On 28/01/2010 20:20, Ian Tickle wrote:
On another point you said you wanted an 'operational' definition of
I(Bragg). I'm not entirely clear what you mean by that. Are you saying
I believe he means something that is relevant to real life where
crystals are small, diffraction weak, and background high, i.e. a
quantity that can realistically be extracted from the crap we get on our
images.
that you want I(Bragg) to be the total background-subtracted integrated
intensity under the peak at the Bragg position, i.e. what I'm calling
I(coherent). If so then it can't be the contribution from the mean
density at the same time! - seems to me that's what everyone means by
I(Bragg) (including you I thought!) so changing the definition will
cause total confusion!
