Solving this for the image positions
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:
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
is given as
In the symmetric case above, the image magnification can be written as (by using the lens equation):
Here we defined
u
as the ``impact parameter'', the angular separation between lens
and source in units of the Einstein radius:
. The magnification of one image (the one inside the Einstein
radius) is negative. This means it has negative parity: It is
mirror-inverted. For
the magnification diverges. In the limit of geometrical optics,
the Einstein ring of a point source has infinite magnification
! The sum of the absolute values of the two image magnifications
is the measurable total magnification
:
Note that this value is (always) larger than one
! The difference between the two image magnifications is
unity:
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Gravitational Lensing in Astronomy
Joachim Wambsganss http://www.livingreviews.org/lrr-1998-12 © Max-Planck-Gesellschaft. ISSN 1433-8351 Problems/Comments to livrev@aei-potsdam.mpg.de |