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seen Jun 12 at 14:49

Jul
2
awarded  Curious
May
27
revised How to specify BRDF measurement
Added information about Zemax capabilities
May
27
comment How to specify BRDF measurement
@CarlWitthoft: If I do not measure polarization, how can I distinguish between high and low reflectivity of s and p polarizations close to Brewster angle? The system is working with a laser, so polarization is relevant there. BTW, thanks for the link, I sent them the request.
May
27
comment How to specify BRDF measurement
@StefanBischof: We need to measure the construction materials, which have been chosen for their mechanical, not optical properties. Moreover, to make things cheep, we do not specify the surface properties to the suppliers. That's why we do not know much about them a priori. An extruded tube can have a very smooth surface while a milled part can be quite rough.
May
26
asked How to specify BRDF measurement
May
7
accepted Anodized aluminium reflectivity at 10.6 um
May
4
comment Anodized aluminium reflectivity at 10.6 um
Thanks for the comments. This does not really answer my question though. And yes, I am talking about a laser application: the body of the system is made from aluminium, and since I am talking about a multi-kilowatt laser beam, I need to model radiation load on the Al parts, as well as the amount of reflected light. That's why I am interested in Al2O3-coated aluminium reflectivity.
May
2
comment Anodized aluminium reflectivity at 10.6 um
I realize now that I did not take surface reflection into account, which is probably responsible for 15 % loss at short wavelengths in your plot: 2 * ((1.75-1)/(1.75+1))^2 = 0.15. But the result (huge absorption at >8 um) stays the same.
May
2
comment Anodized aluminium reflectivity at 10.6 um
From your plot. Transmission is 3 %. $I'=I_0\exp(-\alpha t)$, $0.03 = \exp(-\alpha \times 0.5)$, $\alpha = -2\ln(0.03) = 7.01$.
May
2
comment Anodized aluminium reflectivity at 10.6 um
Interesting. Thanks for explaining about the name, I was confused by that. I've done some math based on your curves. At 7.5 um, for 0.5 mm thickness, transmission is ~3 %. Then the absorption coefficient is ~7 1/mm, leading to a loss of 25 % for double-pass in a 20 um layer. For λ = 10 um it will be even higher, unless there is a local transmission peak there. Do you agree with this understanding?
May
2
asked Anodized aluminium reflectivity at 10.6 um
May
2
comment Light scattering vs roughness
The simulation you linked to is a nice illustration. However it is unclear where they get the results from. It would be nice to see the code producing those curves.
May
2
accepted Light scattering vs roughness
Jun
5
revised Light scattering vs roughness
Added link to Disney's
Jun
5
asked Light scattering vs roughness
Apr
26
comment Software for geometrical optics
@BandGap: Your last comment is actually wrong. Geometrical optics is the same as ray optics (en.wikipedia.org/wiki/Geometrical_optics). It is called so because simple geometry formulae are used to calculate the path of each ray through an optical system. Diffraction effects indeed cannot be treated this way. A wave optics method (there are many of them) would not use the paradigm of a ray at all.
Nov
20
awarded  Editor
Nov
20
revised Are circular polarizations a basis for any light polarization?
Corrected LaTeX code for subscripts
Nov
20
answered Free Optics Simulation Programs
Nov
19
suggested suggested edit on Are circular polarizations a basis for any light polarization?