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Snell's law with a complex refractive index

  1. Mar 14, 2008 #1
    hello everyone

    Consider extinction coefficient k, n becomes N=n-ik.
    the text book says NsinA=N'sinB still holds itself.

    But the sinB,for exmple, may be a complex number, i want to know
    how to get B?
    how to understand this situation, which is the refractive angle?
    the B's real part?
  2. jcsd
  3. Mar 14, 2008 #2


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    The electric field in a plane wave E~exp[i(kx-wt)], where k=w N/c.
    If N=n-ia, then E~exp[i(nw/c)x-wt)exp[-ax].
    This means n is still used by Snell.
    k is usually the wave number, so I used a for Im N.
  4. Mar 14, 2008 #3
    thank you
    and you mean that the refrective angle only depends on Re N--n
    and has nothing to do with Im N--a(using your signal), right?
  5. Mar 19, 2008 #4
    Why you write exp{i[(nw/c)x-wt]}*exp(-ax) and not exp{i[(nw/c)x-wt]}*exp(axw/c) ?
    Last edited: Mar 19, 2008
  6. Mar 20, 2008 #5


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    Sorry. That was a misprint. It should be axw/c.

    "and you mean that the refrective angle only depends on Re N--n"
    I'm afraid I oversimplified. The boundary conditions have to be applied at the interface, and the derivation done from scratch. It gets quite complicated.
  7. Mar 20, 2008 #6
    these days ,i thought further that maybe it's only a kind of appearance to describe the polirized light phase shift at the interface of different media.
    but i wonder which is the refractive angle if k ,ie absorption exist.
    for example. air/silicon,whose N is 1 and 4.4-0.8i, respectively.
    to Snell, 1*sin(AOI)=(4.4-0.8i)*sin(AOR),
    AOR will be a complex number, how to get the real AOR in experiments?
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