Polarization of light -- Determine the thickness of the crystal

[Karl Karlsson]

Homework Statement

A double-breaking material can be used to convert linearly polarized light at right angles to circularly polarized. It is desired to utilize this to change the polarization of light at wavelength 589 nm. Calcite, or its full name, calcium carbonate (CaCO3) is an example of a double-breaking crystal. For the current wavelength, the refractive index is 1.658 for light polarized along the optical axis and 1.486 for light polarized perpendicular to the optical axis. The crystal is oriented so that the optical axis lies in the plane of the surface and the polarization direction of the incident linearly polarized light forms the angle 45 degrees towards the optical axis in the plane of the crystal. Determine the minimum thickness of the crystal required for outgoing light to be circularly polarized.

Relevant Equations

I assume Fresnel's formulas. Circularly polarized light in vector form:
(0, E*cos(kx - wt), E*cos(kx - wt +- pi/2)). Malus law (maybe)

Maybe (I don't know). Bragg condition for interference from an array (since it's a crystal, I don't know if that matters).

I don't even know where to start with this problem. What kind of slit makes linearly polarized light circularly polarized?

The correct answer is d = lambda/(4(n1 - n2)) = 856nm. But how do I get there?

Thanks in beforehand!

 

[Cutter Ketch]

The incoming light polarized at 45 degrees is halfway between the ordinary and extraordinary axes. You can treat it as a sum of two waves aligned along each of the axes. Now, how much phase does each part accumulate passing through a thickness t of the calcite?

 

[Merlin3189]

I don't even know where to start with this problem. What kind of slit makes linearly polarized light circularly polarized?

This is the key. Do you know what circularly polarised light is? How does it differ from linear polarized light, or unpolarized (randomly polarised) light.

 

[Karl Karlsson]

Ooh, i must read wrong i thought i was some slit. Below is my attempt:

However the light is only elliptically polarized and not circular or maybe they meant elliptically polarized light

 

[Cutter Ketch]

However the light is only elliptically polarized and not circular or maybe they meant elliptically polarized light

The light entering is at 45 degrees to the optical axis. It can be treated as a sum of two components. The two components have the same amplitude (because 45 degrees). The two components are eigen-polarizations of the birefringence. The crystal will not convert one to the other. The only thing it can do is change the relative phases. Of course you can get circular polarization.

The only thing I understood on the page was your correct calculation of the thickness based on the path length difference to wind up 1/4 wave out of phase. That is all you need. That is all the problem asked for.

I have no idea what you did at the top. Down at the bottom I have no idea why you think one of the two electric fields needs to be divided by a square root of two. If that is you creating the two parts out of the original 45 degree amplitude, why is only one of them a square root of two smaller than the original?

 

[Merlin3189]

I think your application of Fresnel reflection/transmission is inappropriate here. Since you have normal incidence, it will not discriminate between p and s waves - or IMO there are no p and s, since all polarizations are perpendicular to the surface.
As KK says, just work out the thickness for 1/4 wave difference, as you did. Nothing in the question requires anything more.