= Solution
{c}
In the <thin-lens approximation for gravitational lensing>, consider a ray crossing the lens plane at transverse distance $b$ above the aligned source-observer axis. With the observer at axial position $0$, the <black hole> at $L$ and the source at $D$, the segment joining the lens plane to the source has slope $-b/(D-L)$, while the segment joining the observer to the lens plane has slope $b/L$. The observer sees the ray at apparent angle $\theta\simeq b/L$ to the axis. When following the <photon> toward the observer, the two small angles to the axis have magnitudes $b/(D-L)$ and $b/L$. The physical bend is their sum:
$$
\delta\simeq\frac bL+\frac b{D-L}
=\frac{bD}{L(D-L)}.
$$
Substitute the <Schwarzschild light deflection>, $\delta\simeq4GM/(bc^2)$, and use $b\simeq L\theta$:
$$
b^2\simeq\frac{4GM}{c^2}\frac{L(D-L)}D,
\qquad
\boxed{2\theta\simeq2\sqrt{\frac{4GM}{c^2}\frac{D-L}{LD}}.}
$$
Every azimuth around the axis is equivalent by rotational symmetry. Isotropic emission therefore supplies rays at all these azimuths, and the ideal image is an <Einstein ring> centered on the <black hole>, with <Einstein angular radius> $\theta$ and angular diameter $2\theta$. A finite source size or imperfect alignment broadens or breaks the ideal ring.
The stated expression uses small angles, distant endpoints and <weak-field approximation> $b\gg2GM/c^2$, with ordinary Euclidean distances. For cosmological <gravitational lensing>, use <angular diameter distances> $D_l,D_s,D_{ls}$ in $\theta^2=4GM D_{ls}/(c^2D_lD_s)$; generally $D_{ls}\ne D_s-D_l$.
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