How do you model transparent materials like light guides if transmittance is 90% for 0.5mm of sheet glass but 0.3db/km for fiber optics? I am trying to model a light guide (using a commercial tool - Synopsys Lighttools).  I am getting very low transmittance using glasses in the material library, on the order of 40% loss in the visible spectrum.  Those values seem low considering the performance of high purity glass used in fiber optics - I have a datasheet from Corning showing ~0.3db/km.
I would expect many more reflections in a fiber, and wanted to know how to accurately model the glass in my light guide.  My understanding is the transmittance test is measuring the ratio of light intensity when a sample is inserted between a source and a receiver.  All the sample data is for thin sections, generally under a millimeter whereas the light guide is 100s of millimeters long.
Can someone explain what magnitude to use in optics to get a correct value?
 A: It's not clear what your glass values are. Is it 90% transmission or 40% loss. Are those for 0.5mm and a few hundred mm, respectively?
That obviously doesn't scale, so it must include reflection loss at the surface. Then we have two points and one line:
$$ 0.9 = (1-R) e^{-0.5{\rm mm}/L} $$
$$ 0.4 = (1-R) e^{-200{\rm mm}/L} $$
(Somewhat guessing the numbers...you can adjust as needed).
From that:
$$\frac{0.9}{0.4}=2.25 = e^{195.5/L} $$
so that characteristic loss length is:
$$L = \frac{195.5\,{\rm mm}}{\ln{2.25}} = 241\,{\rm mm}$$
With that:
$$ R = 8.3\,{\%}$$
There is no way to pull 0.3 db/km from those numbers.
The key for fibre optics is that the light is contained by total internal reflection (TIR) at the fiber boundary. If the medium outside the fibre is lossless, than the "T" in TIR is indeed "total". It may be nearly total for actual fiber in-use, and probably included in the 0.3 db/km.
Without clearer data, it's difficult to make further progress.
A: I may be mistaken, but this software (Synopsys Lighttools) does not seem to be adequate for modeling fiber optics because it is based on geometrical optics, whereas you may need Maxwell equations for fiber optics.
