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Physics III course sequence · Interactive lesson

The Michelson Interferometer & Optical Metrology

Split one beam into two copies, send them out and back along separate arms, and let them recombine: every half-wavelength of mirror travel moves the pattern by one fringe. Here that count becomes a displacement, a refractive index and a spectral line separation, and you learn where the counting stops: the coherence length.

UPIII-05.04 · Lesson 4 of 9 in this unit. Position is not a completion score.
Starting model: The Michelson Interferometer & Optical MetrologyOutput intensity I/I₀ against mirror displacement d for a source of wavelength 633 nm and spectral width 100 nm. Fringes repeat every λ/2 = 0.317 μm of travel. The dashed envelope is the visibility of a flat band, which first vanishes where 2d reaches the coherence length l_c = λ²/Δλ = 4.0 μm. The dot marks d = 0.95 μm, N = 3.0 fringes from zero.I / I₀ at the output portλ = 633 nm, Δλ = 100 nmdashed: envelope ±V, lc = λ²/Δλ = 4.0 μmdot: d = 0.95 μm → N = 2d/λ = 3.0 fringes10−20+2mirror displacement d / μm (path difference 2d)
Start with a picture. Change the model in the lab below.
1

Build the model

A length is measured by counting fringes against a wavelength.

The key idea

A Michelson interferometer divides a beam's amplitude at a beam splitter, returns the two halves from separate mirrors and reads the path difference 2d between the arms as a fringe pattern that shifts by one fringe for every λ/2 of mirror travel.

Example. With He-Ne light at 632.8 nm, driving the mirror 0.100 mm sends N = 2 × 0.100 mm ÷ 632.8 nm = 316 fringes past the reference mark: a length of 100 μm measured by counting, against a wavelength known to a part in 10⁶.

The complete physical model and assumptions

Split a wave's amplitude in two at a half-silvered plate, send the halves out and back along two arms, and recombine them: the output holds two copies of one wave with path difference 2d, twice the arm-length difference. Their superposition gives I = (I₀/2)(1 + cos 4πd/λ), so the output cycles through a full fringe each time d changes by λ/2, and the light missing from a dark fringe is not lost but sent back toward the source. That cosine is the whole instrument. Read it forwards and a count of fringes N = 2Δd/λ is a length measured against the wavelength; read it with a gas cell in one arm and N = 2L(n − 1)/λ is a refractive index; read it with two wavelengths at once and the fringe contrast beats with mirror period λ²/(2Δλ), a spectral line separation. The model assumes one wavelength and two clean beams. A real source of bandwidth Δλ makes fringes whose orders drift apart as 2dΔλ/λ², and when that drift reaches one fringe, at the coherence length lc = λ²/Δλ, the pattern washes out: white light survives only two fringes either side of zero path difference, which is exactly why white-light fringes locate d = 0. Glass dispersion in the splitter would spoil even that, so a compensator plate equalises the glass in the two arms. And in a teaching laboratory none of these is the limit: thermal drift, vibration and mirror tilt move the fringes further than the wavelength ever will.

1Amplitude division: one wave, two copies, a virtual film

Young's slits divide a wavefront, sampling it at two points, so the source must be small or the two samples carry unrelated phases. The Michelson divides the amplitude instead: a half-silvered plate sends part of every wavefront down arm 1 and part down arm 2, and the mirrors return both parts to recombine. Each point of the lamp interferes only with its own light, so a broad source gives bright fringes instead of washing them out. Seen from the output, mirror M₂ appears as a virtual image M₂′ parallel to M₁ and a distance d behind it: the instrument is a virtual air film whose thickness d you drive with a screw. A ray leaving at angle θ to the axis returns with path difference 2d cos θ, so bright rings sit where 2d cos θ = mλ. The centre carries the highest order, m₀ = 2d/λ = 3663 for d = 1.00 mm and λ = 546 nm, and the p-th ring out from a bright centre sits near θₚ ≈ √(pλ/d), 1.34° for p = 1, because cos θ ≈ 1 − θ²/2 turns 2d(1 − θ²/2) = (m₀ − p)λ into dθ² = pλ.

2The output intensity, and where the missing light goes

Let the incident intensity be I₀ and the splitter 50:50. Each arm carries I₀/2 out and back, and each returning beam sends half of itself, I₀/4, to the output port. Two equal beams with phase difference δ superpose to I = (I₀/4)(1 + 1 + 2 cos δ) = (I₀/2)(1 + cos δ) = I₀ cos²(δ/2), with δ = (2π/λ)(2d) plus a constant fixed by the reflections. The output swings between I₀ and zero: the interferometer does not destroy energy, it steers it, and at a dark fringe the complementary I₀ sin²(δ/2) leaves through the input port toward the source. Two checks: 4πd/λ is a length over a length, so δ is a pure number as a phase must be; and at d = 0 every wavelength shares the same δ, which is the seed of the white-light fringe. Real beams are never exactly equal, so write I = (I₀/2)[1 + V cos δ] with the visibility V = (Iₘₐₓ − Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ). V = 1 is perfect contrast and V = 0 is no fringes; everything this instrument measures is δ changing by 2π while V stays high enough to see it.

3Counting fringes converts a length into an integer

Moving the mirror by Δd changes the path difference by 2Δd, so δ changes by 2π(2Δd)/λ and N = 2Δd/λ fringes cross a reference mark. Each fringe is λ/2 of mirror travel: 316.4 nm for He-Ne light at 632.8 nm. Drive the mirror 0.100 mm and 316 fringes pass; judge the last one to a quarter and you have measured 100 μm to ±80 nm, about a part in a thousand, with no ruler finer than a lamp. The identity runs backwards too: a calibrated screw and a counted N give λ = 2Δd/N, which is how wavelengths were first tied to the metre. Put a gas cell of length L in one arm and nothing moves; admitting gas raises the index from 1 to n and adds optical path 2L(n − 1), so N = 2L(n − 1)/λ. A 50.0 mm cell filled with room air under sodium light at 589.3 nm passes 2 × 0.0500 m × 2.72 × 10⁻⁴ ÷ 589.3 nm ≈ 46 fringes, and reading N to ±1 fixes n − 1 to about 2%.

4The compensator plate, and why white light needs it

The splitter is a glass plate of thickness t coated on one face. Follow the two beams: the one reflected at the coating first crosses the glass three times in all, the other once, so one arm carries extra optical path 2n(λ)t that the other does not. For a laser this is harmless: absorb the constant into δ and move the mirror. For white light it is fatal, because n depends on λ. In 6.0 mm of crown glass, n(486 nm) − n(656 nm) = 1.5224 − 1.5143 = 0.0081, so the extra path differs between blue and red by 2 × 6.0 mm × 0.0081 = 97 μm; no single mirror position is zero-path-difference for both colours, and coloured fringes never form. The compensator is a second plate of the same glass and thickness, set parallel to the splitter in the other arm, so both beams cross equal glass and the dispersion cancels for every λ at once. Only then does d = 0 mean zero path difference for the whole spectrum: one achromatic central fringe, unmistakable, with a few coloured fringes on either side.

5Coherence length caps the usable path difference

A real source spans a band of width Δλ about λ. Each wavelength makes fringes with its own spacing, and they all agree only at zero path difference. Track the order m = 2d/λ across the band: differentiating, |Δm| = 2dΔλ/λ². While Δm ≪ 1 the colours are in step and the pattern is sharp; when Δm reaches 1 the two edges of the band are a whole fringe apart, every colour between fills the gap, and the fringes vanish. The path difference where that happens is the coherence length lc = λ²/Δλ, and the number of fringes you can count before reaching it is lc/λ = λ/Δλ. White light, λ ≈ 550 nm and Δλ ≈ 300 nm, gives lc ≈ 1 μm and about two fringes each side of zero, which is exactly why white-light fringes locate d = 0 so well. An LED at 630 nm with Δλ = 20 nm allows 20 μm of path; a multimode He-Ne with Δλ ≈ 2 × 10⁻³ nm allows about 20 cm. The sharp cut-off is the flat-band idealisation: a Gaussian line fades smoothly with no revival, and a doublet fades and returns with mirror period λ²/(2Δλ), 0.29 mm for the sodium D lines.

6What actually limits a teaching interferometer

Write the uncertainty budget for a displacement and the wavelength comes last. Counting: ±1 fringe is ±λ/2 = ±0.32 μm, and a photodiode with a counter cuts that to a hundredth of a fringe. Calibration: a lever-and-micrometer drive quoted at 1% turns 0.250 mm of travel into ±2.5 μm, eight times the count error. Temperature: an aluminium arm of 120 mm (α = 23 × 10⁻⁶ K⁻¹) warming by 0.5 K lengthens by 1.4 μm, and the double pass makes that 4.7 spurious fringes at 589 nm, drift of the same order as the effect being measured. Vibration: a footfall that moves a mirror by λ/4 = 160 nm shifts the pattern by half a fringe, and building vibration of a few micrometres at a few hertz smears the fringes into grey unless the table is isolated and the count is fast. Alignment: a tilt ε between M₁ and M₂′ lays 2εD/λ straight fringes across an aperture D, so fewer than one fringe across 20 mm at 633 nm needs ε < λ/(2D) = 1.6 × 10⁻⁵ rad, about three arcseconds. The wavelength enters only as a multiplier, and a laser's part-in-10⁶ uncertainty is invisible beside any of these.

2

Key equations

Read the units, assumptions and limits before substituting.

Output intensity of a Michelson

I = (I₀/2)[1 + V cos δ]δ = 4πd/λ + δ₀

A 50:50 splitter steers the light rather than losing it: the output swings between I₀ and 0 while the complement returns toward the source. Moving a mirror changes δ, never δ₀.

I₀ incident intensity (W m⁻²); d arm-length difference (m); λ wavelength (m); V visibility, 0 to 1; δ₀ fixed by the reflections

Fringe count for a mirror displacement

N = 2Δd/λ

One fringe per λ/2 of travel, 316.4 nm for He-Ne light. Read forwards it measures Δd; read backwards with a calibrated screw it measures λ.

N fringes crossing a mark, not necessarily an integer; Δd mirror displacement (m); λ wavelength in the arm (m)

Refractive index from a gas cell

N = 2L(n − 1)/λ

Nothing moves: the optical path grows by 2L(n − 1) as gas enters the cell. Solve for n − 1 = Nλ/(2L) and read n − 1 to the precision of the count.

L cell length (m); n refractive index of the gas, dimensionless; λ vacuum wavelength (m); the 2 is the double pass

Rings of equal inclination

2d cos θ = mλ

The p-th ring out from a bright centre sits near θₚ ≈ √(pλ/d): shrink d and the rings swell, then drain into the centre one by one.

θ ray angle to the axis (rad); m integer order; the centre carries the largest order, m₀ = 2d/λ

Coherence length of a source

lc = λ²/Δλ

Where the band's fringe orders have spread by one, 2dΔλ/λ² = 1, the pattern washes out; you get about λ/Δλ fringes before it does.

lc largest usable path difference (m); Δλ spectral width of the source (m); the mirror travel to reach it is lc/2

Contrast beats from a doublet

Δdbeat = λ²/(2Δλ)

Two lines slip in and out of step as the mirror moves: 0.29 mm per beat for the sodium D lines, which is how Michelson measured their separation.

Δdbeat mirror travel between successive contrast minima (m); Δλ separation of the two lines (m); λ their mean (m)
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

633 nm
400 to 700 nm; steps of 1.
100 nm
1 to 300 nm; steps of 1.
0.95 μm
−2 to 2 μm; steps of 0.01.

Try this. Set Δλ to 1 nm and the envelope goes flat: a laser-like source keeps full contrast across the whole window, so N = 2d/λ can be counted anywhere. Push Δλ to 300 nm and only about λ/Δλ ≈ 2 fringes survive either side of d = 0: white-light fringes, which is why they locate zero path difference.

Live results

COHERENCE LENGTH λ²/Δλ
4.0 μm
FRINGES FROM ZERO 2d/λ
3.0 fringes
VISIBILITY AT d
0.67
USABLE FRINGES λ/Δλ
6.3 per side

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics III modelOutput intensity I/I₀ against mirror displacement d for a source of wavelength 633 nm and spectral width 100 nm. Fringes repeat every λ/2 = 0.317 μm of travel. The dashed envelope is the visibility of a flat band, which first vanishes where 2d reaches the coherence length l_c = λ²/Δλ = 4.0 μm. The dot marks d = 0.95 μm, N = 3.0 fringes from zero.I / I₀ at the output portλ = 633 nm, Δλ = 100 nmdashed: envelope ±V, lc = λ²/Δλ = 4.0 μmdot: d = 0.95 μm → N = 2d/λ = 3.0 fringes10−20+2mirror displacement d / μm (path difference 2d)
Output intensity I/I₀ against mirror displacement d for a source of wavelength 633 nm and spectral width 100 nm. Fringes repeat every λ/2 = 0.317 μm of travel. The dashed envelope is the visibility of a flat band, which first vanishes where 2d reaches the coherence length lc = λ²/Δλ = 4.0 μm. The dot marks d = 0.95 μm, N = 3.0 fringes from zero.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Centre wavelength λ: 633 nm; Spectral width Δλ: 100 nm; Mirror displacement d: 0.95 μm

Baseline 633; choose 400 to 700, in steps of 1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The first-year error is to count N fringes and write Δd = Nλ. One fringe is one wavelength of path difference, but the light crosses the moving arm twice, out to the mirror and back, so the path difference is 2d and one fringe is only λ/2 of mirror travel. A student who forgets the return trip reports every displacement at twice its true size and, running the identity backwards from a calibrated screw, every wavelength at half its true size. The same double pass sits inside the gas-cell formula: the extra optical path is 2L(n − 1), not L(n − 1), and dropping the 2 doubles n − 1. The check takes one line: a mirror moved through exactly λ/2 sends its beam a full λ further than before, and that is one fringe.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A Michelson interferometer is lit by a He-Ne laser at 632.8 nm. As a micrometer drives the moving mirror, 200 bright fringes cross the reference mark. Find the mirror displacement, and the smallest displacement you could detect if you can judge the pattern to a quarter of a fringe.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A cell of internal length 50.0 mm with plane windows sits in one arm of a Michelson interferometer lit by a sodium lamp, λ = 589.3 nm. It is pumped to a rough vacuum, then air is admitted slowly until it reaches room pressure, and 46 fringes cross the field. Find n − 1 for the air. The count is good to ±1 fringe and the length to ±0.2 mm; find the uncertainty in n − 1 and say which measurement to improve first.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

Under sodium light, a doublet of mean wavelength 589.3 nm, the fringes of a Michelson interferometer fade and recover as the mirror is driven. From one contrast minimum, the mirror is driven 2.910 mm to reach the tenth minimum after it. (a) Show that successive minima are separated by λ²/(2Δλ) and find Δλ. (b) The 120 mm aluminium arm (α = 23 × 10⁻⁶ K⁻¹) warms by 0.5 K during the run: how many spurious fringes is that, and does it threaten the result?

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

An LED at 620 nm with a spectral width of 25 nm lights a Michelson interferometer. Roughly how far can the moving mirror travel away from zero path difference before the fringes fade out?
7

Review & continue

A Michelson interferometer divides a beam's amplitude at a beam splitter, returns the two halves from separate mirrors and reads the path difference 2d between the arms as a fringe pattern that shifts by one fringe for every λ/2 of mirror travel.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.