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What determines which metals will reflect or absorb electromagnetic waves?

  1. Jul 30, 2012 #1
    Why does nickel reflect infrared, but gold does not?

    I've seen devices that shine infrared at a metal surface and measure the reflectance. They're mainly used for looking at thin film coatings.

    For example, you could have a metal part made of nickel, and thinly coated in gold. The gold thin film will not reflect the infrared, but the nickel will. This allows you to see how clean your surface is, because if there are other contaminants on the metal part, the reflectance will be much lower.
     
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  3. Jul 30, 2012 #2

    CWatters

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    I thought gold did? Isn't that why space craft are wrapped in the stuff? To reflect the suns heat?
     
  4. Jul 30, 2012 #3
    I'm guessing it reflects some spectra but not others. I was told that the tool I saw was calibrated to pass through gold but be reflected by nickel.
     
  5. Jul 31, 2012 #4

    DrDu

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    I think the reasoning is different. A nickel surface is covered with nickel oxide, while a gold surface is an atomar surface to which pollutant gasses will only be adsorbed reversibly. Hence you can infer from the change of reflectivity something on the properties of the adsorbed gasses.
     
  6. Jul 31, 2012 #5
    This could explain the spaceship case.

    What about the case I was talking about, with a nickel body covered in a gold thin film? In that instance there wouldn't be much, if any, oxide formation on the nickel. It should basically be a pure nickel source reflecting the infrared. Does the thickness of the metal matter? For example, if the gold were 1 cm thick, instead of one micron thick, would the gold then reflect or absorb the infrared?
     
  7. Jul 31, 2012 #6

    DrDu

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    Probably you are using gold plated nickel because a massive mirror from gold would be too expensive? As long as we don't know better the device you are talking about, it will be difficult to help you more.
     
  8. Jul 31, 2012 #7

    Drakkith

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  9. Aug 1, 2012 #8

    ehild

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    Solid surfaces reflect most that radiation they also absorb. Gold is the best reflector in the infrared. Mirrors for optical instruments are coated by gold.

    Because of the high reflection, only a little part of the incident light energy enters into the metal. Travelling in the metal, the intensity of the wave exponentially decreases with the distance travelled: I=I0e-αd. α, the absorption coefficient is related to the imaginary part of the complex refractive index, κ, and the wavelength λ: κ=αλ/(4π).

    The reflectivity of a surface is determined by the refractive index. If its real part is n, imaginary part is κ:
    R=[(n-1)22]/[(n+1)22]

    Metals have high n and κ values in the infrared range so the surface reflectivity is close to 1.
    The reflectance of gold is less in the visible range, and it changes with wavelength. The optical properties and their wavelength dependence are determined by the electronic structure of the metal and the arrangement of the atoms in the metal crystal or layer.


    ehild
     
    Last edited: Aug 1, 2012
  10. Aug 1, 2012 #9

    DrDu

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    That is not true for highly reflective substances like metals. As most of the light is reflected, hardly any energy is absorbed. In the case of gold, the yellow colour is due to reduced reflection in regions where there are absorption lines due to transitions from d-orbitals to the conduction band.
    In the infrared the absorbance α is very small.
    Stated differently, the value of absorbance α is quite small although κ is large. However the large value of κ is due to the large value of λ (respectively n).
     
  11. Aug 1, 2012 #10

    ehild

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  12. Aug 1, 2012 #11

    DrDu

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    I just wanted to say that absorption will rather lead to a decrease of reflection, specifically in the optical region:
    The reflectivity of gold is described (at least somewhat below the plasma frequency) well by the Drude formula, see e.g.
    http://optics.hanyang.ac.kr/~shsong/27-Metals.pdf
    Parameters can be found here:
    http://www.wave-scattering.com/drudefit.html
    According to this formula the n and κ are well approximated (from the mid IR up to the visible) as
    [itex] \kappa=\omega_\mathrm{P}/\omega[/itex] and [itex]n=\kappa \gamma/\omega [/itex].
    In the mid to far IR, both n and κ are large so that the reflectivity is almost 1.
    However near the visible, κ is not very large but n is very small as long as the parameter γ, which describes damping, is small. Then there is again almost perfect reflection. But if absorption, ie a large value of γ, reflectivity will decrease.
     
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