are inseparable: Michelson spent half a century refining ways to measure light against itself, and the interferometer was the tool he kept returning to — for the ether, for the metre, for the diameter of a star.

Origins · 1 minAlbert Michelson and how this was discovered

Albert Michelson

1852–1931 · precision optics, Chicago & Mount Wilson

Albert Abraham Michelson was born in Strelno, Prussia (now Strzelno, Poland) and emigrated as a child to the mining towns of California and Nevada. Appointed to the US Naval Academy, he first made his name measuring the speed of light with rotating mirrors, then in 1881, on a fellowship in Potsdam, built the first interferometer to hunt for the Earth's motion through the luminiferous ether. With Edward Morley he refined the experiment in Cleveland in 1887; its famous null result — no fringe shift as the apparatus rotated — became a cornerstone of the road to special relativity, though Michelson himself remained wary of abandoning the ether. In 1892–93, working at the International Bureau of Weights and Measures near Paris, he counted the metre against the red cadmium wavelength, fixing a length standard to an atomic constant. In 1907 he became the first American awarded a Nobel Prize in the sciences, "for his optical precision instruments and the spectroscopic and metrological investigations carried out with their aid." In 1920, with Francis Pease, his stellar interferometer on Mount Wilson resolved the disc of Betelgeuse — the first measurement of a star's angular diameter, about $0.047$ arcseconds. A famously exacting experimentalist, he chased the speed of light to the end of his life.

History

1881

— Michelson builds the first interferometer in Potsdam and finds no ether drift

1887

— Michelson and Morley perform the refined ether-drift experiment in Cleveland; the null result stands

1891

— Michelson develops the visibility method to study spectral line structure

1892

— Michelson and Benoît measure the metre in cadmium red wavelengths at Sèvres

1907

— Michelson receives the Nobel Prize in Physics, the first American science laureate

1920

— Michelson and Pease measure the angular diameter of Betelgeuse with a stellar interferometer

2015

— LIGO, a kilometre-scale Michelson, detects gravitational waves from merging black holes

Going further

Texts. Hecht, Optics, 5th ed., §9.4 (the Michelson interferometer) · Born & Wolf, Principles of Optics, §7.5 · Loudon, The Quantum Theory of Light, 3rd ed., Ch. 3 (coherence) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 12 (statistical optics) · Demtröder, Laser Spectroscopy, Vol. 1, §4.2 (interferometers).

Provenance. The fringe law $I = 2I_0[1 + \cos(2\pi\Delta/\lambda)]$ , the round-trip $\Delta = 2(d_2-d_1)$ giving $\lambda/2$ per fringe, and the equal-inclination ring condition $2d\cos\theta = m\lambda$ are standard (Hecht §9.4); the visibility relation $V = |\gamma(\tau)|$ with $\tau = \Delta/c$ and the Wiener–Khinchin route to the spectrum are from Loudon Ch. 3 and Saleh & Teich Ch. 12; the resolving-power limit $\Delta\lambda = \lambda^2/(2\Delta_{\max})$ from Demtröder §4.2. Historical values — the 1892–93 cadmium metre ( $1{,}553{,}163.5$ wavelengths), the 1907 Nobel, the 1920 Betelgeuse angular diameter ( $\sim 0.047''$ ), and the LIGO strain scale — are the accepted figures. Every number is recomputed in verify_aq-optics-michelson-interferometer.py.

Prerequisites. Superposition Principle of Electromagnetic Waves · Two-Wave Interference and the Interference Term · Optical Path Difference and Phase Retardation · Wavefront-Splitting and Amplitude-Splitting Interferometers. Leads to. Temporal Coherence and Coherence Length · Fringe Visibility and Contrast · Fringes of Equal Inclination and Equal Thickness · Thin-Film Interference.

Why is the path difference 2(d₂ − d₁) and not just d₂ − d₁?

Because each arm is traversed twice. Light leaves the beamsplitter, travels to the mirror, and comes all the way back before recombining, so the extra distance in the longer arm is counted on the way out and again on the way back. That is why one interference fringe corresponds to a mirror displacement of $\lambda/2$ , not $\lambda$ : moving the mirror $\Delta d$ lengthens the optical path by $2\,\Delta d$ . It is the same round-trip factor that makes the metre come out to about $1.55$ million cadmium wavelengths.

What exactly does the compensating plate do, and why is it only needed for white light?

The beamsplitter is a glass slab silvered on one face, so one beam passes through the glass three times and the other only once. The compensating plate is an identical unsilvered slab placed in the second arm to equalise the glass thickness each beam sees. For monochromatic light this only shifts the fringe order, which does not matter. But glass is dispersive: different wavelengths acquire different phases, so with white light an unequal glass path smears the coloured fringes and destroys the single black-and-white zero-order fringe you need to locate exactly equal optical paths. The compensator restores it.

If the output can go completely dark, where does the light go — is energy lost?

No. A Michelson has two output ports: one back toward the source and one toward the detector. When the detector port goes dark, all the light is emerging from the other port. The beamsplitter imposes a relative $\pi$ phase between the two ways the beams can recombine, so the two ports are always complementary — bright at one means dark at the other, and their sum is constant. Energy is conserved exactly; it is merely redistributed between the ports as you scan the mirror.

Why do the circular fringes collapse into the centre as I shorten one arm?

The rings obey $2d\cos\theta = m\lambda$ with the highest order $m_0 = 2d/\lambda$ at the centre. As you decrease the effective film thickness $d$ , the central order $m_0$ drops. Each time it falls by one, the central spot cycles from bright to dark and a whole ring shrinks to a point and disappears at the centre — one ring swallowed per $\lambda/2$ of mirror travel. Watching rings vanish at the centre is itself a way to count fringes and measure the mirror displacement.

How is this the same instrument as LIGO?

LIGO is a Michelson interferometer at extreme scale: a beamsplitter sends laser light down two perpendicular four-kilometre arms and recombines it at a photodetector held near a dark fringe. A passing gravitational wave stretches one arm and squeezes the other, changing the path difference and shifting the fringe. To boost sensitivity each arm holds a Fabry–Pérot cavity that makes the light bounce hundreds of times (extending the effective path), and the circulating power and quantum-squeezed light beat down the noise — but the core is exactly the amplitude-splitting, path-difference-reading geometry Michelson built in 1881.

Why does the fringe pattern fade as the arms become very unequal?

Because real light has a finite coherence length. The two arms deliver time-delayed copies of the same field, and interference requires the two copies to still be correlated. Beyond the coherence time $\tau_c \sim 1/\Delta\nu$ (a path difference of the coherence length $\ell_c = c\tau_c$ ) the copies have lost their phase relationship, and the fringe visibility $V = |\gamma(\Delta/c)|$ drops toward zero. A narrow-band laser keeps fringes over metres; a broadband lamp loses them within millimetres. The visibility-versus-path curve is therefore a direct measurement of the source's spectral width.

Can a Michelson measure a refractive index or a gas concentration?

Yes — insert the sample in one arm. A slab of index $n$ and thickness $t$ adds $2(n-1)t$ to that arm's optical path, shifting the fringes by $N = 2(n-1)t/\lambda$ . Counting the fringes that sweep by as a gas cell is evacuated, for instance, gives $n$ to parts per million, since even $n - 1 \sim 3\times10^{-4}$ for air produces hundreds of countable fringes over a few centimetres. This interferometric refractometry is a standard way to measure gas indices, pressures, and flow.