= Solution
The angle in the source ray sketch is inside the gap. Two extra reflections add a full round trip. Comparing adjacent emerging rays on a common outgoing phase plane subtracts the lateral-path contribution, so the <optical path difference> is $2\mu d\cos\theta$, rather than $2\mu d/\cos\theta$. Thus
$$
\boxed{\delta=2k_0\mu d\cos\theta},\qquad \delta=2\pi m
$$
at transmission maxima. The two internal reflection phases add to an integer full cycle in the stated ideal symmetric-coating model. <Snell's law>, $\mu\sin\theta=\mu_0\sin\theta_{\rm out}$, converts to the external direction. A fixed internal angle defines a cone and hence an equal-inclination ring in a <focal plane>.
For identical lossless mirrors the first transmitted amplitude has magnitude $(1-R)$ times the incident amplitude. Each additional round trip multiplies it by $Re^{i\delta}$, because $R$ is an intensity <reflection coefficient> for one mirror but the product of the two amplitude reflection magnitudes. Summing the <geometric series> gives
$$
\boxed{T_e=\left|\frac{1-R}{1-Re^{i\delta}}\right|^2
=\frac{(1-R)^2}{1+R^2-2R\cos\delta}}.
$$
The maximum is one in this symmetric lossless limit; absorption, unequal mirrors or additional incoherent reflections change it. At fixed angle and negligible material dispersion, $m\lambda=2\mu d\cos\theta$ and adjacent orders differ by one. For $m\gg1$,
$$
\boxed{\mathrm{FSR}_\lambda\simeq\frac{\lambda^2}{2\mu d\cos\theta}},\qquad
\delta\lambda\simeq\frac{\mathrm{FSR}_\lambda}{\mathcal F}.
$$
<order-safe angular scanning of a Fabry–Pérot spectrum> should cover the approximately $38\ \mathrm{nm}$ spectral span within one order while making the instrumental width smaller than $1\ \mathrm{nm}$. Near <normal incidence>, requiring the smallest approximate FSR to exceed the span gives $\mu d\lesssim5.3\ \mathrm{\mu m}$, and $\mathcal F$ should exceed roughly $40$. These are design bounds, not a unique optimum. One suitable choice is
$$
\boxed{\mu d=5.0\ \mathrm{\mu m},\qquad R=0.95,\qquad\mathcal F\simeq61.2}.
$$
The physical spacing is $d=5.0\ \mathrm{\mu m}/\mu$; an air gap gives $d\simeq5\ \mathrm{\mu m}$. Approximate FSRs near <normal incidence> are $40.1\ \mathrm{nm}$ at $633\ \mathrm{nm}$ and $44.9\ \mathrm{nm}$ at $670\ \mathrm{nm}$, with widths about $0.65$ and $0.73\ \mathrm{nm}$. For a concrete angular scan choose order $m=14$. The $670\ \mathrm{nm}$ pair lies near internal $\theta=20.3^\circ$ and the $633\ \mathrm{nm}$ line near $27.6^\circ$; their widths within this scan remain below $1\ \mathrm{nm}$. Restricting to this calibrated order window avoids the other-order $633\ \mathrm{nm}$ ring near $18.3^\circ$. Index dispersion and unknown extra source lines would need calibration or order-sorting filters.
The resolving <power> is approximately $m\mathcal F$: many coherent cavity round trips create a much longer effective optical delay than one traversal. A Michelson instrument with the same available physical delay has less resolving <power> unless its mirror travel provides a correspondingly long maximum <optical path difference>. Mirror separation errors spread the round-trip phase; keeping them below a linewidth requires roughly
$$
\boxed{|\Delta d|\ll\frac{\lambda}{2\mu\mathcal F\cos\theta}}.
$$
For this design the scale is about $6\ \mathrm{nm}$ in an air gap. Both surface flatness and parallelism over the illuminated area, as well as angular collimation and coating uniformity, must support the intended <finesse>.
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