= Solution
{c}
A <thin lens> has <ray transfer matrix>
$$
M=\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}.
$$
The <complex beam parameter> transformation gives $q_2=q_1/(1-q_1/f)$, hence
$$
\boxed{\frac1{q_1}-\frac1{q_2}=\frac1f.}
$$
The sign is also obtained by multiplying the incident <wave amplitude> by the <thin lens> <phase> $\exp[-ik\rho^2/(2f)]$.
Choose the <thin lens> plane as $z=0$ and put $a=\pi w_0^2/\lambda$, the incident <Rayleigh range>. Since the incident <beam waist> lies there, $q_1=-ia$. Immediately after the <thin lens>,
$$
q_2=\frac{-ia}{1+ia/f}
=-\frac{fa^2}{f^2+a^2}-i\frac{f^2a}{f^2+a^2}.
$$
Subsequent free propagation adds $z$, so write $q(z)=z-z_f-iz_R'$. Comparing real and imaginary parts gives
$$
\boxed{z_f=\frac{fa^2}{f^2+a^2},\qquad
z_R'=\frac{f^2a}{f^2+a^2},\qquad
w_f^2=\frac{\lambda z_R'}\pi.}
$$
For a positive <focal length>, the new <beam waist> lies downstream but closer than the geometrical focus. At fixed $f,\lambda$, as $w_0\to\infty$,
$$
\boxed{z_f\to f,\quad z_R'\sim\frac{f^2\lambda}{\pi w_0^2}\to0,
\quad w_f\sim\frac{f\lambda}{\pi w_0}.}
$$
This is the formal unbounded-<aperture> scalar limit; eventually finite <aperture> and the <paraxial approximation> limit the extrapolation.
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