There are XRD investigation methods which do need an area correction (it is even mandatory) for a proper exploiting of the data.

One example is the use of a good collimated X-ray beam, like the one from a synchrotron source, in a surface geomtry (grazing incidence, i.e. incident angle is in the several 0.1 deg. range). More details about exploiting such data can be found at

http://www.esrf.eu/computing/scientific/joint_projects/ANA-ROD/

Look then for "manuals", then "From beam time to structure factors" by Elias Vlieg. There you can find a discussion about the active sample area, which needs to be considered for proper data exploiting.

Also, another example, in reflectivity measurements using X-rays (e.g. thin films characterisation), in order to properly describe the reflectivity curve (especially the region of the low incidence), an illuminated sample area corection is mandatory. You can look on the web (ex. google 'x-ray reflectivity'), or an example can be found (for neutrons) at:

http://www.ncnr.nist.gov/reflpak/reflpak/html/reflred/footprint_correction.html

- see also paragraph 3.2 in http://maglab.iphy.ac.cn/XRD%E4%B8%93%E9%A2%98/X-ray%20thin-film%20measurement%20techniques_V_X-ray%20reflectivity%20measurement.pdf

Some biblio (from Wiki page):

Holy, V. et al. Phys. Rev. B. 47, 15896 (1993).

Jens Als-Nielsen, Elements of Modern X-Ray Physics, Wiley, New York, (2001).

J.Daillant, A.Gibaud, X-Ray and Neutron Reflectivity: Principles and Applications. Springer, (1999).

L. G. Parratt, Phys. Rev. 95, 359 (1954).

I believe that the relationship is between the surface area and the half width.

If surface is more intensity will be more as the intensity depends on number of crystals in the sane direction.

I agree with Mr. Yogesh, Surface area more means intensity will be more as intensity is nothing but the energy per unit area.....hence intensity can be a relative measure of the surface area...But i have not seen any articles relating the two..

In a reflection geometry (Bragg-Brentano, more common in in-house diffractometers) the irradiated area of the sample will contribute intensity to the diagram in a linear proportion, so larger area will mean larger intensity. But larger area will also broaden peaks due to a larger amount of x-rays coming from point farther to the point where the bragg condition is fulfilled (or farther from the focusin circle of the diffractometer). In general there are simpler tools than x-ray diffraction to measure an area, that is probably why Prakash Mahakul has seen no articles on this. BTW, keep in mind that X-rays penetrate below the sample surface, depending on the packing density and absroption coefficient of your sample, between ~50-500 micrometers for CuKa radiation, so you cannot interpret your x-rays as coming only from the surface of your sample. This penetration depth may have a big influence on the total intensity since samples with the same area but different packing densities may have different relative intensities.

If you are doing experiments in transmision geometry (Debye-Scherrer or paralell beam transmision geometry) it is the sample volume that scales linearly with the intensity of the pattern if absorption of the x-rays by the sample is negligible, in general absorbtion could be such a problem that very thick samples may give negligible diffracted intensity therefore the intensity/sample volume relation fails.

In general XRD is not used to determine geometric characteristics of the sample that can be measured by many other simpler and cheaper methods.

There are XRD investigation methods which do need an area correction (it is even mandatory) for a proper exploiting of the data.

One example is the use of a good collimated X-ray beam, like the one from a synchrotron source, in a surface geomtry (grazing incidence, i.e. incident angle is in the several 0.1 deg. range). More details about exploiting such data can be found at

http://www.esrf.eu/computing/scientific/joint_projects/ANA-ROD/

Look then for "manuals", then "From beam time to structure factors" by Elias Vlieg. There you can find a discussion about the active sample area, which needs to be considered for proper data exploiting.

Also, another example, in reflectivity measurements using X-rays (e.g. thin films characterisation), in order to properly describe the reflectivity curve (especially the region of the low incidence), an illuminated sample area corection is mandatory. You can look on the web (ex. google 'x-ray reflectivity'), or an example can be found (for neutrons) at:

http://www.ncnr.nist.gov/reflpak/reflpak/html/reflred/footprint_correction.html

- see also paragraph 3.2 in http://maglab.iphy.ac.cn/XRD%E4%B8%93%E9%A2%98/X-ray%20thin-film%20measurement%20techniques_V_X-ray%20reflectivity%20measurement.pdf

Some biblio (from Wiki page):

Holy, V. et al. Phys. Rev. B. 47, 15896 (1993).

Jens Als-Nielsen, Elements of Modern X-Ray Physics, Wiley, New York, (2001).

J.Daillant, A.Gibaud, X-Ray and Neutron Reflectivity: Principles and Applications. Springer, (1999).

L. G. Parratt, Phys. Rev. 95, 359 (1954).

Great basic question! It all depends on the detector & technique used. From your comments, it seems the discussion is focused on a conventional diffractogram acquired using the ubiquitous 0D point or scintillation detector. In such cases the intensity would be dependent on the slits on both ends (source & detector). These slits, source & detector, will control the irradiated and sampling area respectively. Thus the absolute intensity (cps). I strongly suggest taking advantage of the attached reference for a thorough understanding of the fundamentals involved with XRD experimentation.

Here is an example of a real time Debye-Scherrer type camera that was used to acquire this data (NIST LaB6 powder standard) with a 2D detector from the Synchrotron Beam Line X14A @ BNL/NSLS:http://www.youtube.com/watch?v=IU0m4yI7D-k&list=PL7032E2DAF1F3941F

http://www.flickr.com/photos/85210325@N04/10515372183/