Michelson fringes measure doubled paths
Relate mirror motion to optical delay and identify where different fringe families are observed.
On this page
The question for today
What you’ll understand
- Convert mirror displacement to round-trip optical path change.
- Distinguish equal-inclination and equal-thickness fringes.
- Explain broadband fringe localization near matched paths.
A Michelson interferometer splits a field into two arms, reflects each beam, and recombines delayed copies at a detector. The mirror travel is only half the corresponding round-trip path change in air. That factor is the starting point for displacement measurements, but source bandwidth, viewing plane, and mirror alignment decide which fringes are visible. We will separate the central-ray counting rule from the spatial pattern and its coherence envelope.
The round trip supplies the factor of two
Let each arm length be measured from the beamsplitter to its mirror. In the simplest air-path geometry the returning fields have traveled twice those lengths, so their optical path difference is twice the arm-length difference. Moving one mirror along its arm by d changes the optical path difference by 2d. The phase shift is the optical path change divided by the vacuum wavelength, multiplied by a full cycle in radians.
Beamsplitter reflection phases can add a constant offset, and transparent plates can add unequal material dispersion. These affect the location of a chosen bright fringe, especially for broadband light. They do not remove the round-trip factor for a translating mirror in a uniform air arm. In a medium of constant refractive index n, the corresponding path increment is 2nd at normal incidence. A varying or dispersive index must be included through optical path rather than mechanical distance alone.
Alignment determines the fringe family
The returning beams can be described using the virtual images of the two mirrors. When those effective mirror planes are parallel, rays with the same inclination have equal path difference. A lens maps inclination to radius in its focal plane, producing equal-inclination rings. For effective separation h, the ideal angular path difference is 2h cos θ. Expanding the cosine at small angle explains why a broad beam can contain several rings even when its central ray is carefully adjusted.
A slight tilt between the effective mirror planes instead creates an air-wedge geometry. Near the mirror-image plane, local effective separation varies with transverse position, producing equal-thickness fringes that are approximately straight for a small uniform tilt. Saying fringes are localized means their contrast is concentrated near a particular observation plane for an extended source. It does not mean interference exists only on a material surface or that the same viewing plane suits every alignment.
Broadband light selects nearly equal paths
For a broad spectrum, different wavelengths usually place maxima at different mirror settings. At a large delay their interference terms average toward zero. Near matched optical paths they agree sufficiently to produce a visible envelope, often with a central white-light region and colored neighbors. This temporal localization in delay is distinct from spatial fringe localization in an observation plane. Equal geometric arms alone may fail to match the group delays if one arm traverses extra dispersive glass.
The lab can test the phase part of a monochromatic Michelson scan. Reset and set both controls to phase zero degrees and ratio one, then predict constructive addition. Reset again and set phase 180 degrees and ratio one, predicting cancellation; this phase change corresponds to a mirror shift of one quarter wavelength in the ideal air-arm model. Both component waves in the instrument travel in the same direction. It has no mirrors, angular rings, plate dispersion, or bandwidth envelope.
Work it through
Count a calibrated mirror translation
A Michelson interferometer in air uses wavelength 600 nm. One mirror moves 3.0 micrometres along its arm. Find the optical path change, phase change, and number of complete fringe cycles. Assume stable monochromatic illumination.
Double the mechanical travel because the light goes to the mirror and back. The optical path difference changes by 6.0 micrometres.
Divide 6.0 micrometres by 0.600 micrometres to obtain ten wavelengths of path change. Multiply by 2π to obtain phase change 20π radians.
Each full phase cycle moves one bright-to-bright fringe cycle through a fixed observation point. The count is ten; the direction of phase change depends on which arm grows.
The path difference changes by 6.0 micrometres, corresponding to 20π radians and ten complete fringes.
The living notebook
Explore the model
Interactive lab · waves
Convert mirror travel into phase
Reset and set BOTH controls to phase 0° and ratio 1. For a separate run Reset and set phase 180° and ratio 1. Predict the change from constructive to destructive addition, corresponding to λ₀/4 mirror travel.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe phase mapping is useful, but this fixed-frequency lab does not calculate fringe localization or white-light envelopes.
Pause & reason
Check your understanding
Your turn
Try it yourself
With wavelength 500 nm, a mirror displacement produces four complete fringe cycles. Find its travel. Explain whether the count determines the sign of displacement without observing fringe motion or recording phase.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
A complete fringe is a full wavelength of round-trip optical path change.
Compare with the solution
Four cycles mean 2000 nm of optical path change, so mirror travel has magnitude 1000 nm=1.00 micrometre. An unsigned count alone does not say which arm grew or shrank. A direction-sensitive phase measurement or observed motion with a known geometry is needed to assign the sign.
Keep the idea
Take it with you
- Normal mirror travel changes air-arm path difference by twice the travel.
- Equal-inclination and equal-thickness fringes use different observation geometries.
- Spatial localization and a temporal coherence envelope describe different restrictions.
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