Birefringence calculations and questions

In summary, birefringence is the property of a material to split a single light ray into two polarized rays with different refractive indices, which is calculated by taking the difference between the refractive indices of the material for the two polarizations of light passing through it. The equation for calculating birefringence is Δn = n<sub>e</sub> - n<sub>o</sub>, where Δn is the birefringence, n<sub>e</sub> is the refractive index for the extraordinary ray, and n<sub>o</sub> is the refractive index for the ordinary ray. Temperature can affect birefringence in some materials, causing it to increase
  • #1
VortexLattice
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Hi everyone!

I'm trying to do the classic calculation of the coefficients of transmission and reflection at a surface, but for a uniaxial crystal. I'm doing the simplest case, in which the optical axis is normal to the reflecting surface. Here is a simple diagram of what I'm trying to do:

358vwud.png

(Sorry if it's messy or cluttered, I didn't want to leave anything out. We're taking the magnetic permeability to be 1, also. I chose the incident wave's E field to be entirely parallel to the plane of incidence because an incident wave with it at any other angle is just a superposition of this wave and one with the E field perpendicular to the plane of incidence, and that case seems easier)

So, I have a bunch of questions that I can't seem answer. Foremost, I know in the special case of the incident wave being normal to the reflecting surface, there is only one transmitted wave. But in all other cases, will there always be both an ordinary (O) and extraordinary (X) wave?

Second, I was reading in Landau and Lifgarbagez' Electrodynamics of Continuous media, the chapter on anisotropic media, and they say that "The vectors [itex]\vec{D}[/itex] in the O and E waves with the same direction of [itex]\vec{k}[/itex] are perpendicular. Hence the polarization of the ordinary wave is such that E and D lie in a plane perpendicular to the principal section" (the principal section is the plane with S (poynting vector), the optical axis, and k).

But this seems to clash with the boundary conditions I know this problem needs to fulfill. We need to have D perpendicular, B perpendicular, E parallel, and H parallel continuous at the boundary. But if the ordinary wave has its E field perpendicular to the E field of the incident wave... well, that can't add up. So what am I missing?

From what I can tell, that quote may be true if the wave vectors of the O and X waves are the same, but they shouldn't be, right? This is what I'm understanding right now: An incident beam comes in. The O wave refracts completely normally, as if it was refracting in an isotropic medium of index of refraction [itex]ε\bot[/itex]. Because it is refracting normally, its polarization is in the same direction (in this case, in the plane of incidence) as the incident wave, and its wave vector and poynting vector have the same direction.

Then, the wave also refracts into the X wave, which has a different wave vector than the O wave's wave vector. However, the poynting vectors and wave vectors of the O and X waves are still all coplanar, right?

Does that seem correct? Can anyone help me??

Thanks!
 
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  • #2
</code>Yes, that does seem correct. The ordinary wave refracts at the same angle as it would in an isotropic medium of index of refraction ε⊥, and the extraordinary wave refracts at a different angle due to the birefringence of the crystal. The E and D fields of the ordinary wave will be perpendicular, while the E and D fields of the extraordinary wave will be parallel to each other. As for the boundary conditions, you need to make sure that the E, D, H, and B fields are all continuous at the boundary. This means that in the case of the ordinary wave, the E field will be continuous with the E field of the incident wave, while the D field will be perpendicular to both the incident E field and the reflected E field. Similarly, for the extraordinary wave, the E field will be continuous with the E field of the incident wave, while the D field will be parallel to both the incident E field and the reflected E field. Hope this helps!
 

FAQ: Birefringence calculations and questions

1. What is birefringence and how is it calculated?

Birefringence is the property of a material to split a single light ray into two polarized rays with different refractive indices. It is calculated by taking the difference between the refractive indices of the material for the two polarizations of light passing through it.

2. What is the equation for calculating birefringence?

The equation for calculating birefringence is Δn = ne - no, where Δn is the birefringence, ne is the refractive index for the extraordinary ray, and no is the refractive index for the ordinary ray.

3. How does temperature affect birefringence?

Temperature can affect birefringence in some materials, causing it to increase or decrease. For example, in quartz, birefringence decreases as temperature increases. This is due to changes in the crystal lattice structure of the material.

4. What factors can affect the accuracy of birefringence calculations?

Factors that can affect the accuracy of birefringence calculations include imperfect alignment of the material, impurities or defects in the material, and variations in temperature or pressure.

5. What are some applications of birefringence calculations?

Birefringence calculations are used in a variety of fields, including optical mineralogy, polarized light microscopy, and liquid crystal display technology. They can also be used to study the properties of various materials, such as proteins and polymers, and to determine the stress and strain in materials under different conditions.

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