Two images of a point-mass lens (${\theta }_{\pm }=(\beta \pm \sqrt{{\beta }^2+4{\theta }_E^2})/2$)

Choose a signed angular coordinate along the source-lens line. The point-mass gravitational lens becomesFor $\beta \ne 0$, multiplication by $\theta$ gives the quadratic equation ${\theta }^2-\beta \theta -{\theta }_E^2=0$. Its positive discriminant yields the two displayed roots. Their product is ${\theta }_{+}{\theta }_{-}=-{\theta }_E^2$, so they lie on opposite sides of the lens. With $\beta >0$, ${\theta }_{+}>{\theta }_E$ and $-{\theta }_E<{\theta }_{-}<0$: the outer image lies on the source side. For $\beta <0$, the outer image is instead the minus root. These are image directions seen by the observer, not two emitting objects.^[ref]

For a dimensionless example, set $u=\beta /{\theta }_E=0.6$ and $x=\theta /{\theta }_E$. Solving $u=x-1/x$ gives $x_{+}\simeq 1.344$ and $x_{-}\simeq -0.744$, with separation $\sqrt{u^2+4}\simeq 2.088$. The lensing magnification follows from image parity in gravitational lensing: ${\mu }_{+}\simeq 1.442$, ${\mu }_{-}\simeq -0.442$, and total amplification $\simeq 1.884$. Changing $M$ or the distances rescales the angles through the Einstein angular radius, while these dimensionless values remain the same.

For $u\gg 1$, the quadratic equation gives $x_{+}\simeq u+1/u$ and $x_{-}\simeq -1/u$. Using $\mu =1/(1-x^{-4})$, the outer image tends to unit lensing magnification and the inner image has $|{\mu }_{-}|\simeq u^{-4}$. Two mathematical images can therefore look like one at finite sensitivity or angular resolution. Furthermore, the inner impact parameter eventually becomes too small for the stated weak-field approximation or passes through the opaque lens: the ideal formula must then be replaced by an appropriate physical model.