X-ray diffraction pattern from crystal?

ZAF

Z=Atomic Number

A=Absorption

F=Fluorescence

CuKalpha = 8.6keV

Reduced peak intensity principally by absorption as X-ray leaves from inside crystal surface at higher angles. Increased path length increases absorption.

Additional:

Absorption

Absorption phenomena related to fluorescence effects have already been discussed.

Absorption also occurs related to the area of a powder specimen and depth of penetration of

the x-ray beam into the specimen. In general, with a Bragg-Brentano diffractometer, the

larger area of sample irradiated at low 2θ values have less depth of penetration. At higher 2θ

values, the irradiated area will smaller, but depth of penetration greater. In general, these

tend to be offsetting effects as related to diffracted intensity over the angular range of the

data collection. The calculated intensity will include a term for 1/µs where µs is the linear

absorption coefficient of the specimen.

Need to look under Quantitative analysis in X-ray diffraction.

If you think that the above section (quoted from a paper) is what you are looking for I can give the web address to you.

Me: operator of

X-ray and electron micro probe analyser (Jeol) with fully quantitative energy and wavelength dispersive analysis capability.

You are improper. X-rays aren't mirrored from a crystal floor, they're diffracted. That is why it can be referred to as x-ray diffraction, strangely sufficient. Attempt to learn as a minimum the elemental facts before asking silly questions.