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A Emissivity and reflection

  1. Aug 9, 2016 #1
    Hi Guys
    Im studying some thermal properties for my masters and while reading an article a weird thing happened:
    As far as i know, a good emitter must be a good absorber (since emission happens "after" absorption")
    Combining Fresnel equation for a medium that absorbs radiation and Kirchoff law
    ρ= (n1-n2)2+k2/((n1+n2)+k2)
    and 1-ε=ρ (at equilibrium)
    We find that
    ε= 4n/((n+1)+k2) assuming that medium 1 has n=1
    So, in the end i was asking myself, the more medium 2 absorbs (higher K values), smaller the value of ε is. But, higher absoption should increase emissivity
    Can anyone help me to explain this phenomena
    ps: sorry for bad english, im brazilian
    ps2: ε stands for emissivity, ρ for reflection and K, for absorption
     
  2. jcsd
  3. Aug 9, 2016 #2

    Charles Link

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    Materials such as metals are good reflectors and they also quickly attenuate any light/electromagnetic waves that get into their interior. I believe it is their large "k" value that makes them good reflectors. And yes, the metals have low emissivity. I think in general, materials that are good absorbers that have high emissivity are not characterized as being dielectrics, which is what the equations that you are working with assume. ...editing... and what I think is a typo: Your terms in the denominator should be squared: ## (n_1+n_2)^2 ## and ## (n+1)^2 ## . Also, the numerator in the first line for ## \rho ## needs a parenthesis: It should read ## \rho=((n_1-n_2)^2+k^2)/((n_1+n_2)^2+k^2) ##.
     
    Last edited: Aug 9, 2016
  4. Aug 9, 2016 #3

    DrDu

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    Absorption is proportional to ##\epsilon k##, not k alone.
     
  5. Aug 10, 2016 #4
    But the first equation i presented, isn´t it for a perfectly smooth interface between a dieletric and a second medium that can absorb radiation?
    And also, isn´t absorvity β = 4πK/λ assuming there is no scattering event
     
  6. Aug 10, 2016 #5

    DrDu

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    Sorry, my reply wasn't correct. I wanted to say that absorption is proportional to the imaginary part of the relative dielectric constant ##\epsilon_r##. But ##\epsilon_r=(n+ik)^2##, so the imaginary part of ##\epsilon_r## is 2nk. As Charles Link already mentioned, metals are an example of materials with high coefficient of extinction k but low absorbance. A normal metal can't absorb much of the incident light because it reflects most of it. A peculiar exception are gold and copper. These two metals absorb light somewhere in the green region of the spectrum and this has the effect that reflectivity is lower for these wavelength, which gives these metals their specific colour.
     
  7. Aug 10, 2016 #6

    Charles Link

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    Just an additional item or two. At the interface between two dielectrics, the reflections are specular, i.e. angle of incidence=angle of reflection. The equations for reflectivity above are for normal incidence (perpendicular to the surface). The specular reflectivity will vary as a function of incident angle and polarization (and is likely to vary with wavelength). Any emissivities that you compute by using ## \mathcal{E}=1-\rho ## are simply estimates. Many solids are opaque and much of the reflection is of a diffuse nature. A surface scattering and absorption takes place, and the substance typically has a biderectional reflectivity function that is even more detailed than the specular case above and is also a function of wavelength. Again any emissivities that get computed from the reflectivity are normally estimates. Perhaps there is a complex mathematical method that would give the emissivity from the spectral bidrectional reflectivity, but it would be a very complex calculation.
     
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