How to calculate the amplification of images in Schwartzschild metric? In the discussion section of the paper  "Strong field limit of black hole gravitational lensing", the amplification of weak field images in the Schwartzschild metric was given by
$$
\frac{1}{\beta}\sqrt{\frac{2\,D_{LS}}{D_{OL}D_{OS}}}
$$
where $\beta$ is the angular position of
source, $D_{OL}$ is the distance between the lens and the observer,
$D_{LS}$  is the distance between the lens and the projection of the source on the optical axis OL  and $D_{OS} = D_{OL} +D_{LS}$

However the authors did not derive this expression or explain its origin. Does anyone know how to derive this expression or knows where the author got it from?
Any help is appreciated.
Thanks in advance.
 A: This is explained, for example, in Schneider, Ehlers and Falco's Gravitational Lenses, or any other standard source. I'm gonna outline the results.
The first important point is that gravitational lensing conserves surface brightness. What this means is that the lens deflects the light rays but doesn't change their energy, because gravity is not an emitter or absorber, and this static situation doesn't affect the frequency of the photons (for far away sources the cosmological redshift must be taken into account). Therefore, if a bunch of light rays leave the source occupying a solid angle element $d\Omega_S$ and arrive at the observer in a solid angle $d\Omega_O$, the magnification is simply the ratio of the areas:
$$\mu = \frac{d\Omega_O}{d\Omega_S} = \left|\frac{\sin \theta\ d\theta}{\sin \beta\ d\beta}\right| \approx \left|\frac{\theta\ d\theta}{\beta\ d\beta}\right| = \left| \frac{\beta\ d\beta}{\theta\ d\theta}\right| ^{-1}$$
Now we need the weak field lens equation $\beta = \theta - \frac{d_{LS}}{d_{OS}}\alpha$ (all quantities are as in your diagram). The famous Einstein calculation of the deflection angle $\alpha$ gives that $\alpha=4M/r_0$, where $r_0$ is either the impact parameter or the closest approach distance, since they are equal within this approximation. We also have $r_0 = D_{OL}\theta$. We can substitute this into the lens equation and solve for $\theta(\beta)$.
At this point it is convenient to introduce the Einstein radius $\theta_E = \sqrt{\frac{4MD_{LS}}{D_{OS}D_{OL}}}$. The lens equation has two solutions $\theta = \frac12(\beta \pm \sqrt{\beta^2+4\theta_E^2})$, corresponding to the two images. Now it's a matter of taking derivatives and doing some algebra to calculate the magnifications:
$$\mu = \frac14 \left(\frac{\beta}{\sqrt{\beta^2+4\theta_E^2}} + \frac{\sqrt{\beta^2+4\theta_E^2}}{\beta} \pm 2\right)$$
This diverges when $\beta \to 0$ (which is the weak lensing limit), and in this regime both magnifications approach $\theta_E/2\beta$. Summing them both we get the total magnification $\mu = \theta_E/\beta$; setting the Schwarzschild radius $2M$ equal to $1$, we get your expression.
