Orientation of Major Axis for polarized light

[Blanchdog]

Homework Statement

Consider the Jones vector: $$\begin{pmatrix}A \\Be^{i \delta}\end{pmatrix}$$ For the following cases, what is the orientation of the major axis, and
what is the ellipticity of the light? Case I: $A = B = \frac{1}{\sqrt{2}}; \delta = 0;$ Case II: $A = B = \frac{1}{\sqrt{2}}; \delta = \frac{\pi}{2};$ Case III: $A = B = \frac{1}{\sqrt{2}}; \delta = \frac{\pi}{4}$

Relevant Equations

$$\alpha = \frac{1}{2}tan^{-1}(\frac{2 A B cos(\delta)}{A^2-B^2})$$
$$E_{\alpha}=|E_{eff}|\sqrt{A^2 cos^2(\alpha) + B^2 sin^2(\alpha) + 2 A B cos(\delta)sin(2 \alpha)}$$
$$E_{\alpha \pm \frac{pi}{2}}=|E_{eff}|\sqrt{A^2 cos^2(\alpha) + B^2 sin^2(\alpha) - 2 A B cos(\delta)sin(2 \alpha)}$$

Case 1 worked out great, I found it to be linearly polarized light at an angle $\alpha = \frac{\pi}{4}$, but Case 2 is giving me trouble. As best I can tell, $\alpha$ is undefined in case 2. How do I solve case 2?

 

[Blanchdog]

I believe I figured it out, though I would love confirmation. Since $cos(\delta) = cos(\frac{\pi}{2})=0$ and $A = B$, we end up with $E_\alpha = E_{\alpha_\pm+\frac{pi}{2}}$. That means we have circularly polarized light! So of course $\alpha$ is undefined; a circle has no determined axes!

 

[stephen8686]

You are correct. Your professor might also want you to say if it is right-hand or left-hand circular polarized. More information can be found here https://en.wikipedia.org/wiki/Jones_calculus
And you can use euler's equation to make the exponential into trig functions and plug in the angle.

 

[Blanchdog]

You are correct. Your professor might also want you to say if it is right-hand or left-hand circular polarized. More information can be found here https://en.wikipedia.org/wiki/Jones_calculus
And you can use euler's equation to make the exponential into trig functions and plug in the angle.

How can I tell the handedness?