Wavelength of photon density wave?

Question

How would you calculate the wavelength of a photon density wave if:

f (frequency)=200 MHz
ua (absorption coefficient)=0.1 cm^(-1)
us' (reduced scattering coefficient)=10 cm^(-1)
n (index of refraction)=1.37

I've look everywhere. I just need something to get me started.

Thanks!

Answer 1

i saw this question yesterday as well, and noted sincere efforts of answerer - 2 (yesterday - perhaps this person solved -3 -2 -1 0 number question faster than me).

I think for this question you go into the details of Bose Einstien condensation of radiation. Photon wave densities have been dealt with therein only. Your reduced scattering coeffi is just extinction coeffi in new terminology - note perhaps word REDUCED is used for h-cross (ie parameter scattering coeff divided by 2 pi).

One beer - law, i guess, directly answers your question. Link below. along with some links to BE condensation.

yesterday, I had searched one good article on net about complex refrative index m' + i m" - it is around this only

http://en.wikipedia.org/wiki/Extinction_coefficient


http://en.wikipedia.org/wiki/Beer-Lambert_law

http://www.fisk.edu/~aburger/Published03_06/Introduction/Optical/Extinction_coefficient/extinction_coefficient.html

Answer 2

Hey, didn't I try to answer this exact question for you yesterday? The problem's still the same: don't know what a "photon density wave" actually is. Today I called an old friend who's a physics professor at Cal Tech. and he doesn't know what it is. Perhaps there's a more common (older) term for this phenomenon? It seems obvious we're talking about an electromagnetic wave propagating through some sort of media.....In quantum mechanics there's a "probability density wave" but this isn't even close because the variables you've listed are from classical electrodynamics...........I went through my 17 classical electrodynamic texts a second time and still got nowhere.

Here's the best I've got (same as yesterday):
Dividing the speed of light in vacuum by the index of refraction gives the speed if light in the media:
(2.9979245 X 10^8) / 1.37 = 2.18826 x 10^8 m/s
The frequency remains unchanged regardless of which media the wave passes through, so the wavelength is found by simple division:
((2.18826 x 10^8) / (200 x 10^6) = 1.094133 meters

While the absorbtion coefficient and the scattering coefficient do effect the wave's energy losses in the media, they don't effect the wave's wavelength or frequency. For this reason, I stand by my calculations until the proper meaning of "photon density wave" becomes known. It could easily be that photon density wave is "new-speak" for wave-packet or some such term. If so, my answer is correct. It's also possible that I'm wrong, because I'm only human and don't know everything.............

Answer 3

A photon is a quantum of an EM wave...a "packet of potential" in touch in making an electromagnetic wave what that's. EM waves as you recognize them are superpositions of multiple photons. Even in simple terms one wavelength. there won't be any answer to "what does a photon appear like? "...in simple terms think of of it like the metallic ball version of light.

Answer 4

Have you tried your physics book or class notes ?