3.4 Image positions and magnifications

The lens equation (6View Equation) can be re-formulated in the case of a single point lens:
2 β = 𝜃 − 𝜃E. (18 ) 𝜃
Solving this for the image positions 𝜃 one finds that an isolated point source always produces two images of a background source. The positions of the images are given by the two solutions:
1 ( ∘---------) 𝜃1,2 = -- β ± β2 + 4 𝜃2E . (19 ) 2
The magnification of an image is defined by the ratio between the solid angles of the image and the source, since the surface brightness is conserved. Hence the magnification μ is given as
𝜃-d𝜃- μ = β dβ. (20 )
In the symmetric case above, the image magnification can be written as (by using the lens equation):
( ) [ 𝜃 ]4 −1 u2 + 2 1 μ1,2 = (1 − -E-- ) = ---√-------± -. (21 ) 𝜃1,2 2u u2 + 4 2
Here we defined u as the “impact parameter”, the angular separation between lens and source in units of the Einstein radius: u = β∕𝜃E. The magnification of one image (the one inside the Einstein radius) is negative. This means it has negative parity: It is mirror-inverted. For β → 0 the magnification diverges. In the limit of geometrical optics, the Einstein ring of a point source has infinite magnification4! The sum of the absolute values of the two image magnifications is the measurable total magnification μ:
u2 + 2 μ = |μ1 | + |μ2| =-√--2----. (22 ) u u + 4
Note that this value is (always) larger than one5! The difference between the two image magnifications is unity:
μ1 + μ2 = 1. (23 )

  Go to previous page Go up Go to next page