Is the thickness of a sample related to the intensity of x-ray diffraction? I understand that in general if we're adding more planes of atoms (increasing thickness of sample) then the intensity would increase because we have more constructive interference. But isn't there a breaking point for this? Shouldn't there be a finite thickness past which the intensity decreases? 
 A: Yes, there is such a point. The precise formula varies as a function of the scattering geometry, but if we consider a special case:


*

*normal incidence on a flat sample and

*small scattering/diffraction angle


it is quite simple: the scattered intensity is proportional with the sample thickness $d$ but it gets attenuated as $\exp(-\mu d)$ (the Beer-Lambert law with $\mu$ the absorbance).
The strongest signal is thus obtained at the maximum of $I_S \sim d \exp(-\mu d)$, which occurs at $d = 1/\mu$, i.e. a transmission of $1/\text{e}$. Beyond this value, the attenuation (an exponential effect) dominates the increase of scattering volume (a linear effect).
A: Each layer contributes an increase proportional to the intensity of the original beam and a decrease proportional to the intensity of the diffracted beam (from multiple diffractions and extinction). In the approximation that the original beam passes through the sample with minimal losses (and the diffracted beams escape with minimal losses), the signal increases linearly with thickness. If the sample is thick enough that the losses become significant, then you will need to use a more involved expression.
