# Direction of extraordinary and ordinary waves (birefringence).

1. Mar 26, 2012

### Silversonic

1. The problem statement, all variables and given/known data

A linearly polarized wave travelling in the y-direction falls on to a plate of uniaxial material of thickness d, and the optical axis is in the z-direction. Which ones of the components $E_x$ or $E_z$ become the ordinary and extraordinary waves inside the plate?

2. Relevant equations

Ordinary wave information

$k \circ E = 0$

$\frac{E_x}{E_y} = -\frac{k_y}{k_x}$

$E_z = 0$

Extraordinary wave information

$\frac{E_x}{E_y} = \frac{k_x}{k_y}$

$E_z = - (\frac {n_1}{n_3})^2 \frac{k_x^2 + k_y^2}{k_x k_z} E_x$

$E_z = - (\frac {n_1}{n_3})^2 \frac{k_x^2 + k_y^2}{k_y k_z} E_y$

$n_1 , n_3$ are the indices of refraction, the latter of which is for the optical axis.

$k_x, k_y, k_z$ are the wavenumbers in the respective directions for the ordinary/extraordinary waves.

3. The attempt at a solution

I've looked it up online, and found that when the incident wave is perpendicular to the ordinary axis (the z-axis), the ordinary and extraordinary wave both propagate in the original incident wave's direction, but with different speeds. The problem I'm having is, with the information in the equation section, how does the maths show this is the case (the directions, not the speeds)?

The ordinary wave has no z-component. And, since k.E = 0 we have that the electric field is perpendicular to the ordinary axis and the direction of propagation. But how does this suggests the ordinary wave propagates in the direction of the incident wave i.e. the y-direction? If that is the case, it means $E_x$ becomes the ordinary wave.

The extraordinary wave, I'm not sure how I can pull anything out with the information provided.

There appears to be no way to know $k_x, k_y, k_z$ for the ordinary or extraordinary waves - and thus I can't determine anything about the electric fields. What am I missing?

Last edited: Mar 26, 2012