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Dec
29
comment Fermi level in equilibrium and non-equilibrium situations
Possible duplicate: physics.stackexchange.com/questions/68162/…
Dec
19
answered Understanding Wikipedia's “Semiconductor Band Structure” diagram where the bandgap appears to increase with increasing density of states
Dec
5
comment Lasers and Collimation
You need the correct definition, not just an unambiguous definition. For example, if you say "'neutron' is a term for any neutral particle", it's an unambiguous definition, but it's not the correct definition. :-P You can't call something a "plane wave" unless it has planar wavefronts (note the plural; one planar wavefront is not enough) AND (related to that) it has constant intensity in the lateral directions. The Naik quote is consistent with that. Again, you can look it up in any textbook or ask any expert what "plane wave" means.
Dec
5
comment Lasers and Collimation
"Plane wave" is a technical term with a standard definition that you can (and should) look up in any textbook. You'll see that the definition of "plane wave" is not what you think it is. (Sure, if you look around enough, you can find someone who has misused the term "plane wave". But I promise, the term "plane wave" is completely standard and unambiguous.) I'm glad you seemingly understand the physics of light propagation, but you do NOT know the terminology for talking about it, e.g. the standard definitions of "plane wave" and "collimated".
Dec
3
comment Lasers and Collimation
I hope you are aware that a Gaussian beam is not a plane wave anywhere, full stop. It is not a plane wave at the beam waist, it is not a plane wave anywhere else. Nor is it "collimated" at the beam waist, or anywhere else. I think you're confused about the definition of "plane wave", and "collimated".
Dec
1
comment Lasers and Collimation
I wrote "it's impossible to put 100% of [a laser's optical] energy into a perfect gaussian beam". This is obviously true because lasers have finite lateral size while gaussian beams do not. Are you really arguing that this statement is not true? Your comment above ("...not part of the beam") sounds like "100% of the energy goes into the gaussian beam, if you ignore the energy that does not go into the gaussian beam". Are you really arguing something so silly? If not, I don't understand you. The paper you linked to references a ~99.3%-efficient coupling, which they rounded to 100%.
Dec
1
comment Lasers and Collimation
@LukeBurgess - As you go away from the center of a gaussian beam, the intensity reduces exponentially, but never reaches zero in a finite distance. Therefore a finite-sized source or cavity CANNOT be 100% mode-matched to a gaussian beam. Maybe you can find a source that puts 99.99% of the energy into a perfect gaussian beam, but not 100%.
Dec
1
comment Lasers and Collimation
@LukeBurgess -- You have a funny definition of "collimated" if a wave can be "collimated at a single z-position". The whole point of collimating a beam is that it is supposed to stay collimated as it travels, until it hits the next lens. A beam of finite size can never be perfectly collimated. You can prove it by thinking about photons, or by working through Maxwell's equations, or by thinking about Huygens' principle, or by thinking about diffraction, or by thinking about fourier transforms and plane-wave decompositions. (Actually, these all amount to the same thing.)
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Nov
27
revised Definition of mean free time in the Drude model
deleted 89 characters in body
Nov
27
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Nov
27
comment Photons from stars--how do they fill in such large angular distances?
@ArmanSchwarz - OK, I put the equation
Nov
27
revised Photons from stars--how do they fill in such large angular distances?
add equation
Nov
27
answered Definition of mean free time in the Drude model
Nov
26
awarded  Nice Answer
Nov
26
comment Photons from stars--how do they fill in such large angular distances?
This answer is sort of missing the point. It is correct to calculate a starlight photon flux, multiply it by the photon absorption cross-section of rhodopsin, and thereby calculate rhodopsin's photon absorption rate (as if the photons were bullets, randomly distributed in all directions). You are insinuating that there's something wrong with this procedure, but there's not. It gives the right answer! (I mean, the right answer for calculating light absorption rate.)
Nov
26
answered Photons from stars--how do they fill in such large angular distances?
Nov
24
awarded  Good Answer