Reflection of light in metallic surface and glass

  • #1
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Hi All,
Due to the differences in their conductivities, but considering that both the metallic and the glass plane surfaces produce mirror images of objects, is there any difference in the explanation of the mechanism of reflection of light in these materials?
 

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  • #2
Charles Link
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At normal incidence, there is the formula for the Fresnel reflection coefficient ## \rho=\frac{n_1-n_2}{n_1+n_2} ## and the energy reflection coefficient ##R=|\rho|^2 ##. (Here ## n_1 \approx 1.0 ## for air). I once had an Optics professor tell us the high reflectivity of metals is due to a very large imaginary part in the index of refraction ## \tilde{n}_2=n_{2r}+in_{2i} ##. (This imaginary part is responsible for the rapid attenuation in the material, so that the metal is almost completely non-transmissive). The index ## \tilde{n}_2 =\sqrt{\epsilon} ## where ## \epsilon ## is the dielectric constant ( normally frequency dependent). (And the dielectric constant ## \epsilon ## can also be a complex number). ## \\ ## Perhaps @vanhees71 can add to this explanation, because I think a complete answer requires some additional fine detail.
 
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  • #3
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Thank you, Charles.
Would it be possible to build an explantion having as basic elements the conduction band, free electrons, valence band and the expected behavior/response of electrons when an EM wave enters the material? I guess this language I am proposing now is more inclined to the quantum optics side, isn´t is?
 
  • #4
Charles Link
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Thank you, Charles.
Would it be possible to build an explantion having as basic elements the conduction band, free electrons, valence band and the expected behavior/response of electrons when an EM wave enters the material? I guess this language I am proposing now is more inclined to the quantum optics side, isn´t is?
I would imagine it would be quite possible and has already been done, by working with the dielectric function ## \epsilon(\omega) ## from a quantum optics viewpoint, but this is beyond my level of expertise.
 
  • #5
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Thank you anyway, Charles.
 
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  • #6
vanhees71
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Well, you need just classical electrodynamics with the usual constitutive relations to evaluate the transfer matrix that relates the incoming beam to the outgoing (reflected and transmitted) beams. They also hold to an excellent approximation for single photons, i.e., you don't need full-blown in-medium QED to treat at least dielectrics. Some aspects of metals and also plasmas, however may need a quantum treatment.

For the classical theory see

M. Born, E. Wolf, Principles of Optics, Cambridge University Press

A very nice summary of how to use the transfer matrix in quantum optics (however without giving the detailed derivation of it for concrete realizations of beam splitters) can be found here:

http://www.qolah.org/papers/leshouches.preprint.pdf
 
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  • #7
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I appreciate very much your contribution, Vanhees71. I have the Born and Wolf (but I just have read few pages of it) and some of the leshouches material is also at hand. Going a bit further and daring to explore you a little more, is there any consistent explanation that could be given to a college audience?
 
  • #8
vanhees71
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Hm, I'm not so sure about the college-level of sophistication concerning the math. You need the plane-wave solution in and out of the medium and the boundary conditions at the boundary surface of the two media to get the Fresnel equations to finally get the coefficients for the reflected and transmitted parts of the wave. I'd consider this a pretty tough task. It should be suited more for advanced undergrad physics majors.
 
  • #9
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Let me pose the question in another way. Would it be correct to say that, regardless of being a conductor or a dieletric, the electrons of the atoms of the first layer of both surfaces (plane ones) are able to aborb and re-emit the incoming light wave in such a way that it permits the production of an image with practically no distortions?
 
  • #10
sophiecentaur
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aborb and re-emit
That, on its own, would not give coherent reflection (no proper image would be formed). An individual electron, absorbing and re-emitting a photon, would introduce a random delay so the sum of all re-emitted electrons would not have phase coherence. The interaction has to be with the structure as a whole. You cannot get coherent reflection at the boundary of a gas where the atoms are all independent.
 
  • #11
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That, on its own, would not give coherent reflection (no proper image would be formed). An individual electron, absorbing and re-emitting a photon, would introduce a random delay so the sum of all re-emitted electrons would not have phase coherence. The interaction has to be with the structure as a whole. You cannot get coherent reflection at the boundary of a gas where the atoms are all independent.
But wouldn´t it be correct to say that once they are forced by an incoming EM field they will respond with the same frequency of the incoming patern? And, from this accelerative response, an EM field would be generated by these electrons possessing the same information content of the source (frequency paterns, phase).
 

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