Finesse, Free Spectral Range, and Resolving Power

A Fabry–Pérot resonator does something an ordinary interferometer cannot: it sorts light into transmission peaks so narrow that two colours differing by one part in a million stand cleanly apart. But a sharp instrument is only useful if you also know how far apart its peaks sit and how finely it can split a pair of lines — and those three questions have three clean answers. This article owns them. The sharpness is the finesse; the spacing is the free spectral range; the splitting power is the resolving power, and each is a compact formula you can read off from just the mirror reflectance and the plate separation. The remarkable thing is how cleanly the labour divides. The free spectral range is a clock: it is fixed by the round-trip time $2n\ell/c$ and nothing else, so the mirrors never touch it. The finesse is a mirror property: $\mathcal{F}=\pi\sqrt{R}/(1-R)$ , a pure function of the reflectance, blind to how far apart the plates sit. Multiply the finesse by the interference order and you get the resolving power, $\lambda/\delta\lambda=m\mathcal{F}$ , which is why a centimetre-thick etalon reaches $10^6$ where a hand-held grating struggles to reach $10^4$ . These are the figures of merit recalled by every later cavity — the grating spectrometer's own free spectral range and the laser cavity's mode spacing both come back to this page by name. Underneath all three sits the Airy transmission function, whose peaks this article inherits from aq-optics-multiple-beam-interference-airy and whose resonator it inherits from aq-optics-fabry-perot-interferometer.

Key takeaways

Three numbers grade a Fabry–Pérot resonator: how sharp its peaks are (finesse), how far apart they sit (free spectral range), and how finely it splits two wavelengths (resolving power).
Free spectral range is purely geometric: $\Delta\nu_{\text{FSR}}=c/(2n\ell)$ — set by the round-trip time, independent of the mirrors.
Finesse is purely a mirror property: $\mathcal{F}=\pi\sqrt{R}/(1-R)=\Delta\nu_{\text{FSR}}/\delta\nu_{1/2}$ — set by reflectance, independent of the spacing.
Resolving power is their product with the order: $\lambda/\delta\lambda=m\mathcal{F}$ , so thick, high-finesse etalons reach $10^6$ – $10^8$ — at the price of a small free spectral range.
Real etalons fall short of the reflectivity finesse: mirror flatness and aperture add reciprocally-in-square, and a $\lambda/100$ surface caps the finesse near 50 no matter how good the coatings.
Key idea: finesse (from reflectance) sets peak width, free spectral range (from length) sets peak spacing, and their product with the order, $\lambda/\delta\lambda=m\mathcal{F}$ , sets how finely the etalon resolves — the whole instrument in three formulas.

The core insight

For an ideal two-mirror cavity the finesse depends only on the reflectance, $\mathcal{F}=\pi\sqrt{R}/(1-R)$ ; the free spectral range depends only on the optical length, $\Delta\nu_{\text{FSR}}=c/(2n\ell)$ ; and their product with the interference order gives the resolving power, $\lambda/\delta\lambda=m\mathcal{F}$ . Sharpness, spacing, and discrimination — three independent knobs, three formulas.

Where this lives today

The three figures of merit on this page are the working vocabulary of modern precision optics. In a single-frequency laser, an intracavity etalon whose free spectral range exceeds the gain bandwidth and whose finesse rejects neighbouring longitudinal modes forces the laser onto one frequency; the same numbers, scaled up to a finesse of $10^{5}$ – $10^{6}$ in an ultra-stable reference cavity, provide the razor-thin discriminator that Pound–Drever–Hall locking uses to stabilise the lasers behind optical atomic clocks. High-resolution spectroscopy still leans on $\lambda/\delta\lambda=m\mathcal{F}$ reaching $10^{6}$ – $10^{8}$ to split hyperfine structure and isotope shifts, handling the small free spectral range by pre-dispersing the light with a grating or by piezo-scanning the plates. The concepts also scale to the very large and the very small. LIGO's arm cavities are Fabry–Pérot resonators kilometres long: their finesse sets the photon storage time $\tau_c=n\ell\mathcal{F}/(\pi c)$ , folding the light path hundreds of times to multiply the phase shift a passing gravitational wave imprints. At the other end, thin-film and fibre Fabry–Pérot filters select single channels in dense wavelength-division multiplexing, free spectral range matched to the channel grid and finesse setting the isolation — the same three numbers, from a spectroscopy bench to the backbone of the internet.

The physical picture

Picture a wave trapped between two highly reflecting mirrors, leaking a little through each pass. On resonance every round trip returns the light in step with itself, the trapped field piles up, and a strong beam leaks out the far side — a transmission peak. A hair off resonance the many partial waves fall out of step and cancel, and the cavity reflects almost everything. Crank the reflectance up and the light survives more round trips before escaping, more partial waves add into the sum, and the resonance grows narrower. That single dial — reflectance — controls the sharpness, and the finesse is just the number that measures it: it counts how many resolvable peaks fit between one order and the next.

Figure

Transmission comb of a Fabry–Pérot etalon: sharp Airy peaks spaced by the free spectral range Δν_FSR, each of width δν₁ᐟ₂ = FSR ⁄ 𝓕 — finesse is their ratio

Finesse has a beautifully concrete meaning: up to a factor of $2\pi$ it is the number of round trips a photon makes before it leaks out — the storage number. A finesse of $300$ means the light bounces roughly $300/\pi\approx95$ times, sampling the mirror separation that many times over, which is exactly why such a cavity measures length (or frequency) so precisely. The free spectral range, by contrast, is not a ruler but a clock: fixed by the round-trip time $2n\ell/c$ , it is untouched by the mirrors. Change the coatings and you change sharpness but never spacing; change the spacing and you change spacing and resolving power but never the finesse. That clean separation is what makes the etalon such an intuitive instrument — and it is also the source of its one hard tradeoff, drawn next.

Figure

The punchline: finesse 𝓕 = π√R ⁄ (1−R) diverges as R → 1 — reflectance 0.90 gives 𝓕 ≈ 29.8, 0.95 gives 61.2, 0.99 gives 313

The curve tells the whole story of high-reflectance cavities. Finesse is nearly flat until $R\approx0.9$ , then it turns almost vertical: the last percent of reflectance is where all the performance lives, because $\mathcal{F}\to\pi/(1-R)$ near $R=1$ . Push $R$ from $0.95$ to $0.99$ and the finesse leaps from $61$ to $313$ ; push it to $0.9999$ and you reach the tens of thousands used to stabilise lasers. This is why the reference cavities behind optical clocks obsess over the last fraction of a percent of mirror loss — and why, past a point, real cavities stop being limited by reflectance at all and start being limited by how flat their mirrors are.

The core equations

Everything flows from the Airy transmission function of the resonator — the shape this article inherits from aq-optics-multiple-beam-interference-airy. It is the intensity transmitted as a function of the round-trip phase $\delta$ , and its peaks are the resonances whose width and spacing we are about to name.

T(\delta)=\frac{1}{1+\left(\dfrac{2\mathcal{F}}{\pi}\right)^{2}\sin^{2}\!\left(\dfrac{\delta}{2}\right)},\qquad \left(\frac{2\mathcal{F}}{\pi}\right)^{2}=\frac{4R}{(1-R)^{2}} \qquad (1)

where: $T$ — fractional transmitted intensity (unity on resonance for lossless mirrors); $\delta$ — round-trip phase, $\delta=(4\pi n\ell\cos\theta)/\lambda$ ; $\mathcal{F}$ — the finesse; $R$ — intensity reflectance of each mirror. The quantity $4R/(1-R)^2$ is the coefficient of finesse; it, and hence the whole peak shape, is fixed by $R$ alone. The free spectral range answers the spacing question. Transmission maxima recur every time the round-trip phase advances by $2\pi$ , and since $\delta$ is proportional to frequency, those maxima are equally spaced in frequency. Converting one full $2\pi$ of phase into a frequency interval gives a spacing set entirely by the optical length.

\Delta\nu_{\text{FSR}}=\frac{c}{2n\ell},\qquad \Delta\lambda_{\text{FSR}}=\frac{\lambda^{2}}{2n\ell} \qquad (2)

where: $c$ — vacuum speed of light; $n$ — index of the medium between the mirrors; $\ell$ — mirror separation; $n\ell$ — one-way optical path. Notice what is absent: no reflectance. Two etalons of the same length have the same free spectral range whether their mirrors are dull or nearly perfect. The finesse answers the sharpness question. It is defined as the ratio of the free spectral range to the full width at half maximum of a single peak, and for ideal mirrors that ratio collapses to a pure function of the reflectance — the curve plotted above.

\mathcal{F}\equiv\frac{\Delta\nu_{\text{FSR}}}{\delta\nu_{1/2}}=\frac{\pi\sqrt{R}}{1-R} \qquad (3)

where: $\delta\nu_{1/2}$ — full width at half maximum of a transmission peak; $\mathcal{F}$ — finesse. Read it two ways: as a definition (spacing over width, a dimensionless count of resolvable peaks per order) and as a formula (what that count equals for reflectance $R$ ). The resolving power answers the discrimination question by combining the other two. The smallest resolvable wavelength difference is one linewidth, and dividing the wavelength by it turns out to be the finesse multiplied by the interference order.

\frac{\lambda}{\delta\lambda}=m\,\mathcal{F},\qquad m=\frac{2n\ell\cos\theta}{\lambda} \qquad (4)

where: $m$ — interference order (an integer at a peak, counting how many half-waves fit in the round trip); $\delta\lambda$ — minimum resolvable wavelength difference; $\lambda/\delta\lambda$ — chromatic resolving power. Because $m$ can be tens of thousands, even a modest finesse of $60$ buys a resolving power in the millions.

Derivations

Derivation AFinesse as the ratio of free spectral range to linewidth

The goal is to turn the definition of finesse — spacing divided by width — into the reflectance formula $\mathcal{F}=\pi\sqrt{R}/(1-R)$ , and in doing so to show that the coefficient buried in the Airy function is not an arbitrary abbreviation but is precisely the peaks-per-order count. We start from the Airy transmission $T(\delta)$ inherited from aq-optics-multiple-beam-interference-airy, find the phase width of a peak at half its height, and convert that phase width into a frequency width using the free spectral range as the phase-to-frequency dictionary.

Step 1. Sit at a transmission maximum, where $\delta=2\pi m$ , and write nearby phases as $\delta=2\pi m+\phi$ with $\phi$ small. Then $\sin(\delta/2)=\sin(\pi m+\phi/2)=\pm\sin(\phi/2)\approx\pm\phi/2$ , so near the peak the Airy function becomes $T\approx 1/\bigl[1+(2\mathcal{F}/\pi)^{2}(\phi/2)^{2}\bigr]$ . Why: a sharp peak is entirely shaped by its immediate neighbourhood, and the small-angle form of the sine is the whole content of "sharp".

Step 2. Find the half-power points by setting $T=\tfrac12$ , which requires $(2\mathcal{F}/\pi)^{2}(\phi/2)^{2}=1$ . Solving gives $\phi/2=\pm\pi/(2\mathcal{F})$ , so the full phase width between the two half-power points is $\phi_{1/2}=2\pi/\mathcal{F}$ . Why: half maximum is where the denominator has doubled, and the two symmetric solutions bracket the peak — their separation is the linewidth in phase units.

Step 3. Convert phase width to frequency width. One full free spectral range corresponds to a phase advance of exactly $\Delta\delta=2\pi$ , so a phase interval $\phi_{1/2}$ maps to a frequency interval $\delta\nu_{1/2}=(\phi_{1/2}/2\pi)\,\Delta\nu_{\text{FSR}}=\Delta\nu_{\text{FSR}}/\mathcal{F}$ . Why: the map from phase to frequency is linear and its scale is fixed by "one order = $2\pi$ = one FSR", the only conversion factor available.

Step 4. Read off the definition and the formula together. The ratio $\Delta\nu_{\text{FSR}}/\delta\nu_{1/2}=\mathcal{F}$ is the definition; and because the coefficient of finesse in the Airy function is $(2\mathcal{F}/\pi)^2=4R/(1-R)^2$ , the same $\mathcal{F}$ equals $\pi\sqrt{R}/(1-R)$ . Why: the number that sets the peak's width in the Airy function must be the number that equals spacing-over-width — self-consistency ties the shape parameter to the reflectance. *Verified symbolically ( $4R/(1-R)^2=(2\mathcal{F}/\pi)^2\Rightarrow\mathcal{F}=\pi\sqrt{R}/(1-R)$ ) and numerically ( $\mathcal{F}=29.8,61.2,312.6$ at $R=0.90,0.95,0.99$ ) in verify_aq-optics-finesse-free-spectral-range.py.*

Go deeper · Derivation B · 1 minFree spectral range from the round-trip phase

Derivation BFree spectral range from the round-trip phase

The aim here is to pin the peak spacing to the geometry alone and prove that the free spectral range carries no imprint of the mirrors. The engine is the round-trip phase $\delta$ , which is what the light accumulates on one there-and-back pass between the plates; resonances occur whenever it is a whole multiple of $2\pi$ , so the spacing between resonances is whatever frequency change advances $\delta$ by exactly $2\pi$ .

Step 1. Write the round-trip phase in terms of frequency at normal incidence ( $\cos\theta=1$ ): $\delta=(4\pi n\ell/\lambda)=(4\pi n\ell/c)\,\nu$ . Why: expressing $\delta$ through $\nu$ rather than $\lambda$ makes the peak spacing come out directly in the frequency units a spectrometer actually reads.

Step 2. Demand that adjacent maxima differ by one full turn of phase, $\Delta\delta=2\pi$ . Substituting the linear relation gives $(4\pi n\ell/c)\,\Delta\nu_{\text{FSR}}=2\pi$ . Why: successive transmission peaks are, by construction, one order apart, and one order is one $2\pi$ of accumulated round-trip phase.

Step 3. Solve for the frequency spacing: $\Delta\nu_{\text{FSR}}=c/(2n\ell)$ ; equivalently, since $\Delta\nu=\, -\,(c/\lambda^{2})\,\Delta\lambda$ , the wavelength spacing is $\Delta\lambda_{\text{FSR}}=\lambda^{2}/(2n\ell)$ . Why: only the optical length $n\ell$ survives — the reflectance dropped out at Step 1, so spacing is geometry and clock, never mirror quality. *Verified: $\Delta\nu_{\text{FSR}}=c/2n\ell=14.99$ GHz for a $1$ cm air gap, and the thickness-scaling ( $\ell\to2\ell$ halves the FSR) in verify_aq-optics-finesse-free-spectral-range.py.*

Go deeper · Derivation C · 1 minResolving power as order times finesse

Derivation CResolving power as order times finesse

This derivation shows why a Fabry–Pérot outguns a grating: its resolving power is the interference order — which can be tens of thousands — multiplied by the finesse. We adopt the natural resolution criterion that two spectral lines are just resolved when their transmission peaks are separated by one full width at half maximum, then translate that frequency criterion into a wavelength ratio using the peak condition.

Step 1. Take two just-resolved lines separated by exactly one linewidth, $\delta\nu=\delta\nu_{1/2}=\Delta\nu_{\text{FSR}}/\mathcal{F}$ . Why: one FWHM is the standard "just distinguishable" separation for Lorentzian-like peaks — closer than that and the two peaks blur into one.

Step 2. Relate fractional wavelength and frequency shifts. From the peak condition $m\lambda=2n\ell\cos\theta$ at fixed order and geometry, a wavelength change maps to a frequency change of the same fractional size, $\delta\lambda/\lambda=\delta\nu/\nu$ . Why: $\nu\lambda=c$ makes fractional shifts equal and opposite, so the resolving power is the same whether phrased in $\lambda$ or $\nu$ .

Step 3. Assemble the ratio. Since the frequency $\nu=m\,\Delta\nu_{\text{FSR}}$ — the order literally counts how many free spectral ranges fit into the optical frequency — we get $\lambda/\delta\lambda=\nu/\delta\nu_{1/2}=(m\,\Delta\nu_{\text{FSR}})/(\Delta\nu_{\text{FSR}}/\mathcal{F})=m\mathcal{F}$ . Why: the free spectral range cancels, leaving the two things that actually cost money — a large order (thick, precisely spaced plates) and a high finesse (excellent mirrors).

Step 4. Note the tradeoff this exposes. Doubling $\ell$ doubles $m$ and hence the resolving power, but halves $\Delta\nu_{\text{FSR}}$ and so the unambiguous bandwidth — the finesse, the number of resolvable channels, stays fixed. Why: finesse is the true information capacity per order; you can spend it on fine resolution over a narrow range or coarse resolution over a wide one, but not both at once. *Verified: $m=2n\ell/\lambda\approx33\,939$ and $\lambda/\delta\lambda=m\mathcal{F}=2.08\times10^{6}$ for a $1$ cm air etalon at $589.3$ nm with $R=0.95$ , cross-checked against $\nu/\delta\nu_{1/2}$ (the $Q$ -factor) in verify_aq-optics-finesse-free-spectral-range.py.*

Limiting cases

Case	Result	Meaning
High reflectance $R\to1$	$\mathcal{F}\approx\pi/(1-R)\to\infty$	peaks become arbitrarily sharp; storage time and resolving power diverge — until defects take over
Low reflectance $R\to0$	$\mathcal{F}\to0$ , Airy $\to$ shallow cosine	the resonator degrades into ordinary two-beam interference with poor contrast
On resonance, lossless	$T_{\max}=1$	a symmetric lossless etalon transmits 100% at a peak regardless of $R$ — high $R$ sharpens, never dims
Anti-resonance	$T_{\min}=(1-R)^2/(1+R)^2$	the floor between peaks; $R=0.95$ gives $T_{\min}\approx6.6\times10^{-4}$ , deep dark rejection
Thicker plates ( $\ell\to2\ell$ )	$\mathcal{F}$ fixed, FSR halves, $R_p$ doubles	length buys resolution and costs unambiguous range; the mirrors are untouched
Flatness-limited	$1/\mathcal{F}_{\text{eff}}^2=\sum 1/\mathcal{F}_i^2$	a $\lambda/100$ surface caps $\mathcal{F}_{\text{flat}}$ near $50$ ; then better coatings buy nothing

Common trapFinesse and resolving power are the same thing — just make the mirrors better

They are not the same, and conflating them is the classic Fabry–Pérot mistake. Finesse is spacing over width: it is set by the mirror reflectance and is completely blind to how far apart the plates sit. Resolving power is $m\mathcal{F}$ : it also carries the interference order, which grows with the plate separation. So a centimetre etalon and a millimetre etalon with identical mirrors have identical finesse, yet the thicker one resolves ten times finer — because it operates at ten times the order. Turn this around and it becomes a design rule: to sharpen the peaks, improve the mirrors (raise $\mathcal{F}$ ); to split finer lines at fixed mirrors, separate the plates (raise $m$ ), accepting the smaller free spectral range that comes with it. And beware the ceiling: once $R$ is very high, the finesse you actually get is capped by mirror flatness and aperture, not by the coatings — so quoting $\pi\sqrt{R}/(1-R)$ for a real etalon overstates it, sometimes by a large factor.

Worked examples

1 · Full figures of merit for a 1 cm air-spaced etalon (confidence builder)

An air-spaced Fabry–Pérot etalon has mirror separation $\ell=1$ cm and mirror reflectance $R=0.95$ , used with sodium light at $\lambda=589.3$ nm. Find its free spectral range, finesse, peak linewidth, interference order, resolving power, and the smallest wavelength difference it can resolve.

Step 1. Free spectral range from the geometry alone: $\Delta\nu_{\text{FSR}}=c/(2n\ell)=(2.998\times10^{8})/(2\times1\times0.01)=1.499\times10^{10}$ Hz $=\mathbf{14.99\ GHz}$ (≈ 15 GHz). This is the unambiguous window: features farther apart than this overlap into neighbouring orders.

Step 2. Finesse from the reflectance alone: $\mathcal{F}=\pi\sqrt{0.95}/(1-0.95)=\pi(0.9747)/0.05=\mathbf{61.2}$ . So the peaks are about sixty times narrower than their spacing.

Step 3. Peak linewidth is the ratio of the two: $\delta\nu_{1/2}=\Delta\nu_{\text{FSR}}/\mathcal{F}=14.99\ \text{GHz}/61.2=\mathbf{245\ MHz}$ . That is already narrow enough to resolve Doppler-broadened atomic lines.

Step 4. Interference order and resolving power: $m=2n\ell/\lambda=(2\times0.01)/(589.3\times10^{-9})\approx\mathbf{33\,939}$ , so $\lambda/\delta\lambda=m\mathcal{F}=33\,939\times61.2=\mathbf{2.08\times10^{6}}$ . The minimum resolvable wavelength difference is $\delta\lambda=\lambda/(m\mathcal{F})=589.3\ \text{nm}/2.08\times10^{6}=\mathbf{0.284\ pm}$ — under a picometre, hundreds of times finer than a good grating.

2 · Why a real etalon never reaches its reflectivity finesse (exam level)

The same mirrors ( $R=0.95$ , reflectivity finesse $61.2$ ) are used, but the plates are flat only to $\lambda/100$ peak-to-valley. Estimate the flatness-limited finesse and the effective finesse of the real instrument.

Step 1. A surface flat to $\lambda/M$ spreads the round-trip phase across the beam by an amount that lets only $M/2$ peaks fit between orders, so the flatness finesse is $\mathcal{F}_{\text{flat}}\approx M/2$ . For $M=100$ , $\mathcal{F}_{\text{flat}}=\mathbf{50}$ .

Step 2. The independent broadening mechanisms add reciprocally-in-square: $1/\mathcal{F}_{\text{eff}}^{2}=1/\mathcal{F}_{R}^{2}+1/\mathcal{F}_{\text{flat}}^{2}=1/61.2^{2}+1/50^{2}=2.67\times10^{-4}+4.00\times10^{-4}=6.67\times10^{-4}$ .

Step 3. Invert: $\mathcal{F}_{\text{eff}}=1/\sqrt{6.67\times10^{-4}}=\mathbf{38.7}$ . The real finesse is well below both the reflectivity value ( $61.2$ ) and even the flatness value ( $50$ ) — the two defects together are worse than either alone, and the worst term dominates.

Step 4. The lesson: with $\lambda/100$ plates, upgrading the coatings from $R=0.95$ to $R=0.99$ (reflectivity finesse $313$ ) would raise the effective finesse only from $38.7$ to about $49$ — almost all of it wasted on a surface that cannot keep up. Real etalons are flatness-limited; polishing, not coating, is usually the binding constraint.

Physics of the portrait

The finesse is one of those rare figures of merit that is at once a definition, a number you can compute from a single material property, and a direct count of a physical process — the round trips a photon survives. It deserves a portrait of its own.

Origins · 1 minfinesse and how this was discovered

finesse

figure of merit · Fabry–Pérot spectroscopy, early 1900s · resonator optics

Finesse entered optics with the Fabry–Pérot interferometer around 1900, as spectroscopists needed one number for how "fine" an etalon's fringes were — hence the French finesse, sharpness. Its genius is that it separates cleanly from everything else: it is the ratio of fringe spacing to fringe width, and for ideal mirrors it reduces to $\pi\sqrt{R}/(1-R)$ , independent of wavelength, plate spacing, or order. That independence lets an engineer reason about a cavity in modular pieces — length sets the free spectral range, reflectance sets the finesse, their product with the order sets the resolving power. The concept travelled far beyond spectroscopy: the same finesse grades a laser cavity's mode selectivity, a gravitational-wave detector's photon storage, and an optical clock's reference resonator, where finesses above $100\,000$ are now routine. Whenever a later article needs to say how sharply a resonance is defined, it reaches back for the finesse defined here.

History

1899

— Charles Fabry and Alfred Pérot introduce the interferometer that bears their names, and with it the language of finesse and free spectral range

1901

— Fabry and Pérot establish the interferometer as a standard tool for high-resolution spectroscopy and wavelength metrology

1948

— Samuel Tolansky develops multiple-beam interferometry, pushing surface-flatness metrology to fractions of a wavelength

1960

s — Fabry–Pérot etalons become intracavity mode selectors and scanning spectrometers with the rise of the laser

2015

— kilometre-scale Fabry–Pérot arm cavities, finesse of order a few hundred, help LIGO detect gravitational waves

Going further

Texts. Hecht, Optics, 5th ed., §9.6 (The Fabry–Perot Interferometer) · Foot, Atomic Physics, §A.1 (Fabry–Perot etalon and interferometer) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §2.5 and §11.1 (optical resonators) · Demtröder, Laser Spectroscopy, Vol. 1, §4.2 (interferometers) · Loudon, The Quantum Theory of Light, 3rd ed., Ch. 2 (optical cavities).

Provenance. The Airy transmission function and the three figures of merit follow Hecht §9.6, Saleh & Teich §2.5/§11.1, and Foot Appendix A. verify_aq-optics-finesse-free-spectral-range.py recomputes: the geometric-series sum giving $T=(1-R)^2/[(1-R)^2+4R\sin^2(\delta/2)]$ ; $\mathcal{F}=\pi\sqrt{R}/(1-R)$ from the FWHM condition, with $\mathcal{F}\approx61.2$ at $R=0.95$ and $\approx155$ at $R=0.98$ ; $\Delta\nu_{\text{FSR}}=c/2n\ell$ from $\Delta\delta=2\pi$ ; and $\lambda/\delta\lambda=m\mathcal{F}$ , with the worked etalon giving $m\approx33\,939$ and $\lambda/\delta\lambda=2.08\times10^{6}$ . Owns the finesse / free-spectral-range / resolving-power concept; recalls the Airy function from aq-optics-multiple-beam-interference-airy and the resonator from aq-optics-fabry-perot-interferometer by name.

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

What is the difference between the coefficient of finesse and the finesse?

They are two names for the sharpness, related but not equal. The coefficient of finesse $F=4R/(1-R)^2$ is the number that appears inside the Airy function, $T=1/(1+F\sin^2(\delta/2))$ ; it also equals the peak-to-valley contrast ratio minus one. The finesse $\mathcal{F}=\pi\sqrt{R}/(1-R)$ is the ratio of free spectral range to linewidth — a direct count of resolvable peaks per order. They are linked by $(2\mathcal{F}/\pi)^2=F$ , so at $R=0.95$ the coefficient of finesse is $1520$ while the finesse is $61.2$ . When someone says "the finesse is $60$ " they mean the peaks-per-order count; when a formula shows $F$ inside a $\sin^2$ , that is the coefficient.

Why does a lossless etalon transmit 100% on resonance even with 99% mirrors?

It seems paradoxical — surely mirrors that reflect 99% should block almost everything. The resolution is resonant build-up. On resonance the field inside the cavity grows until the small fraction leaking back out the front mirror exactly cancels the directly reflected beam by destructive interference; with nowhere else for the energy to go, all of it exits the far mirror. The Airy function makes this exact: $T_{\max}=1$ whenever the mirrors are lossless, independent of $R$ . Raising the reflectance does not lower the peak — it narrows it, so you must tune more precisely to sit on top, but the summit is always at 100%. Real absorption and scatter in the coatings are what pull the peak below unity.

How is finesse related to the number of bounces a photon makes?

Very directly: the finesse is $2\pi$ times the average number of round trips a photon survives before leaking out, sometimes called the storage number. A finesse of $300$ means light bounces roughly $300/\pi\approx95$ times, so it samples the mirror separation ninety-odd times over — which is exactly why the resolving power is so high, since each pass re-measures the same length. The same idea appears as the cavity photon lifetime $\tau_c=n\ell\mathcal{F}/(\pi c)$ and the quality factor $Q=m\mathcal{F}=\nu/\delta\nu_{1/2}$ : high finesse means long storage, sharp resonance, and fine spectral discrimination — three faces of one number.

If I want to resolve two very close spectral lines, do I raise the finesse or the plate spacing?

Either raises the resolving power $m\mathcal{F}$ , but they cost different things. Raising the finesse (better mirrors) sharpens every peak and improves resolution without shrinking the free spectral range — the clean win, limited by how good and how flat your mirrors are. Raising the plate spacing increases the order $m$ and so the resolving power in proportion, but it shrinks the free spectral range by the same factor, so your unambiguous window narrows and neighbouring orders may overlap the lines you care about. In practice you push the finesse as high as flatness allows, then set the spacing to reach the resolution you need while keeping the free spectral range wide enough to stay unambiguous — often with a grating pre-filter to remove order overlap.

Why do real etalons fall short of the π√R/(1−R) finesse?

Because that formula assumes perfect, parallel, uniformly illuminated mirrors, and no real etalon is. Two extra effects broaden the peaks. Surface flatness: if the plates deviate from flat by $\lambda/M$ across the beam, different parts of the aperture resonate at slightly different phases, giving a flatness finesse of about $M/2$ . Aperture: a finite illumination cone spreads $\cos\theta$ and hence the resonance frequency. These combine with the reflectivity finesse reciprocally-in-square, $1/\mathcal{F}_{\text{eff}}^2=1/\mathcal{F}_R^2+1/\mathcal{F}_{\text{flat}}^2+1/\mathcal{F}_{\text{ap}}^2$ , so the smallest finesse dominates. A $\lambda/100$ surface caps you near $50$ whatever the coatings — which is why serious etalons are polished to $\lambda/200$ or better before anyone worries about reflectance.

Is the etalon's resolving power really higher than a diffraction grating's?

Yes, often by orders of magnitude, but with a catch. A grating's resolving power is $mN$ (order times illuminated grooves), typically $10^4$ – $10^5$ , spread over a wide, easy-to-use free spectral range. A Fabry–Pérot reaches $m\mathcal{F}$ with $m\sim10^4$ and $\mathcal{F}\sim10^2$ , so $10^6$ – $10^8$ — but over a tiny free spectral range, so only a narrow window is unambiguous. That is why the two are combined: a grating pre-disperses the light to isolate a band, and the etalon resolves the fine structure within it. The etalon buys the resolution; the grating buys the range to use it.