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At normal incidence, with no optical dispersion and zero round-trip reflection wave phase modulo $2\pi$, the resonance condition in Fabry–Pérot transmission from multiple reflections isHere $\nu$ is frequency and $c$ is the speed of light in vacuum. A change of $\nu$ by one free spectral range advances the round-trip wave phase by $2\pi$. At fixed internal angle of incidence $\theta$, the same nondispersive model gives $c/(2nd\cos \theta )$. Holding the external angle of incidence fixed in a dispersive medium need not hold the internal $\theta$ fixed, because of Snell's law.

For stipulated $n=1$ and $d=10\mathrm{mm}$, the free spectral range is $14.9896229\mathrm{GHz}$ using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. A frequency difference of $0.10\Delta {\nu }_{\mathrm{FSR}}$ is therefore $1.49896229\mathrm{GHz}$. Transmission repeats every free spectral range: one isolated scan cannot distinguish that separation from $1.10\Delta {\nu }_{\mathrm{FSR}}$. Independent knowledge of the optical spectrum, an additional filter, or a changed mirror separation is needed to assign orders.

Equal spacing in frequency does not imply equal spacing in wavelength. At normal incidence the exact adjacent resonance wavelengths obeyFor example, choosing $2nd=10{\lambda }_{10}$ makes the exact gap to order 11 equal to ${\lambda }_{10}/11$, while the high-order estimate gives ${\lambda }_{10}/10$. The approximation is already ten percent larger at order 10. With optical dispersion or frequency-dependent reflection wave phase, obtain local spacings from the total phase difference instead: $\Delta \nu \simeq 2\pi /(d\delta /d\nu )$ when its derivative varies little over one order.

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 $ϵ=\delta -2\pi m$. Solving $T=1/2$ givesand, since one free spectral range advances the wave phase by $2\pi$,These are exact for the ideal nondispersive model with constant $R$ and linear frequency dependence of $\delta$. The inverse sine requires $(1-R)/(2\sqrt{R})\le 1$, or $R\ge 3-2\sqrt{2}\simeq 0.171573$. 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 $\arcsin x\simeq x$ givesThis 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.

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.^[ref]

Let $\mathcal{R}=|r{|}^2$ be the single-mirror power reflectivity and ${\theta }_g$ the propagation angle inside the gas. Define $\lambda$ as the vacuum wavelength. Successive transmitted complex amplitudes differ by a round-trip factor $\mathcal{R}{e}^{i\delta }$, where, for real dielectric reflection coefficients whose product has positive sign,The cosine arises by comparing successive emerging rays on the same outgoing wavefront; it is not the literal zigzag distance $2d/\cos {\theta }_g$. Equivalently, the round-trip phase is $2k_{z}d$. If the source's $\theta$ is the angle in external air, Snell's law gives $n\sin {\theta }_g=\sin \theta$ andIf $\theta$ denotes the internal angle, simply set ${\theta }_g=\theta$.

For identical lossless mirrors, their power transmission is $1-\mathcal{R}$. In an energy-normalized amplitude convention this is $|t{|}^2=1-\mathcal{R}$; more general interface amplitudes require the corresponding wave impedance factors. The geometric series of transmitted amplitudes givesHenceThis is Fabry–Pérot transmission from multiple reflections.^[ref]

At normal incidence, for nondispersive $n$ and zero net reflection phase modulo $2\pi$, the resonance condition is $\delta =2\pi m$. The resonance frequencies and free spectral range areFor finite $\mathcal{R}<1$ the resonances have finite width, and transmission away from them is nonzero. Thus “only discrete frequencies” is an ideal high-reflectivity description of the selected resonances, not an exact statement about the support of the transmitted spectrum. For example, $\mathcal{R}=0.9$ at $\delta =\pi$ still gives $I_t/I_0=(0.1/1.9{)}^2\simeq 0.00277$.

At half the unit peak, $Q{\sin }^2(\delta /2)=1$. Measuring the phase displacement from a resonance yieldsThe last expression is the high-reflectivity approximation, with finesse $\pi \sqrt{\mathcal{R}}/(1-\mathcal{R})$. At $\mathcal{R}=0.9$, the full phase width is about $0.2109$ radians and $\Delta {\nu }_{\mathrm{FWHM}}/\Delta {\nu }_{\mathrm{FSR}}\simeq 0.03357$. At $\mathcal{R}=0.1$, the minimum transmission is $0.6694$, so the trace never reaches half the peak and this full width at half maximum is not defined. The exact half-maximum condition requires $\mathcal{R}\ge 3-2\sqrt{2}$.

For the absorbing Fabry–Pérot mirror phase and transmission, write $r=\sqrt{\mathcal{R}}{e}^{i{\varphi }_r}$ and let $\mathcal{T}$ be the power transmission of one mirror. Optical absorption makes $\mathcal{R}+\mathcal{T}<1$. The transmitted intensity becomesHere the phase formula is at normal incidence. The peak is reduced below unity, while a change in $\mathcal{R}$ changes the width. The complex reflection phase also moves the resonances: if ${\varphi }_r$ is approximately constant,If either $n$ or ${\varphi }_r$ depends on frequency, this is an implicit condition and the free spectral range need not be constant. No numerical shift is determined without the mirrors' optical response.

The wave impedance is the travelling-wave field ratio $Z=E/H$; for the stated nonmagnetic media it is $Z_0/n$. A reflection coefficient is a reflected-to-incident complex amplitude ratio with an explicitly chosen field orientation. With this reflection-amplitude sign convention, using the question's formula for incidence inside silicon givesWith a common tangential electric-field direction, the usual electric-field Fresnel equations instead give $r_E=(Z_2-Z_1)/(Z_2+Z_1)=+5/9$ for this incidence. The printed formula corresponds to the opposite reflected-field sign convention, for example the magnetic amplitude convention. It gives the same reflectance and the same round-trip factor $r^2$.

Each extra internal round trip multiplies the transmitted amplitude by $r^2{e}^{i\delta }$, with $\delta =4\pi nd/\lambda$. A geometric series therefore givesThis is Fabry–Pérot transmission from multiple reflections. Assuming lossless silicon with wavelength-independent $n$, the full intensity transmittance between air regions isMaxima have $2nd/\lambda =j$, minima $2nd/\lambda =j+\frac{1}{2}$. Here $2nd=7000\mathrm{nm}$, so within the specified tuning range the only maximum is $1400\mathrm{nm}$, and the two minima are $1555.6\mathrm{nm}$ and $1272.7\mathrm{nm}$. The contrast is

Identical highly reflecting coatings increase the finesse: transmission maxima become narrower and the intensity between them falls. Ideal symmetric lossless coatings retain $T_{\max }=1$; absorption or unequal mirrors reduce it. The illustrated maxima stay fixed only because additional reflection wave phases are held unchanged. Real coating wave phases can also shift the resonances.