 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: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.