Fabry–Pérot interferometer

Example

A Fabry–Pérot interferometer is a pair of parallel, partially reflecting mirrors at distance $\ell$, with a medium of refractive index $n$ between them. Assume the mirrors are lossless, with intensity reflectance $R \in (0, 1)$ and transmittance $1 - R$. Light of vacuum wavelength $\lambda$ crosses the gap at angle $\theta$ to the normal. Let $\delta = \frac{2\pi}{\lambda} \, 2 n \ell \cos\theta$ be the phase delay of one round trip, from the optical path length. The transmitted intensity, as a fraction of the incident one, is $$T(\delta) = \frac{1}{1 + F \sin^2(\delta / 2)}, \qquad F = \frac{4R}{(1 - R)^2}.$$

Check: the transmitted wave is the sum of the beams that bounce back and forth $0, 1, 2, \dots$ times. Each round trip multiplies the amplitude by $R \, e^{i\delta}$, and the two passes through the mirrors give the factor $1 - R$. The geometric series gives the amplitude $(1 - R) / (1 - R e^{i\delta})$. Its squared modulus is $(1 - R)^2 / ((1 - R)^2 + 4R \sin^2(\delta/2))$, which equals $T(\delta)$.

So $T = 1$ exactly when $\delta \in 2\pi\Z$, that is when $2 n \ell \cos \theta = m \lambda$ for an integer $m$: all the beams interfere constructively. Between these peaks $T$ drops to $1/(1 + F)$, which is small when $R$ is close to $1$. At normal incidence the peaks are spaced in frequency by the free spectral range $c / (2 n \ell)$, where $c$ is the speed of light in vacuum. The ratio of this spacing to the width of a peak at half height is the finesse, approximately $\pi \sqrt{R} / (1 - R)$ when $R$ is close to $1$.