= Fabry–Pérot resonance spacing
{c}
{title2=$\Delta\nu_{\rm FSR}=c/(2nd)$}
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> is
$$
2nd=m\lambda_m,\qquad
\nu_m=\frac{mc}{2nd},\qquad
\Delta\nu_{\rm FSR}=\nu_{m+1}-\nu_m=\frac{c}{2nd}.
$$
Here $\nu$ 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 $c=299792458\,\mathrm{m\,s^{-1}}$. A <frequency> difference of $0.10\Delta\nu_{\rm FSR}$ is therefore $1.49896229\,\mathrm{GHz}$. Transmission repeats every <free spectral range>: one isolated scan cannot distinguish that separation from $1.10\Delta\nu_{\rm FSR}$. 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> obey
$$
\lambda_m-\lambda_{m+1}=\frac{2nd}{m(m+1)}
=\frac{\lambda_m}{m+1}
\simeq\frac{\lambda_m^2}{2nd}\quad(m\gg1).
$$
For 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: $\Delta\nu\simeq2\pi/(d\delta/d\nu)$ when its <derivative> varies little over one order.
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