I am interested to know if in a materials which has no absorbance at a wavelength "L", but with strong Mie resonance at this wavelength "L" due to adequate surface structuring, where does the photon ultimately go? Is it somewhat absorbed? Or will it exit the system? Thank you for your suggestions.
In the case that material has some absorption part could be absorbed. Generally, scattering on particles cause that electromagnetic waves (photons) change direction only. Particle shape, and size and index of refraction determines the spatial distribution for EM wave with selected wavelength. Following Mie theory the scattering is "resonant" if particle size is close to wavelength. The consequence is attenuation of the EM wave along desired propagation direction. Observer (receiver) always has limited field of view so scattered photons are lost form the field of view only.
Dear Prof. Livada,
Thank you for taking the time. My apologies for asking further clarifications. Do you mean that the resonator does not influence the absorption/trapping of photons, but "redirect" them, which can be interpreted as a decrease in both the transmission and reflection in, let's say, the observation direction.
According to Mie model of Light scattering on a particles system, the light with a wavelength comparable to the particle size is diffused in a non-isotropic way, and most of the light is not back-reflected towards the emitter. Therefore a measurement of backscatterd light close to the emitter location gives a low (or very low) intensity, but no absorption has occurred by the target particles.
Dear Victor,
I am not involved in academic (teaching) so do not have title "Prof.". My answer is based on some experience regarding radiation attenuation on particles and its influence on propagation and did not go deep in theory about interaction of radiation and matter, so I am just telling how Mie scattering is sensed from practical point of view or what is usable consequence. Scattering caused direction changing is dominant process viewable for observer so absorption part, if any, could be neglected.
It seems that you are interested in deep understanding what happens with Photon (EM wave) and that is out of my competence. Hope that someone more competent will appear to discuss this topic.
Best regards,
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