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Really simple and quick question about optics

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data
    ZnSe is a II-VI semiconductor and a very useful optical material used in various applications such as optical windows (especially high power laser windows), lenses, and prisms. It transmits over 0.50-19 μm. Refractive index, n in the 1-11 μm range is described by a Cauchy expression of the form
    z3IGQU6.png
    in which λ is in μm. What are the n-2, n0, n2, and n4 coefficients?

    2. Relevant equations
    z3IGQU6.png
    ucgUq65.png

    3. The attempt at a solution
    Given the index of refraction by the following Cauchy equation relationship
    z3IGQU6.png
    with λ given in micrometers, I am to find the Cauchy coefficients
    n-2, n0, n2, and n4 such that
    ucgUq65.png
    Using
    8Iwtk2o.png
    I managed to get the first expression to look like
    p4U3bnJ.png
    Since n is dimensionless λ can't have a dimension either... but the fact that it is given in micrometers is important, clearly, because a 1 meter value λ would require me to plug in
    λ = 1000000​
    into the very first expression for n.

    Typically hυ is given in eV (electron volts). For n to be dimensionless, wouldn't hc have to be equal to eV (a unit of energy) as well? But obviously energy has units of hc/λ. So something is wrong. How do I take into account the fact that λ is given in micrometers in finding the coefficients?

    I know I am close, but I can't finish it up. :(
     
  2. jcsd
  3. Oct 20, 2014 #2

    BvU

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    Hello log, and welcome to PF :)

    Don't know what I'm talking about, but as a physicist I expect your Cauchy coefficients to have dimensions: 0.0485 ##\mu m^2##, etc..
     
    Last edited: Oct 20, 2014
  4. Oct 20, 2014 #3
    Why would you expect them to have dimensions of length squared?

    I'm confused what to do with, for example, what to do with (hc)^2?

    Which units should I use? How do I make this equal to eV^2 as the Cauchy coefficient n-2 is normally quoted as?
     
  5. Oct 20, 2014 #4

    BvU

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    As I said, don't know what I'm talking about. But your coefficients all have different dimensions. The correct expression for n is$$ n= 2.4365 + {0.0485\over (\lambda/1\mu m)^2 }+ {0.0061\over (\lambda/1\mu m)^4 } - 0.0003 (\lambda/1\mu m)^2$$

    My interpretation of the problem is that the coefficients have to be found for ##h\nu## expressed in eV. You kind of indicate that clarification of the OP in your post # 3.

    So start with: how many eV corresponds to 1 ##\mu m##.
     
  6. Oct 20, 2014 #5
    Thanks for the response, BvU.

    I see what you mean about the differing dimensions. So, 1 eV corresponds to 1.23984193 electron volts.

    How did I do this? hc/λ for λ = 1 μm.
     
  7. Oct 21, 2014 #6

    BvU

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    You correctly pulled in e too, but in "1 eV corresponds to 1.23984193 electron volts." the 1 eV is probaby 1 ##\mu m## :) .

    So now think about what happens to the various coefficients. You can easily check by filling in e.g. ##\lambda = 5\; \mu m## in the one and ##h\nu=6.2## eV in the other
     
  8. Oct 23, 2014 #7
    Thanks. I got how it works because of you! :)
     
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