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Polarization and crystallography

  1. Oct 31, 2010 #1
    Can any one explain what is principal section and optic axis of a crystal?
    Any help wud be highly appreciated.
    Thank You
  2. jcsd
  3. Oct 31, 2010 #2

    Andy Resnick

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    Off the top of my head, the optic axis is the direction (for a biaxial crystal) along which n_e = n_o. I'd have to look up the other one.
  4. Oct 31, 2010 #3
    You will have to do a little reading to fully understand it if you haven't yet seen the tensor representation of crystal properties and optical properties in particular. But the essential point is that refractive index can vary for different polarizations in a material. For a uniaxial crystal, which is the most typical case, it turns out that you can represent the refractive index by drawing an ellipsoid with a=b =/= c if a,b,c are the three semimajor axes. Then if you draw a vector from the center to any point on the surface, that length will be inversely related to the value of the refractive index felt for a wave with polarization pointing in that direction. So, the optic axis of the material is the long axis of this ellipse, and if the wave is propagating along the optic axis, it will feel the same refractive index regardless of polarization. Therefore there is no birefringence experienced by the wave.

    For any other direction of propagation, the refractive index will depend on the polarization state (draw a vector in the direction of propagation, and draw the ellipse normal to it which intersects the ellipsoid). In this case, two orthogonal polarization components will experience two refractive indices - called the ordinary and extraordinary index. If the wave impinges on the crystal at an angle, the different refractive indices will cause the two polarization components to refract at different angles, leading to two spatially separated beams, the visible manifestation of birefringence.

    Again, you'll have to read through the math to see why the refractive index can be represented by this ellipse. It's not difficult but you just have to go through it.

    So if one central intersection of the ellipsoid is a circle, the section orthogonal to that is the principal section. That's just terminology, I had to look that up as well. But the principal section will tell you the wave vector which experiences the most pronounced birefringence.
  5. Oct 31, 2010 #4
  6. Nov 1, 2010 #5
    thank you for replying but it wud be better if u wud make it more specific....because i cudnt get evrything of wat u wrote...i found it tough to understand!:confused:
  7. Nov 1, 2010 #6

    Look at the picture on the right. The three arrows coming out of the ellipsoid are the propogation directions of three different waves. Each has a plane normal to the direction which intersects the ellipsoid, forming an ellipse (or circle). The polarization can lie anywhere in the plane of that ellipse. It's length is related to the refractive index. So you can see that the wave travelling straight up (ie, along the optic axis) has the same refractive index for all polarizations, since the intersection with the ellipsoid is a circle. Any other direction, the two perpendicular polarization components will experience a different refractive index.

    I can't be more specific. Maybe I shouldn't be trying to explain the optical indicatrix to you, but if you want to understand it on this level you will have to read about it. The simplest answer is just that, travelling along the optic axis, light will travel at one speed. Travelling in any other direction, light will split into two beams with slightly different speeds.
  8. Nov 8, 2010 #7
    thank u vry much got it....:D
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