Interferometers

Two-beam interference gave you fringes from a fixed double slit or a fixed thin film — pretty, but not adjustable. An interferometer turns that same physics into an instrument you can dial: split one beam into two paths with a beamsplitter, send each down its own arm to a mirror, and recombine them so that moving a mirror by a fraction of a wavelength visibly shifts the fringes. The Michelson interferometer is the archetype and the sole quantitative focus of this module — Fabry–Pérot and other multiple-beam devices are out of scope here. By the end you should be able to sketch the Michelson layout, derive its path-difference formula, connect a mirror displacement to a fringe count in either direction — measuring length given wavelength, or measuring wavelength given a calibrated displacement — and explain why an extended source produces circular fringes.

The core insight

A Michelson interferometer splits light into two arms with a beamsplitter and recombines it after reflection off two mirrors, so moving one mirror by $\delta L$ changes the round-trip path difference by $2\,\delta L$ and sweeps $N=2\,\delta L/\lambda$ fringes past the detector.

The core idea

Construction. A Michelson interferometer has a monochromatic source, a beamsplitter (a half-silvered mirror) set at $45^\circ$ to the incoming beam, two fully-reflecting plane mirrors $M_1$ and $M_2$ placed along the two arms the beamsplitter creates, and a detector. The beamsplitter divides the incoming beam in two: one part travels to $M_1$ and back, the other to $M_2$ and back. Both return to the beamsplitter, recombine, and travel on to the detector, where they interfere.

Path difference between the arms. Let the two arm lengths, beamsplitter to mirror, be $L_1$ and $L_2$ . Each beam makes a round trip along its arm, so the optical path difference between the recombined beams is

\Delta = 2(L_1 - L_2)

where: $\Delta$ is the path difference at the detector and $L_1, L_2$ are the beamsplitter-to-mirror distances of the two arms. A compensating plate is placed in the shorter arm in practice, purely to equalize dispersion through the beamsplitter glass for white-light use; it does not enter this formula. The interference conditions are the familiar ones from two-beam interference generally — bright centre when $\Delta = m\lambda$ , dark when $\Delta = (m+\tfrac12)\lambda$ — now controlled by mechanically sliding a mirror rather than by fixed slit or film geometry.

Fringe shift on moving a mirror. Move $M_2$ by a small distance $\delta L$ along the beam. Because the light traverses that arm twice, the path difference changes by $2\,\delta L$ . Each time $\Delta$ changes by one full wavelength, exactly one fringe sweeps past a fixed point in the pattern, passing through one dark state on the way from one bright state to the next. Counting $N$ fringes as the mirror moves gives the working relation of this entire topic, derived below.

Circular fringes and alignment. With an extended source, the fringes usually appear as concentric circles, the same visual family as Newton's rings, because rays leaving the source at slightly different angles pick up slightly different path differences. When $M_1$ and the virtual image of $M_2$ formed by the beamsplitter are exactly parallel, the circular pattern is centred on the optical axis; tilting a mirror breaks that circular symmetry and the fringes distort or become nearly straight lines localized near the mirror.

Figure

A Michelson interferometer with a source beam hitting a beamsplitter that sends light to mirror one and mirror two, both beams returning and recombining toward a detector where fringes are shown

Derivation AFringe count from a mirror displacement

The goal is to connect a small, physically produced mirror displacement $\delta L$ in one arm of a Michelson interferometer to the integer number of fringes $N$ that sweep past a fixed point on the detector, so that the relation can be read in either direction: measuring an unknown displacement from a counted fringe number and a known wavelength, or measuring an unknown wavelength from a counted fringe number and a precisely calibrated displacement.

Step 1. Start from the path-difference formula. With arm lengths $L_1$ and $L_2$ , the path difference is $\Delta = 2(L_1-L_2)$ . Hold $M_1$ and $L_1$ fixed and let only $M_2$ move, so $L_2 \to L_2+\delta L$ .

Step 2. Find the change in path difference. The new path difference is $\Delta' = 2(L_1 - L_2 - \delta L) = \Delta - 2\,\delta L$ , so moving the mirror by $\delta L$ changes the path difference by $2\,\delta L$ in magnitude — the factor of 2 is the round trip, since the light travels to the mirror and back through the same displaced distance.

Step 3. Apply the fringe condition. Successive bright fringes at the centre of the pattern occur whenever $\Delta$ increases by exactly one wavelength $\lambda$ . If the path difference change $2\,\delta L$ corresponds to $N$ such full-wavelength steps, then

2\,\delta L = N\lambda.

Step 4. Solve for whichever quantity is unknown. If $\lambda$ is known and $N$ is counted, $\delta L = N\lambda/2$ gives the displacement to sub-micron precision because $\lambda$ itself is only hundreds of nanometres. If instead $\delta L$ is set by a calibrated micrometer screw and $N$ is counted, $\lambda = 2\,\delta L/N$ gives the wavelength — this is the working principle behind using a Michelson interferometer as a length gauge or as a wavelength meter. Check: setting $N=1$ gives $\delta L = \lambda/2$ , meaning one full fringe cycle corresponds to moving the mirror by only half a wavelength — a useful sanity figure to remember for the common traps below.

Worked example

A Michelson interferometer is illuminated by a helium–neon laser, $\lambda = 632.8\ \text{nm}$ . Mirror $M_2$ is moved through $\delta L = 0.1\ \text{mm}$ on a precision translation stage.

Step 1. Apply $2\,\delta L = N\lambda$ , so $N = 2\,\delta L/\lambda$ .

Step 2. Substitute: $N = \dfrac{2(0.1\times10^{-3})}{632.8\times10^{-9}} = \dfrac{2.0\times10^{-4}}{6.328\times10^{-7}} \approx 316$ .

About 316 fringes sweep past during the 0.1 mm displacement. A second student runs the reverse experiment: moving the mirror by a carefully calibrated $\delta L = 25.00\ \mu\text{m}$ , they count $N = 79$ fringes and want the source wavelength, plus the uncertainty if the fringe count is only good to $\pm1$ .

Step 3. Use $\lambda = 2\,\delta L/N = \dfrac{2(25.00\times10^{-6})}{79} = \dfrac{5.000\times10^{-5}}{79} \approx 6.329\times10^{-7}\ \text{m} = 632.9\ \text{nm}$ , consistent with a He–Ne laser.

Step 4. Since $\lambda \propto 1/N$ with $\delta L$ held exact, $\dfrac{\Delta\lambda}{\lambda} = \dfrac{\Delta N}{N} = \dfrac{1}{79} \approx 1.3\%$ .

The wavelength is 632.9 nm, with about a 1.3 percent fractional uncertainty from a plus-or-minus-one fringe counting error — precision work uses as large an $N$ as practical to dilute this fixed counting error.

Common traps

Forgetting the round-trip factor of 2 — the path-difference change is $2\,\delta L$ , not $\delta L$ , because light travels to the mirror and back in that arm.
Confusing "fringes counted" with "wavelengths of mirror motion" — one full wavelength of path-difference change corresponds to only $\delta L = \lambda/2$ of actual mirror motion, half as much.
Treating the Michelson interferometer as different physics from Young's double slit — it is still ordinary two-beam interference, with the two mirror images (formed via the beamsplitter) playing the role of the two slits.
Overstating the compensating plate's role — it balances dispersion in glass between the arms so white-light fringes stay visible; it does not change the basic $\Delta = 2(L_1-L_2)$ relation for monochromatic light.

Ask the tutor: if you want to see how this fringe-shift idea historically underpins the Michelson–Morley experiment, or want the compensating plate's role worked through in more detail, ask the AI tutor to walk through it.

Why is it 2 delta L and not just delta L in the fringe formula?

Because light in a Michelson interferometer does not just travel to the mirror once — it reflects and comes straight back along the same arm to the beamsplitter. If you slide the mirror away by $\delta L$ , the light in that arm now has to cover an extra $\delta L$ going out and another $\delta L$ coming back, so the round-trip path length grows by $2\,\delta L$ . That factor of 2 is purely geometric, not something special to interference, and it is the single most common place marks are lost in this topic.

Is the Michelson interferometer really the same physics as Young's double slit?

Yes, and recognizing this is the fastest way to feel confident with the topic. Both are two-beam interference: two coherent beams travel different path lengths and recombine, producing bright and dark fringes according to whether the path difference is a whole or half number of wavelengths. In Young's experiment the two beams originate from two physical slits; in the Michelson interferometer the beamsplitter creates two virtual sources — the real mirror $M_1$ and the virtual image of $M_2$ seen through the beamsplitter — and the same $\Delta = m\lambda$ and $\Delta=(m+\tfrac12)\lambda$ conditions apply.

Why do we bother counting hundreds of fringes instead of just measuring delta L directly with a ruler?

Because a ruler cannot resolve motion on the scale of a wavelength of light, a few hundred nanometres, while fringe counting can. Each fringe transition corresponds to a mirror motion of only $\lambda/2$ , so counting fringes turns an easily observed, discrete event, a bright spot becoming dark and bright again, into a measurement with sub-micron precision. This is exactly why interferometers, not rulers or verniers, are used whenever displacement or wavelength must be pinned down to a fraction of a micron.

Interferometers

The core idea

Worked example

Common traps

Before this

Where this leads