Determination of the polarization of a light wave

[Khamey]

Homework Statement

The polarization of a light wave is described by two complex parameters.

[itex]\lambda[/itex] = cos[itex]\Theta[/itex]e^(i[itex]\delta[/itex]x)

[itex]\mu[/itex] = sin[itex]\Theta[/itex]e^(i[itex]\delta[/itex]y)

Satisfying |[itex]\lambda[/itex]|^2 + |[itex]\mu[/itex]|^2 = 1

More explicitly, the electric field is

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

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

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

Homework Equations

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?

 

[vela]

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*}

 

[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[itex]\Theta[/itex]e^i[itex]\delta[/itex]x e^-iwt


Ey(t) = Eo Re (sin[itex]\Theta[/itex]e^i[itex]\delta[/itex]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?

 

[vela]

Look at a simple case where [itex]\theta = \pi/4[/itex], so that the x and y components have the same amplitude, and where [itex]\delta_x = 0[/itex] and [itex]\delta_y = \pi/2[/itex], 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 [itex]E_0/\sqrt{2}[/itex] in the counter-clockwise direction.

What happens when you vary [itex]\theta[/itex]? 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.

 

[paranormal]

I would approach this problem by first understanding the concept of polarization of light. Polarization refers to the direction of the electric field vector of a light wave as it travels through space. It can be described by two complex parameters, \lambda and \mu, which satisfy the equation |\lambda|^2 + |\mu|^2 = 1.

To determine the axes of the ellipse traced by the tip of the electric field vector, we can use the Jones calculus method. This method involves using matrices to represent the polarization of light, and then using matrix multiplication to determine the resulting polarization after passing through different optical elements.

In this case, the two complex parameters \lambda and \mu can be represented by a 2x2 matrix:

\begin{bmatrix} \lambda & 0 \\ 0 & \mu \end{bmatrix}

We can then use this matrix to determine the polarization of the light wave after it passes through a polarizer, which is an optical element that only allows light with a specific polarization to pass through. The angle of the polarizer will determine the axis of the ellipse traced by the electric field vector.

To determine the direction in which the ellipse is traced, we can use the concept of elliptical polarization. This occurs when both \lambda and \mu have non-zero values and are not equal in magnitude. The direction of the major axis of the ellipse will be determined by the phase difference between \lambda and \mu, which is represented by the complex parameters \deltax and \deltay. The angle of the major axis can be calculated using the equation tan(2\alpha) = 2|Re(\deltax-\deltay)/(\lambda-\mu)|.

In summary, to determine the axes of the ellipse traced by the tip of the electric field vector and the direction in which it is traced, we can use the Jones calculus method and the concept of elliptical polarization. This approach allows us to mathematically determine the characteristics of the light wave's polarization without the need for complex equations or multiple coordinate systems.