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The Einstein angular radius is the angular form of the Einstein radius: $R_E=D_l{\theta }_E$ in the lens plane. Its displayed value applies to a point-mass gravitational lens with the angular diameter distances defined above. It depends on source distance as well as lens mass and position. At fixed distances, multiplying $M$ by four doubles ${\theta }_E$.

At exact alignment, the vector thin gravitational lens equation is $0=(1-{\theta }_E^2/r^2)\mathbf{\theta }$. For a nonzero image direction it requires $r={\theta }_E$ and leaves the direction unrestricted: all azimuths form an Einstein ring. The signed equation's roots $\theta =\pm {\theta }_E$ are only a diameter cut through that circle. A source of finite angular size makes an annulus or arcs with finite received light; the infinite point-source lensing magnification at alignment is an idealization. An Einstein ring expresses rotational symmetry and source alignment, not a circular arrangement of emitting stars.

Gravitational lensing is the deflection and focusing of light by spacetime curvature along the line of sight.

A useful first model is a point-mass gravitational lens: one displaced source gives two images of a point-mass lens, while exact alignment gives an Einstein ring. The Einstein angular radius sets the angular scale, and image parity in gravitational lensing explains why the two images have different handedness. The spectroscopic test of the Twin Quasar lens shows how optical spectra can test whether two apparent quasars represent one source.

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.