DATA TREATMENT - Two-Dimensional X-Ray Diffraction - page 177

FIGURE 6.11 Illustration of geometric relationship between the monochromator and
detector in the laboratory coordinates, X L Y L Z L .

In two-dimensional X-ray diffraction the diffractionvectors of the monochromator
diffraction and sample crystal diffraction are not necessarily in the same plane or
perpendicular planes. Therefore, the overall polarization factor is a function of both
2u and g. Figure 6.11 illustrates the geometric relationship between the monochro-
mator and detector in the laboratory coordinates, X L Y L Z L . The monochromator is
located at the coordinates X L Y 0 L Z 0 L , which is a translation of the laboratory
coordinates along the X L -axis in the negative direction. The monochromator crystal
is rotated above the Z 0 L -axis for an angle of u M . The X-rays from the source reach the
monochromator with an angle of 2u M from the negative X L direction. The diffracted
beam from the monochromator propagates along the X L direction. This is the incident
beam to the sample located at the instrument center O. The incident plane, containing
the monochromator diffraction vector H M , is coincident with the diffractometer plane
X L -Y L . The area detector location is given by the sample-to-detector distance D and
swing angle a. The pixel P(x, y) represents an arbitrary pixel on the detector. The 2u
and g are the corresponding diffraction space parameters for the pixel. The diffraction
vector H P represents the scattered X-rays collected by the pixel P(x, y). The angle r is
the angle between the diffraction plane containing the diffraction vector H P and the
incident plane containing the monochromator diffraction vector. In this particular
case, the angle r is also the angle between the diffraction plane and the diffractometer
plane X L -Y L . Since a monochromator or other beam conditioning optics can only be
used on the incident beam for two-dimensional diffraction system, the general
polarization factor is given as [24,25].

P G ¼ ðcos 2 2u cos 2 r þ sin 2 rÞcos 2 2u M þ cos 2 2usin 2 r þ cos 2 r
1þ cos 2 2u M

ð6:34Þ

where 2u angle for each pixel is given by Eq. (6.16), and cos 2 r and sin 2 r can be
calculated from 2u and g. The g angle for each pixel is given by Eq. (6.17).