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Sorry for the late reply. I don't monitor this forum and your question would have been more apparent had it been in the 'optical photon physics' forum.
> I'm wondering what's the phisical meaning of REFLECTIVITY and EFFICENCY?
From the Application Developers Manual:
In the case of an interface between a dielectric and a metal, the photon can be absorbed by the metal or reflected back into the dielectric. If the photon is absorbed it can be detected according to the photoelectron efficiency of the metal.
But I admit this can easily be missed. Therefore, the efficiency doesn't change anything about the reflectivity. It just flags a fraction of the absorbed photons (1-reflectivity) as having been 'detected'. Please, see also Figure 5.1 in the manual.
> I thought the reflectivity represents the fraction of incident > electromagnetic power that is reflected at an interface
Yes, for dielectric_metal surface. For a dielectric_dielectric surface, strictly speaking the reflectivity is not the reflection coefficient but 1 minus the absorption coefficient.
> Then the larger the > reflectivity, more photons will goes through the surface.
The larger the reflectivity the more photons will be reflected (and not absorbed).
> When I set the reflectivity to 1, the result shows maximum, and it seems > the reflectivity and transmission is calculated by the given > RefractiveIndex of the two material.
When you set the reflectivity to exactly ONE (for a dielectric_dielectric surface) - which is the code's way of telling that no user specified reflectivity has been provided - then the program will NOT use it. You should be setting it to anything but exactly ONE if you want it to overwrite the default Snell's Law.
Hope this helps, Peter