Scott,
Let me try again.
I think I understand you now, and I think I agree completely.
First of all, perhaps you missed that after my first post, I acknowledged that my statement in the first post was mistaken? In fact, I later said:
So there's nothing wrong with the arguments in Lu and Stern, Scarrow, etc.--it's the notion I had that we use multiple layers of tape to improve uniformity that's mistaken.
The Lu and Stern paper, which I now agree with, comes up with an equation I have attached as a png file.
That equation is derived for spheres of diameter D in a monolayer.
I think it's not that important, but Equations 1 through 6 in Lu and Stern discuss the absorption by a single spherical particle, not a monolayer of particles. Absorption by a monolayer would need to include finite packing fraction (pinholes), which this section does not discuss. It only discusses the effect of the varying thickness of a sphere. Incomplete coverage of a layer is in section 4. I am not entirely sure I understand their section 4, and the notation a bit confusing ("f" is not given in Section 2 -- I assume f=Xeff/Xreal, though I am not sure).
Lu and Stern then make the following claim:
Finally, the attenuation in N layers is given by (I/I0)^N, where I is the transmitted intensity through one layer. Chi_eff for N layers is then the same as for a single layer since N will cancel in the final result.
That's the statement I initially thought was incorrect, and thought was equivalent to a non-random stacking of layers. Since I'm now convinced that the statement is correct, I don't want to get bogged down in a discussion of why I still find it a bit puzzling.
Other than my first post, no one else in this thread has yet challenged the Lu and Stern statement, and some, such as Matthew Marcus, have supported it.
The statement can be rephrased to say that, for a given particle diameter D, the ratio of chi_eff to chi_real is independent of the number of layers; that is, the distortion due to nonuniformity is not reduced by adding additional identical layers.
I read their "attenuation of N layers .." to simply being saying that the integrated attenuation over a layer is multiplicative. And, indeed, that multiplication of exponentials is what makes the transmission independent of the number of layers. But as far as I understand, this section is all in the "thin layer" limit. So their statement (and your rephrasing) may be correct only over a limited range of mu and particle size. Again, I think this section is the most confusing part of the manuscript.
I found this conclusion so startling that at first I rejected it. Several others in this thread indicated that they too tended to think that additional identical layers would reduce distortions in the spectrum due to nonuniformity.
I agree. I definitely thought this. I now agree that although the thickness is made more uniform by adding layers, the transmission through that thickness is not decreased, and the measured mu is independent of the number of layers. For a given thickness, a sample of many thin layers not only has a more uniform thickness than a sample with fewer, thicker layers, but also gives less (and so better) transmission.
Take a sample that has a fraction p of pinholes, with the remaining 1-p of the sample being of uniform thickness x. In that case,
I = p I_o + (1-p) I_o exp(-mu x)
mu_eff = ln (I/I_o) = ln [p + (1-p) exp(-mu x)]
Now take two layers of that type, with the pinholes randomly distributed. A fraction p^2 will still be pinhole, a fraction 2p(1-p) will have thickness x, and a fraction (1-p)^2 will have thickness x^2.
So now
I/I_o = p^2 + 2p(1-p) exp(-mu x) + (1-p)^2 exp(-2 mu x)
mu_eff = ln[p^2 + 2p(1-p) exp(-mu x) + (1-p)^2 exp(-2 mu x)] = ln[p+(1-p)exp(-mu x)]^2 = 2ln[p + (1-p) exp(-mu x)]
And that is exactly twice the mu_eff of the single layer--i.e., after normalization it will be identical. Any distortions caused by pinholes will be identical, despite the fraction of sample having pinholes all the way through dropping from p to p^2.
OK, I agree. I think I understand what your point is now. I'm pretty sure you wouldn't disagree, but I just want to reiterate that in the important comparison of 2 layers of thickness t and 1 layer of thickness 2t, there is definitely an important difference. By reducing the thickness of each layer, one reduces the mu dependence of the measured mu. I think you would say is that "many thin layers" versus "few thick layers" is "thickness effect", and should be thought of separately from pinhole effects. I guess I never really had clearly separated those. --Matt PS: If I had calculated mu in my earlier simulations, this would have been more obvious. Attached is a revised simulation code, that does calculate i1_measured and the effective mu. Typical results are: ### total thickness ~= 1, 10% pinholes: ### #nlayers | % holes | t_ave | i1_meas | mu_eff |mu_eff/t_ave| # 1 | 10.0 | 0.900 | 0.4311 | 0.8414 | 0.9349 | # 2 | 10.0 | 0.900 | 0.4171 | 0.8744 | 0.9715 | # 4 | 10.0 | 0.900 | 0.4115 | 0.8879 | 0.9865 | # 10 | 10.0 | 0.900 | 0.4085 | 0.8954 | 0.9949 | ### layer thickness ~= 0.1, 10% pinholes: ### #nlayers | % holes | t_ave | i1_meas | mu_eff |mu_eff/t_ave| # 1 | 10.0 | 0.090 | 0.9144 | 0.0895 | 0.9949 | # 2 | 10.0 | 0.180 | 0.8360 | 0.1791 | 0.9949 | # 4 | 10.0 | 0.360 | 0.6990 | 0.3582 | 0.9949 | # 10 | 10.0 | 0.900 | 0.4085 | 0.8954 | 0.9949 | Here, the ideal mu with no pinholes should be 1. Scott's point is reflected here in the fact that adding layers does not improve mu_eff / t_ave, even though it does improve mu_eff. I think the most important point is that one wants thin layers.