Birefringent filter, optical path length difference? In 'The Light Fantastic' by Kenyon, I.R. (p424), it is said that for a birefringent material inclined at Brewster's angle and who's optical axis lies in the plane of the plate, we have an optical path length difference between the ordinary and extraordinary waves of:
$$\Delta s= \frac{\Delta n t}{\sin(\phi_B)}$$
Where $\Delta n=n_0-n_e$, $t$ is the thickness of the plate and $\phi_B$ is Brewster's angle. My question is where does this equation come from and have any assumptions been made deriving it? It seems to be assuming that no refraction of either the ordinary or extraordinary wave occurs at the surface of the birefringent material, when infact I think they should refract by different amount.
 A: Note I am the OP.
The first and foremost thing to note about this equation is that it is not exact and is based on approximations.
For the situation described above, once the light has entered the filter it splits into two, one that experiences the ordinary refractive index $n_0$ and the other that experiences the refractive index $n_e$ (which may be different from actual extraordinary refractive index of the crystal). We assume that the two beams follow the same path through the beam, at an angle dictated by:
$$\sin(\phi_B)=n\sin(\theta)$$
Where $n$ is the mean refractive index experienced by the two beams. The optical path length therefore between the two beams after passing through the filter is:
$$\Delta s=(n_0-n_e) \frac{t}{\cos(\theta)}$$
We now note that at Brewster's angle we have:
$$\theta+\phi=\pi/2$$
which therefore gives us:
$$\Delta s= \frac{\Delta n t}{\sin(\phi_B)}$$
As required.
Reference
Svelto, O. 2010. Principles of Lasers. 5th ed. Translated from Italian by D.C.Hanna. New York: Springer (p286-287)
