Reflection of light

  • #1
As I understand that the electrons absorb the radiation and then re-radiate a part (or color) towards the incoming radiation is called reflection. However, if the frequency of radiation absorbed and re-radiated is dependent on the electron state, how does a faithful reflection of all visible frequencies result as in the shiny metal surface?
 

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  • #2
HallsofIvy
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Because, in a metal, many of the electrons are 'free', not tied to a particular electron state nor even to a particular atom. That is also why metals are good conductors of electricity.
 
  • #3
Andy Resnick
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As I understand that the electrons absorb the radiation and then re-radiate a part (or color) towards the incoming radiation is called reflection. However, if the frequency of radiation absorbed and re-radiated is dependent on the electron state, how does a faithful reflection of all visible frequencies result as in the shiny metal surface?
I have a real problem with the concepts used here- light propogation should not be considered a sequence of absorption/re-emission events, even though that conceit is commonly used.

Your confusion stems, fundamentally, from this poor conceptual analysis. Absorption is an incoherent process, and destroys information contained in the electromagnetic field. Thus, your question follows naturally.

Scattering of radiation is a better concept to use. It can either be a coherent or incoherent process- coherent scattering is how to describe reflection. Scattering theory uses the index of refraction to describe the process. Metals, or conductors in general, have a complex refractive index. Matching the boundary conditions for E/D and B/H at the interface leads to a strongly reflected component and vanishing transmission component. The presence of free electrons in conductors means the refractive index is very broadly peaked, leading to a wideband reflectivity.

Reflections can also be specular or diffuse (or a combination). Diffuse scattering is typically attributed to inhomogeneities in the surface- a rough surface, usually.
 
  • #4
Matching the boundary conditions for E/D and B/H at the interface leads to a strongly reflected component and vanishing transmission component.
While I am trying to understand other statements, could you please explain the above.
 
  • #5
Andy Resnick
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The boundary conditions are:

1) the normal components of eE and B are continuous across an interface
2) the tengential components of B/m and E are continuous across an interface

where e is the permittivity, m the permeability. The refractive index is constructed from e and m. Look up "fresnel coefficients" to see a complete derivation (typically for dielectrics- put in complex e and m to look at conductors)
 

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