 Solution

Use the small-angle, thin gravitational lens equation geometry. An undeflected ray seen at angular position $\theta$ would have transverse source position $D_s\theta$. Bending that ray through physical angle $\hat{\alpha }$ at the lens changes its source intercept by $D_{ds}\hat{\alpha }$. Thusand the required physical deflection isAlso $\beta =\eta /D_s$. Define the reduced deflection $\alpha$ as the angular difference $\theta -\beta$. Dividing the first equation by $D_s$ givesThese are angular diameter distances; in a cosmological spacetime one must not replace $D_s$ by $D_d+D_{ds}$ as if the distances were ordinary collinear Euclidean lengths.