= Solution
Let $\alpha$ be the reduced, observed angular displacement between source and image. In the small-angle lens geometry, the physical bending angle is weighted by the lens-to-source and observer-to-source <angular diameter distances>:
$$
\boxed{\alpha=\frac{D_{LS}}{D_S}\,\xi.}
$$
In a Euclidean diagram this follows because the transverse displacement at the source plane is $D_{LS}\xi$, whose observed angle is that displacement divided by $D_S$. In cosmology the same thin-lens expression uses the corresponding <angular diameter distances>. For a signed one-dimensional coordinate, the direction is supplied by the image's side of the lens.
Back to article page
