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Absorbing an electromagnetic wave

  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data.
    On the course, I was asked a question: What is less absorbing electromagnetic wave: metal or glass? Explain why.

    3. The attempt at a solution
    Now, i'm not sure if i did it correctly. I answered that glass is less absorbing electromagnetic wave due to it's structure and that metal is a good conductor, so it will absorb the wave and transform it into the heat. But still i'm not certain about my answer, because it is also said that electromagnetic wave hardly penteratres material.

    I'll be very grateful for every answer.

  2. jcsd
  3. Feb 1, 2017 #2


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    You can see through a piece of glass, but you cannot see through a piece of metal. Why?
    You can see an image of yourself very well in a silvered mirror, but only faintly in a piece of glass. Why?
  4. Feb 1, 2017 #3
    First, even in the most ideal description the absorption of both metals and dielectrics varies from zero to 100% depending on the wavelength. Ok, so we make an assumption that they mean in a band where the glass is transparent and the metal reflects. In those bands in the most ideal description neither absorbs much light. The dielectric transmits and the metal reflects. However an ideal dielectric absorbs no light below the band gap (no available states for the energy) and a Drude metal absorbs absolutely no light only at absolute zero. So it is reasonable to say a metal absorbs more because there are states the electrons can occupy upon absorption. This is the answer you gave, and to tell the truth I think it is the right answer and pretty much true in the real world.

    Nevertheless, this simple idea shouldn't be carried too far. Real dielectrics and metals are not ideal. The absorption in real glasses and metals is often dominated by the non-ideal parts particularly at specific wavelengths. Real metals in particular are notoriously far from any ideal description. But even dielectrics have a lot going on beyond that the simple description.
  5. Feb 1, 2017 #4
    Actually I shouldn't say an ideal dielectric absorbs no light. The Fermi distribution determines the population of the conduction band. Even in an ideal case there will be some absorption, but the population is very small.
  6. Feb 1, 2017 #5
    "You can see an image of yourself very well in a silvered mirror, but only faintly in a piece of glass. Why?"
    Because metal better reflects magnetic waves? I'm not sure.
    That's the point that worry me the most, because i don't know if reflection is kind of absorption or it's completely opposite phenomenon.

  7. Feb 1, 2017 #6


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    Yes, metal reflects EM waves much better than glass and transmits none of it as long as it is reasonably thin. Glass transmits much more radiation than it reflects. However, reflection and transmission are not absorption because they involve radiation that leaves the material one way or another. Absorption has to do with radiation that stays in the material. The energy of that radiation is mostly converted to heat. Bottom line: If you place a piece of shiny, unpainted metal, e.g. aluminum foil, and a piece of glass in sunlight, do you think will get hotter?
  8. Feb 1, 2017 #7
    Reflection and absorption are not the same. In reflection the energy is still in the field propagating as light. It just reversed direction. Absorption is when the light goes away and is converted to some other energy in the medium.

    When an electromagnetic wave encounters a boundary between media it will transmit and reflect and be absorbed all at the same time and to differing degrees depending on how the medium changes across the boundary. What changes is how much the charge will polarize in response to the applied field. We characterize that with the dielectric constant. In metal where the charge is nearly free to respond to the field the dielectric constant is very high. In a gas like air where the density is low so the polarization is weak the dielectric constant is practically 1. How much of the wave reflects and how much continues across the boundary is a simple function of the two dielectric constants. Look up Fresnel coefficients.

    If you are more familiar with EE this is the same as an impedance mismatch. You can also do the same thing with water waves. Make waves in a water tank that goes abruptly from deep to shallow, and the waves will demonstrate all the properties of Fresnel reflection and refraction including the Fresnel splitting between transmission and reflection, Snell's law of refraction, the change in speed of the waves which is the same as the dielectric constant, etc.
  9. Feb 1, 2017 #8
    Ok, now i can see the difference. I think that foil will get hotter, because it absorbs better the electromagnetic waves.
    Thanks for your help.
  10. Feb 1, 2017 #9


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    you sure about that ?
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