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Negative index of refraction with anisotropy, quick question

  1. May 16, 2012 #1
    Hey all,

    I'm trying to solve a problem in which I have a uniaxial (uniform in two directions, different in the other) crystal with the optical axis normal to the plane of reflection. I call the permittivity parallel to the optical axis [itex]ε_par[/itex] and the permittivity in the other two directions (in the plane perpendicular to the optical axis [itex]ε_perp[/itex]. The magnetic permeability is isotropic and uniform.

    Now, I want to investigate what happens when the index of refraction is negative in ONE direction, not both. This means that we need to say the magnetic permeability is negative (say -1 here) and the permittivity in the direction we want the negative refractive index is also negative. Then we take the negative square root: [itex] n = \sqrt(εμ)[/itex] and everything is dandy.

    But that leaves the index in the other direction -- This one has positive ε and negative μ, so its refractive index is imaginary, which usually means it is absorbed by the material. And a paper I've been reading does say this, as long as the it is polarized in this direction. But what if it's propagating in between these two axes? It seems like the component of the electric field parallel to the imaginary refractive index should just die, but the other component should happily continue.

    I'm really confused. Can anyone help??

    Thanks!
     
  2. jcsd
  3. May 16, 2012 #2

    DrDu

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    Already in an uniaxial crystal with normal reflection like calc spar an arbitrarily polarized ray splits up into an ordinary and an extraordinary ray, so I don't think it is astonishing that the same happens for negative or imaginary index of refraction. In fact this is how linear polarising foils act: One polarization gets heavily damped and the other is transmitted.
     
  4. May 16, 2012 #3
    Thanks for the response!

    Ok, but that still leaves me with a few questions... I know how the splitting up happens in a calcite crystal, but here, it seems like it's more continuous and not really "splitting" it up at all.

    Let's say I have the imaginary index of refraction (IoR) parallel to the optical axis and the ones perpendicular to it be the negative ones. Then, the only wave that should be able to propagate without getting damped is along the optical axis, because its E field "sees" only a negative IoR and no imaginary IoR at all.

    So what if we change the angle of that propagating wave just ever so slightly away from that axis? It still mostly "sees" the negative IoR, but now it also sees a tiny bit of the imaginary one. It seems like it should almost propagate like the other but should be dying out a bit as it goes...

    I'm trying to show analytically that in the case I mentioned above, propagating waves die out in every case but when they are going along the optical axis.

    Any ideas?

    Thanks!
     
  5. May 16, 2012 #4

    DrDu

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    I still think you have an ordinary an an extraordinary ray. If I understand your setup correctly, you have a crystal face perpendicular to the optical axis and a ray entering under an oblique angle.
    Then the ray which is polarised perpendicular to the plane of incidence (the plane spanned by the k vector and the normal to the crystal surface) is the normal ray which doesn't get damped.
    The extraordinary ray, which is polarized in the incidence plane, will see some mixture of the two indices of refraction. For perpendicular incidence it shouldn't get damped at all while with increasing incidence angle the damping increases.
    Landau Lifgarbagez should be useful to calculate the effective refractive index as a function of phi.

    PS: Also in a Calcite crystal the effective n depends on the angle for the extraordinary ray.
     
  6. May 16, 2012 #5
    Thanks so much!

    Yeah, you understand my setup correctly. You're right, there are two types of incident waves to look at: the S wave (polarized perpendicular to the plane of incidence) and the P wave (polarized parallel to the plane of incidence). Then I have two cases to worry about: when [itex]n_x[/itex] = [itex]n _y[/itex] < 0 and [itex]n _z[/itex] is imaginary, and the other case in which [itex]n _x[/itex] = [itex]n _y[/itex] is imaginary, and [itex]n _z[/itex] < 0. (Here I'm saying z is the optical axis.)

    The S wave is simple in both cases because it only "sees" [itex]n _x[/itex] and [itex]n _y[/itex], so either in is simple isotropic refraction with a negative index, or it is dies out quickly like a conductor.

    The P wave is the trickier part like you mentioned. Like you also said, it sees some mixture of the two indices. Also like you said, I know in each case it will be able to propagate without damping along the axis of the imaginary index and will be the most damped out when propagating perpendicular to that axis.

    I have L&L and I'm trying to apply it here but I'm running into problems. When I take the wave vector surface (the equation that relates [itex]k_x[/itex] and [itex]k_z[/itex] for a given set of indices):

    [tex]\frac{k_x ^2}{n_z ^2} + \frac{k_z ^2}{n_x ^2} = \frac{\omega^2}{c^2} = k^2[/tex]

    and plug in an imaginary n_x (and negative n_z, though it doesn't change the equation here), I get [itex]k_z = i \left| n_x \right| \sqrt{1 - sin(\theta)^2/n_z ^2}[/itex]. This actually kind of works at first glance because for θ = 0 (traveling along the optical axis), it is the most complex it will be and when we plug k_z into [itex]e^{i \vec{k} \cdot \vec{x}}[/itex], the i*k_z*z term will just be a falling exponential, which represents the damping. But this doesn't quite work, because the wave should be able to travel in the x or y direction without damping, and when I plug in [itex]\theta = \pi/2[/itex], it still gives an imaginary k_z, whereas it should give k_z = 0 (because it's not propagating in the z direction at this angle).

    It gets even worse if I try it for the other case... any ideas?

    Thanks so much!
     
  7. May 16, 2012 #6

    DrDu

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    I have at home a copy of Agranovich, Crystal Optics. Maybe I find something useful.
     
  8. May 17, 2012 #7

    DrDu

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    According to Agranovich and Ginzburg, the wave becomes inhomogeneous when entering an absorbing medium, i.e. the real and imaginary part of k point in different directions.
     
  9. May 17, 2012 #8

    DrDu

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    Up to a k missing on the right, I think this formula is correct. The case theta=90 deg corresponds to a surface plasmon on a metal, no? I think the field strength decays exponentially for these in the metal.
     
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