Finesse (source code)
= Finesse
{title2=$\mathcal F=\Delta\nu_{\rm FSR}/\Delta\nu_{\rm FWHM}$}
= Optical finesse
{synonym}
The <finesse> of a <Fabry–Pérot interferometer> is its <free spectral range> divided by a transmission peak's <full width at half maximum>. Large <finesse> means a narrow accepted <frequency> interval compared with the spacing between repeated orders. Here half maximum means $T=1/2$ of the ideal unit peak, with zero detector background subtracted; it does not mean halfway between the transmission minimum and maximum.
Write $A=4R/(1-R)^2$ for the coefficient multiplying $\sin^2(\delta/2)$ in <Fabry–Pérot transmission from multiple reflections>. This coefficient is sometimes called the coefficient of finesse; it is not the <finesse> $\mathcal F$. About a resonance $\delta=2\pi m$, put $\epsilon=\delta-2\pi m$. Solving $T=1/2$ gives
$$
\sin^2(\epsilon/2)=\frac1A,\qquad
\Delta\delta_{\rm FWHM}=4\arcsin\!\left(\frac{1-R}{2\sqrt R}\right),
$$
and, since one <free spectral range> advances the <wave phase> by $2\pi$,
$$
\frac{\Delta\nu_{\rm FWHM}}{\Delta\nu_{\rm FSR}}
=\frac2\pi\arcsin\!\left(\frac{1-R}{2\sqrt R}\right),\qquad
\mathcal F=\frac{\pi}{2\arcsin[(1-R)/(2\sqrt R)]}.
$$
These are exact for the ideal nondispersive model with constant $R$ and linear <frequency> dependence of $\delta$. The <inverse sine> requires $(1-R)/(2\sqrt R)\leq1$, or $R\geq3-2\sqrt2\simeq0.171573$. At equality, neighbouring peaks merely touch half maximum at their shared minima: the nominal width is one full <free spectral range> and $\mathcal F=1$. Isolated half-maximum peaks require the strict inequality. Below this threshold the transmission never falls to $1/2$, so this peak-width definition supplies no <finesse>; at $R=0$ the transmission is flat.
For large reflectivity, the <small-angle approximation> $\arcsin x\simeq x$ gives
$$
\mathcal F\simeq\frac{\pi\sqrt R}{1-R},\qquad
\frac{\Delta\nu_{\rm FWHM}}{\Delta\nu_{\rm FSR}}\simeq\frac{1-R}{\pi\sqrt R}.
$$
This is a high-reflectivity approximation, not an identity for every $R$. The original dimensionless examples below have exact <full width at half maximum> values obtained from the <inverse sine> formula.
|| Mirror intensity reflectivity $R$
|| Exact $\Delta\nu_{\rm FWHM}/\Delta\nu_{\rm FSR}$
|| Exact <finesse>
|| Approximate <finesse> $\pi\sqrt R/(1-R)$
| $0.25$
| $0.53989$
| $1.85222$
| $2.09440$
| $0.75$
| $0.09221$
| $10.84478$
| $10.88280$
| $0.95$
| $0.01633$
| $61.23420$
| $61.24092$
\Image[/fabry-perot-normalized-transmission-peaks.png]
{title=Increasing mirror reflectivity narrows <Fabry–Pérot interferometer> transmission peaks}
{description=Original stipulated dimensionless model for $R=0.25,0.75,0.95$, with $x=(\nu-\nu_m)/\Delta\nu_{\rm FSR}$ and $\delta=2\pi x$ modulo $2\pi$. All curves peak at unity and repeat after $\Delta x=1$. The second panel enlarges the central resonance; coloured segments at $T=1/2$ mark each exact <full width at half maximum>. The inset compares exact <finesse> with the high-reflectivity approximation. These are calculations, not measured scans.}
The beauty of the narrowing is visible directly in the <geometric series>: large $R$ preserves contributions from many round trips, so a small <phase difference> accumulates across many terms and spoils their addition. For the stipulated $10\,\mathrm{mm}$ vacuum gap, $R=0.95$ gives an ideal <full width at half maximum> of about $244.792\,\mathrm{MHz}$. This instrumental width is not a measured <spectral line> width. A real scan also depends on the source profile, accepted <angles>, mirror shape, alignment, and scan stability. The University of Washington handout separately discusses mirror-reflectivity, mirror-flatness, and detector-pinhole contributions; increasing $R$ alone cannot remove the others.<Fabry-Perot Interferometer, University of Washington Physics 331>[Finesse discussion, equations (5)–(8), and §2.3 Determine the Finesse]{ref}
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