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Refraction index

  1. Jan 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Within a certain material, an EM wave with = 1 mm is attenuated to
    10% of its original intensity after propagating 10 cm. Determine the imaginary part of the index
    of refraction ni

    2. Relevant equations
    3. The attempt at a solution

    so [tex]n_i=\sqrt{\frac{\epsilon\mu}{\epsilon_0\mu_0}}[/tex]
    but I still don't get the imagery part...
     
  2. jcsd
  3. Jan 21, 2016 #2

    ehild

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    Gold Member

    What have you learnt about EM wave propagation in absorbing media?
    IN complex notation, the electric field moving in the + x direction and having angular frequency ω is described with the function ##E = E_0 e^{i(\frac{2π}{λ_0}Nx-ωt)}##, where λ0 is the vacuum wavelength and N is the refractive index.
    The complex refractive index has both real and imaginary parts, N=n+iκ. Replacing it into the wave formula, you get
    [tex]E = E_0 e^{i(\frac{2π}{λ_0}(n+iκ)x-ωt)}=E_0 e^{-\frac{2π}{λ_0}κx+i(\frac{2π}{λ_0}nx-ωt)}=\left (E_0 e^{-\frac{2π}{λ_0}κx} \right) e^{i(\frac{2π}{λ_0}nx-ωt)}[/tex],
    which is a wave with exponentially decreasing amplitude.
    The intensity is proportional to the square of the magnitude of the electric field I ∝|E|2. From here, you get the change of intensity with the distance and the imaginary part of the refractive index.
     
  4. Jan 24, 2016 #3
    Thanks! That is helpful!
     
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