= Solution
{c}
Use the <complex beam parameter> convention
$$
\frac1q=\frac1{R_b}-i\frac{\lambda}{\pi w^2},\qquad q_{\rm waist}=iz_R.
$$
Here $R_b$ is the <wavefront> curvature, distinguished from the mirror curvature $R$. In an unfolded propagation coordinate, a concave mirror has the <ray transfer matrix> $\left(\begin{smallmatrix}1&0\\-2/R&1\end{smallmatrix}\right)$: a ray at height $x$ changes direction by $-2x/R$. The return to the original plane is therefore
$$
M=\begin{pmatrix}1&L\\0&1\end{pmatrix}
\begin{pmatrix}1&0\\-2/R&1\end{pmatrix}
\begin{pmatrix}1&L\\0&1\end{pmatrix}
=\begin{pmatrix}1-2L/R&2L(1-L/R)\\-2/R&1-2L/R\end{pmatrix}.
$$
The <Gaussian beam> transformation $q'=(Aq+B)/(Cq+D)$ gives
$$
\boxed{q'=\frac{(1-2L/R)iz_R+2L(1-L/R)}{1-2L/R-2iz_R/R}.}
$$
For $R=2L$, $M=\left(\begin{smallmatrix}0&L\\-1/L&0\end{smallmatrix}\right)$ and $q'=-L^2/q=iL^2/z_R$. The returning <Gaussian beam> consequently has a <beam waist> at the same plane, with
$$
\boxed{z_R'=L^2/z_R.}
$$
This alone does not make it the same <Gaussian beam>; equal <Rayleigh ranges> require $z_R=L$.
Adding the plane mirror requires a self-consistent <optical resonator> <normal mode>, $q'=q=iz_R$. Since $A=D$, the fixed-point equation reduces to $Cq^2=B$, giving
$$
\boxed{z_R=\sqrt{L(R-L)}=\sqrt{LR(1-L/R)}.}
$$
For a finite stable <Gaussian beam>, $0<L<R$. The same formula remains valid at $R=2L$, where self-consistency selects $z_R=L$. At the geometrical hemispherical limit $R=L$ it predicts $z_R=0$; that marginal boundary has no finite-width stable fundamental <Gaussian beam>. Thus the source's word “hemispherical” is used for the plano-concave family, including the near-hemispherical case below.
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