# Determination of the polarization of a light wave

1. Sep 13, 2011

### Khamey

1. The problem statement, all variables and given/known data
The polarization of a light wave is described by two complex parameters.

$\lambda$ = cos$\Theta$e^(i$\delta$x)

$\mu$ = sin$\Theta$e^(i$\delta$y)

Satisfying |$\lambda$|^2 + |$\mu$|^2 = 1

More explicitly, the electric field is

Ex(t)=Eocos$\Theta$cos($\omega/itex]t-[itex]\delta$x)

Ey(t)=Eosin$\Theta$cos($\omega/itex]t-[itex]\delta$y)

Determine the axes of the ellipse traced by the tip of the electric field vector and the direction in which it is traced.

2. Relevant equations

3. The attempt at a solution

I've tried to create new axis which the ellipse falls on x' and y' and is at an angle of alpha from the original coordinates. It leaves a nasty equation with a lot of variables. Talking to my professor he briefly went over ten pages of work where he lost me around page two and even in the end said he knew his answer was not correct. Any help to ease this problem?

2. Sep 13, 2011

### vela

Staff Emeritus
Try using something like
\begin{align*}
\frac{E_x}{E_0} &= \mathrm{Re}(\cos\theta\ e^{i(\omega t - \delta_x)}) \\
\frac{E_y}{E_0} &= \mathrm{Re}(\sin\theta\ e^{i(\omega t - \delta_y)})
\end{align*}

3. Sep 13, 2011

### Khamey

I have no idea how to get this started.

I have something similar to what vela posted given in the problem.

Ex(t) = Eo Re (cos$\Theta$e^i$\delta$x e^-iwt

Ey(t) = Eo Re (sin$\Theta$e^i$\delta$y e^-iwt

As well given in the problem. What do you mean by try using that? How does this get me started in finding the axes traced by the e field vector?

4. Sep 15, 2011

### vela

Staff Emeritus
Look at a simple case where $\theta = \pi/4$, so that the x and y components have the same amplitude, and where $\delta_x = 0$ and $\delta_y = \pi/2$, so that the two components differ in phase by 90 degrees. Then you have
\begin{align*}
E_x &= (E_0/\sqrt{2}) \mathrm{Re}(e^{-i\omega t}) = (E_0/\sqrt{2}) \cos \omega t \\
E_y &= (E_0/\sqrt{2}) \mathrm{Re}(ie^{-i\omega t}) = (E_0/\sqrt{2}) \sin \omega t
\end{align*}
So the tip of the electric field traces out a circle of radius $E_0/\sqrt{2}$ in the counter-clockwise direction.

What happens when you vary $\theta$? What if the phases are swapped? What happens when the relative phase is 0?

I don't know how the whole thing works out, this should at least get you an idea of how you might analyze the situation.