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Does a red-hot mirror reflect light the same way?

  1. Jan 6, 2012 #1

    cmb

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    If you heat a reflective surface to incandescence, let's say it's emission is peaking at 600nm, will it then reflect 600nm light in the same way it did when it was cold?
     
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  3. Jan 6, 2012 #2

    xts

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    If it is a good mirror (reflecting nearly all light) it won't glow - it's emission/absorption coefficient is close to zero. And it will still reflect the light as previously.

    The only cause to change reflection coffeicient may be changes to mirror surface, but, as I guess it was not your concern. As long, as the surface is intact, the mirror works.

    Actually, there are lots of mirrors working for deep infrared at room temperature.
     
  4. Jan 6, 2012 #3
    Optical properties of materials, generally speaking, depend on temperature. For example, reflectivity of metals, as far as I know, decrease at temperature rise. Concerning multi-layer dielectric mirrors, their reflectivity may also change at high temperature. First of all due to change of optical thickness of the layers, but also it may be due to absorption of initially transparent layers and may be (theoretically) due to crystalline modification of the material.
     
  5. Jan 7, 2012 #4

    Andy Resnick

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    Start with conservation of energy: reflection+ transmission + absorption = 1 = reflection + emission, for a perfectly opaque material (and absorption = emission).

    If the coefficients do not depend on temperature, then you can see that the reflective surface will reflect the same way regardless. If the coefficients do vary with temperature, then YMMV.

    In practical terms, if the emitted light occupies the same spectral region as reflected light, the reflection will have a noise component due to the thermally emitted light.
     
  6. Jan 8, 2012 #5

    cmb

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    Yes, naturally I recognise the issues regarding actual modifications of the surfaces due to the heat, and of course the 's/n' will change, but I was principally wondering whether the process of reflection, itself, is 'interfered' with when an atom gets hot enough to generate its own light. The assumption is therefore where 'all else being equal'.

    The point about a highly reflective surface not radiating light is interesting. I had not thought about emissivity as an issue in this question. This begs a far more interesting question - if a piece of rough steel (say) can be heated to incandescence where it glows fiercely then I am presuming an equivalent piece of the same steel but polished to a mirror shine won't glow as intensely at the same temperature?

    If so, this shows up big holes in my comprehension of black body radiation - [why] is the surface finish of a black body a factor in the intensity of its radiation?
     
  7. Jan 8, 2012 #6

    Andy Resnick

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    This is an excellent question. The best way I can think of to answer is to point out that the surface is an interface between the field within the object and the free-space field. Thus, the properties of the surface control how the two fields couple together- either efficiently (transmission) or inefficiently (reflection).
     
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