= Solution
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,
$$
\delta=\frac{4\pi nd\cos\theta_g}{\lambda}.
$$
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 $2k_zd$. If the source's $\theta$ is the angle in external air, <Snell's law> gives $n\sin\theta_g=\sin\theta$ and
$$
\delta=\frac{4\pi d}{\lambda}\sqrt{n^2-\sin^2\theta}.
$$
If $\theta$ 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 gives
$$
\frac{I_t}{I_0}=\frac{|t|^4}{|1-\mathcal R e^{i\delta}|^2}
=\frac{(1-\mathcal R)^2}{(1-\mathcal R)^2+4\mathcal R\sin^2(\delta/2)}.
$$
Hence
$$
\boxed{I_t=\frac{I_0}{1+Q\sin^2(\delta/2)},\qquad Q=\frac{4\mathcal R}{(1-\mathcal R)^2}=\frac{4|r|^2}{(1-|r|^2)^2}.}
$$
This is <Fabry–Pérot transmission from multiple reflections>.<Fabry-Perot Interferometer, University of Washington Physics 331>[Theory equations (1)–(2), including the coefficient of finesse]{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> are
$$
\boxed{\nu_m=\frac{mc}{2nd},\qquad \Delta\nu_{\rm FSR}=\frac{c}{2nd}.}
$$
For 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 $I_t/I_0=(0.1/1.9)^2\simeq0.00277$.
At half the unit peak, $Q\sin^2(\delta/2)=1$. Measuring the phase displacement from a resonance yields
$$
\Delta\delta_{\rm FWHM}=4\arcsin\frac1{\sqrt Q},\qquad
\boxed{\Delta\nu_{\rm FWHM}=\frac{2\Delta\nu_{\rm FSR}}\pi\arcsin\left(\frac{1-\mathcal R}{2\sqrt{\mathcal R}}\right)
\simeq\frac{c}{2nd}\frac{1-\mathcal R}{\pi\sqrt{\mathcal R}}.}
$$
The 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 $\Delta\nu_{\rm FWHM}/\Delta\nu_{\rm FSR}\simeq0.03357$. 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 $\mathcal R\ge3-2\sqrt2$.
\Image[/past-exam-of-the-electromagnetism-and-optics-course-of-the-university-of-oxford/2016-fabry-perot-transmission.png]
{title=Fabry–Pérot transmission for mirror reflectivities 0.1 and 0.9}
{description=Original ideal <Fabry–Pérot interferometer> calculation. The broad low-reflectivity curve stays above half maximum; the right panel resolves the high-reflectivity peak and its full width at half maximum.}
For the <absorbing Fabry–Pérot mirror phase and transmission>, write $r=\sqrt{\mathcal R}e^{i\phi_r}$ and let $\mathcal T$ be the power transmission of one mirror. <Optical absorption> makes $\mathcal R+\mathcal T<1$. The transmitted intensity becomes
$$
\boxed{\frac{I_t}{I_0}=\frac{\mathcal T^2}{(1-\mathcal R)^2}\frac1{1+\frac{4\mathcal R}{(1-\mathcal R)^2}\sin^2(\delta_{\rm tot}/2)},\qquad
\delta_{\rm tot}=\frac{4\pi\nu nd}{c}+2\phi_r.}
$$
Here 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 $\phi_r$ is approximately constant,
$$
\boxed{\nu_m=\frac{c}{2nd}\left(m-\frac{\phi_r}{\pi}\right).}
$$
If either $n$ or $\phi_r$ 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.
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