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Birefringence deviation angle?

  1. Oct 9, 2015 #1
    The best known effect of birefringence is the lateral displacement of the extraordinary image. Why is this effect rarely quantified? I couldn't find a table of materials specifying the deviation angle δ of the extraordinary ray (say, for an angle of incidence equal to zero). Birefringence seems to be quantified exclusively by Δn = ne-no, and δ cannot be derived from Δn.
     
  2. jcsd
  3. Oct 9, 2015 #2
    Well, for normal incidence there is no deviation so what will be the point to make tables with a bunch of zeros? :smile:
    The deviation depends on the angle and the geometry of the setup. You have to calculate it itself for your specific problem.
    The birefringence value (Δn) allows you to do it.
     
  4. Oct 9, 2015 #3
    Well, for normal incidence the extraordinary ray does have a deviation. That's why it is extraordinary, it doesn't obey Snell's law, in contrast to the ordinary ray. If you happen to have a calcite crystal you could see it with your own eyes. It is also mentioned in many places, for example Wikipedia: "So even in the case of normal incidence, where the angle of refraction is zero (according to Snell's law, regardless of effective index of refraction), the energy of the extraordinary ray may be propagated at an angle." (click)
     
  5. Oct 9, 2015 #4

    blue_leaf77

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  6. Oct 9, 2015 #5
    I know there are tables for those two refractive indices. However, again, if a laser beam enters a calcite crystal, at an angle of incidence equal to zero, it is still split into two rays. The ordinary ray travels along the normal. The extraordinary ray travels at angle δ with the normal. You cannot explain that lateral deviation by any value of ne and no.
     
  7. Oct 9, 2015 #6

    blue_leaf77

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    You can, provided you also know the relative orientation of the crystal axis to the surface of the material on which your laser is shone. The theory has long been developed, which can be found in almost any photonics textbook.
     
  8. Oct 9, 2015 #7
    Yes, you are right.
     
  9. Oct 11, 2015 #8
    Maybe someone knows a simple equation for the lateral displacement without a photonics book?
     
  10. Oct 11, 2015 #9

    blue_leaf77

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    Again, you need to know the crystal's optic axis orientation with respect to the incoming ray. For an arrangement such that the input plane is parallel to the so-called principal plane of the (uniaxial) crystal (like the one in the picture below), one can prove that the deviation angle ##\theta_s## between e- and o-rays is given by
    $$
    \cos \theta_s = \left( \frac{\cos^2\alpha}{n_o^2} + \frac{\sin^2\alpha}{n_e^2} \right) \left( \frac{\cos^2\alpha}{n_o^4} + \frac{\sin^2\alpha}{n_e^4} \right)^{-1/2}
    $$
    where ##\alpha## is the angle subtended by the incoming ray and optic axis. If you want to know more about how optics axis direction is defined in commercial crystals I suggest that you look up the section for Birefringence in "Optics" by Eugene Hecht.
     

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    Last edited: Dec 11, 2015
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