= Solution
Write $C_R=4R/(1-R)^2$, keeping this coefficient distinct from the <finesse> $\mathcal F$. The transmission has maxima at $\delta=2\pi m$. At a displacement $\epsilon$ from a maximum, its ratio to peak transmission is $[1+C_R\sin^2(\epsilon/2)]^{-1}$. Half maximum therefore requires
$$
\sin^2(\delta_{1/2}/2)=\frac{(1-R)^2}{4R},\qquad
\boxed{\delta_{1/2}=2\arcsin\!\left(\frac{1-R}{2\sqrt R}\right).}
$$
This is the half-width in <wave phase>; the <full width at half maximum> is $2\delta_{1/2}$. Successive resonances are $2\pi$ apart, so the exact ideal-model result is
$$
\boxed{\mathcal F=\frac{\pi}{2\arcsin[(1-R)/(2\sqrt R)]}.}
$$
For narrow peaks, $(1-R)/(2\sqrt R)\ll1$, the <small-angle approximation> to the inverse sine gives
$$
\delta_{1/2}\simeq\frac{1-R}{\sqrt R},\qquad
\boxed{\mathcal F\simeq\frac{\pi\sqrt R}{1-R}.}
$$
Thus the formula requested in the paper is the high-reflectivity approximation, not an exact identity at arbitrary $R$. Ordinary half-maximum peaks exist only when $(1-R)/(2\sqrt R)\leq1$, or $R\geq3-2\sqrt2$; below that threshold the transmission minimum never reaches half the peak. The prefactor $T^2/(1-R)^2$ cancels from this relative-width calculation.
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