- #1
skrat
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Homework Statement
From an uniaxial crystal, a 0.7 mm thick plate is cut out in a way that the optical axis is parallel to the surface. A light falls perpendicular to the surface (circular polarization) with 500 nm. After the crossing of the plate, the light has elliptical polarization and the ratio of the amplitudes is 0.5 . What is the difference between ordinary and extraordinary refractive index?
Homework Equations
The Attempt at a Solution
The idea is:
Initial Jones vector is $$J_i=\binom{1}{i}$$ while the output is $$J_f=\binom{e^{ik_od}}{ie^{ik_ed}},$$ if "o" stands for "ordinary", "e" for "extraordinary" and ##d## is the thickness of the plate. Final Jones vector can be rewritten as $$J_f=e^{ik_ed}\binom{e^{i(k_o-k_e)d}}{i}\approx \binom{e^{i\delta }}{i}$$ for ##\delta=(k_o-k_e)d=k_0(n_o-n_e)d##.
Now the idea is that from the fact that after the crossing of the plate, the light has elliptical polarization with ratio of amplitudes ##1/2##, I am supposed to somehow find out how much ##\delta ## is. But have no idea how to do that. I tried to use the form of final Jones vector: $$E_x=e^{i\delta}e^{-i\omega t}$$ and $$E_y=ie^{-i\omega t}.$$ If I consider only the real parts and forget about the imaginary parts, this simplifies to $$E_x=\cos\delta\cos\omega t+\sin\delta\sin\omega t$$ and $$E_y=\sin\omega t.$$ Inserting the last identity into the previous one leaves me with general equation for ellipse $$\frac{E_x^2}{\cos^2 \delta}+\frac{E_y^2}{\sin^2 \delta}-\frac{2E_xE_x\sin\delta}{\cos^2\delta}=1$$ and I am not really sure how this helps me. I wanted to somehow express ##\cos\delta## or ##\sin\delta## with the amplitudes ##E_x## and ##E_y##...
Any help will make my day!