Einstein ring

Example

Let a point mass $M$, the lens, lie at distance $D_L$ from an observer. Let a point source lie at distance $D_S > D_L$ exactly behind it on the same line. Let $D_{LS} = D_S - D_L$ be the distance from lens to source, and work with small angles. By symmetry, the image of the source is a circle around the lens, the Einstein ring, of angular radius $$\theta_E = \sqrt{\frac{4 G M}{c^2} \, \frac{D_{LS}}{D_L D_S}}.$$ Check: a ray seen at angle $\theta$ from the lens passes it at impact parameter $b = D_L \theta$. By the gravitational deflection of light, it is bent by the angle $\alpha = 4GM/(c^2 b)$. For the ray to reach the source, which sits at angle $0$, the geometry of small angles requires $\theta D_S = \alpha D_{LS}$. So $\theta D_S = 4GM D_{LS}/(c^2 D_L \theta)$, which gives $\theta^2 = \theta_E^2$.