In cases of strongly scattering or absorbing samples, the detected scattering intensity is lower than the 'true' scattering. Moreover, for oddly-shaped samples, the extinction of scattering may be anisotropic: some parts of the detector image are more attenuated than others because of the differing path-lengths through the sample.
Formulation
In the following, we assume a transmission-scattering (TSAXS) experiment.
The measured scattering,
, at a particular scattering angle
(where
is the full (
) scattering angle between the scattered ray and the incident beam, and
is the azimuthal angle:
corresponds to the
axis, whereas
is along the
axis) can be computed by summing the scattering contributions for all the elements along the beam path through the sample.
We define a realspace coordinate system
where z points vertically, y points along the beam direction, and x points horizontally with respect to the sample. Let the sample size along the beam direction be L. Defining the point where the beam first enters the sample as
we write:

The scattering from a particular location within the sample is affected by two attenuation effects:
- The beam flux within the sample decreases due to absorption/scattering, such that the flux at position
is not the incident flux,
but attenuated to:

- where
is (Beer-Lambert like) extinction coefficient. If the 'true' scattering probability is given by
(i.e.
is the scattering observed in the absence of attenuation), then the scattering at
is:

- The scattered radiation is itself attenuated as it passes through the sample. Let this path-length (from scattering location
until it exits the sample along the direction
) be denoted
. In such a case, the scattering that exits the sample is:

The measured scattering is thus:

The integral of course depends on the form of
which depends on the sample shape. Note that in the limiting case of weak attenuation (
), we obtain the very simple result:

As expected, scattering intensity scales with the scattering volume.
Normalization
To normalize-out the effect of attenuation, one can simply divide by the integral:

Of course in the case of weak attenuation the integral is simply L, and we are normalizing by the beam-path through the sample.
Coordinates
For a vector that starts at
and terminates at
, pointing along direction
, the full length is:

We can consider triangles in various planes:
- xz plane (looking along beam):






- xy plane (looking from above):



- yz plane (looking from side):



- plane of beam elevation:



- plane of full scattering angle:



Height Z
If the vector's final point is at height
:

This has a minimum of
when
.
Depth L
If the vector's final position is at depth
:

This has a minimum of
when
.
Width X
If the vector's final position is at width
:

This has a minimum of
when
.
Rectangular prism
If the sample is a rectangular prism with dimensions
and the beam falls upon the center of the xz front-face, then the beam travels a distance L through the sample, and the scattered radiation in any quadrant passes through the rectangular prism of size
. The distance from
to the exit-point from the sample is the distance to the closest sample face:

The distances are:


