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Optics: Why not polished surfaces reflect less than polished ones?

  1. Apr 3, 2014 #1
    Hello, I would like to ask why, when a surface is not polished, it reflects less.

    I understand that when the surface is not polished, microscopically it presents a lot of irregularities so that when the light strikes the surface it gets reflected in all directions and instead of getting specular reflection, one gets diffuse reflection.

    The problem is that then the surface is not reflecting less, it's just you that are collecting less light.

    I'm working with an integrating sphere (http://en.wikipedia.org/wiki/Integrating_sphere), so theoretically, it doesn't matter that the light get scattered and diffuse in all direction because anyway it cannot get out of the sphere.

    In practice, when I measure not polished metallic surfaces, I get values of reflectivity which are up to 70% less than polished ones.

    Obviously the light gets absorbed (I can't see any other option) but I can't justify why so much.
    Especially I don't understand why this phenomena would depend so much on wavelenght (the same sample (copper, not polished), reflectivity reduced by 65% around 400 nm, reduced by 5% around 900 nm).
     
    Last edited: Apr 3, 2014
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  3. Apr 3, 2014 #2

    Andy Resnick

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    Interesting question. Can you provide more information about your samples? For example, how are they roughened, and could there be contamination or oxidation present? Have you done any instrument calibration with say, Spectralon or Infragold surfaces? For that matter, have you actually measured the reflectivity of a pure and polished Cu surface with your instrument?
     
  4. Apr 3, 2014 #3

    UltrafastPED

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  5. Apr 3, 2014 #4
    AndyResnick: Unfortunately not, I only know the samples were generally roughened but I think there was no contamination present. The instrument (coated in Spectralon) has been correctly calibrated.
    The data from polished copper (and other metals) are just theoretical.

    UltrafastPED: I'm sorry, i didn't understand what's the connection with emissivity. I'm considering visible wavelenght, emissivity should be negligible.
     
  6. Apr 3, 2014 #5

    D H

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    In general, an unpolished surface does reflect less than does a polished surface.

    Even a highly polished metal surface is not quite optically smooth. You have to use some technique other than polishing (e.g., deposition) to get an optically smooth surface. At the microscopic level, that highly polished metal surface still looks like a mountain range. An unpolished surface? They make the Himalaya look rather flat.

    The reason you get diffuse reflection is because incoming light bounces around in that mountainous terrain multiple times before coming out. This "bouncing around" drastically increases the chance that a photon will be absorbed rather than reflected. Diffuse reflection is diffuse in two senses: it is spread around rather than concentrated and the intensity is diminished.
     
  7. Apr 3, 2014 #6
    I thought about this but i wasn't sure. This means that an unpolished surface gets warmer than a polished one?
     
  8. Apr 3, 2014 #7

    Nugatory

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    All else being the same, if it reflects less it absorbs more... and the absorbed light ends up as heat. So yes, the less reflective surface gets warmer.

    Set a chunk of black cast iron next to a piece of shiny polished cast iron in full sunlight and the difference in temperature will be quite noticeable.
     
  9. Apr 3, 2014 #8

    D H

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    Not necessarily. For a given object, absorptivity and emissivity at a given frequency must be equal. However, most substances have frequency-dependent absorptivity and emissivity. Consider a polished metal surface versus one with a matte finish. Suppose the object with a polished surface has a reflectivity of 0.8 at optical wavelengths, >0.9 at thermal infrared wavelengths. If the object with a rough finish has a reflectivity of 0.6 at both frequencies, it's the object with a polished surface that will get warmer, not the one with a matte finish. The shiny object only absorbs half as much light as the dull one, but the shiny object can't dump heat. It's too reflective in the IR.
     
    Last edited: Apr 3, 2014
  10. Apr 11, 2014 #9
    I have had to think about this issue quite a lot. I can tell you my thoughts, but I don't know how they'd be tested.

    As you've said it is generally thought that roughening the surface will just change the angular distribution of the light. However I think that it depends on the nature of the roughening. A common trick when using pyrometers it to drill a hole in the sample to get an emmisivity of one. The question is, when does a surface defect become a hole?

    It is possible to imagine that if the surface finish resemebles lots of holes it will have a reduced reflectivity as light that enters a cavity will be reflected more times and more of it will get absorbed. Also, depending on the size of these features, different wavelengths will be affected differently.

    I admit it is uninutitive that multiple reflections could have such a large effect on reflectivity!

    Edit: here is a link to the hole drilling stuff http://www.reliableplant.com/Read/1...-difference-between-apparent,-actual-ir-temps
     
    Last edited: Apr 11, 2014
  11. Apr 12, 2014 #10

    stevendaryl

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    Could it be just that in the case of an irregular surface, some of the light is reflected multiple times by the surface? That is, the light hits one part of the surface, and instead of being reflected away, it is reflected onto another part of the surface? If the same light hits the surface multiple times, it has more opportunity to get absorbed.
     
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