Hi Jeremy, Anatoly, Thanks, you're absolutely right -- I got the parentheses wrong. Anatoly,
However, thickness is present in mu*t only because of the total number of absorbers. There are 1/2 absorbers in the foil with 50% holes, and thus (t*mu)_measured is equal to (1/2) * t*mu_measured.
I agree, but I think that you'd have to know your sample was 50% holes to make that work. I was considering 't*mu' to be a single thing quantity (and really meant that to stand for -ln(I_t/I_0) in the sense of the log of intensities sampled by ion chambers, not absolute fluxes). I think it should be (starting with I_t = I_0 * exp(-tmu) ) that a half full / half empty sample will have: I_t = (I_0 + I_0 * exp(-tmu) ) / 2 = I_0 * (1 + exp(-tmu) ) / 2 so that tmu_measured = -ln (I_t / I_0) = -ln( (1 + exp(-tmu))/2) = ln(2) - ln(1+ exp(-tmu)) As for whether there is a reduction in chi(k) of a factor of 2 or not, I think this would depend on the sample thickness (or, the reltive size of tmu to 1) in the portion of the sample that was non-empty. Using this corrected formula on cu foil data (that actually has an edge jump ~= 2.3, so is probably on the thick side, but is still decent data), I do see a reduction in chi(k) that is a little more than a factor of 2, with some k-dependence. Attached is an Athena project of original and "half empty" data. Am I a pessimist for not calling it "half full"? --Matt PS: I did this to make the half empty data, then read in the data file into athena. read_data(cu.xmu, group =good) set pinhole.energy = good.energy set pinhole.xmu = -ln( (1 + exp(-good.xmu) )/2) write_data(file=half_empty.xmu, pinhole.energy, pinhole.xmu)