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1. The problem statement, all variables and given/known data

I am utterly confused with the mathematics in a section of my notes, I swear that it's wrong - and it related to a piece of homework which I cannot understand.

The electric field vector in a z-directed monochromatic EM wave is given by;

http://img213.imageshack.us/img213/4768/58814275.jpg [Broken]

It says that if [itex]\alpha[/itex] = [itex]\pi[/itex]/2, [itex]\theta[/itex] = 0 and [itex]E_{0,x}[/itex] = [itex]E_{0,y}[/itex]

Then

[itex]E_{x}[/itex] = [itex]E_{R}[/itex]cos(kz-wt),

[itex]E_{y}[/itex] = [itex]E_{R}[/itex]sin(kz-wt)

and [itex]E_{R}[/itex] = [itex]\frac{1}{\sqrt{2}}E_{0,x}[/itex]

3. The attempt at a solution

I do not see how the component in y direction is right. Taking the parameters [itex]\alpha[/itex] = [itex]\pi[/itex]/2, [itex]\theta[/itex] = 0, [itex]E_{0,x}[/itex] = [itex]E_{0,y}[/itex]

we get

[itex]E_{0} = \frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + E_{0,x} e^{i(\pi/2)} \hat{y}][/itex]

[itex]e^{i(\pi/2)} = i[/itex]

[itex]E_{0} = \frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + iE_{0,x} \hat{y}][/itex]

Subbing this in the equation for overall E.

[itex] E = Re(\frac{1}{\sqrt{2}}[E_{0,x} \hat{x} + iE_{0,x} \hat{y}]e^{i(kz-wt)}) [/itex]

[itex]E_{R}[/itex] = [itex]\frac{1}{\sqrt{2}}E_{0,x}[/itex]

[itex] E = Re(E_{R}[\hat{x} + i\hat{y}]e^{i(kz-wt)}) [/itex]

[itex] E = Re(E_{R}[\hat{x} + i\hat{y}][cos(kz-wt)+isin(kz-wt)]) [/itex]

[itex] E = E_{R}[\hat{x} cos(kz-wt) - \hat{y}sin(kz-wt)] [/itex]

So then surely for [itex]\alpha[/itex] = [itex]\pi[/itex]/2

We have

[itex]E_{x}[/itex] = [itex]E_{R}cos(kz-wt)[/itex]

[itex]E_{y}[/itex] = -[itex]E_{R}sin(kz-wt)[/itex]

So the [itex]E_{y}[/itex] direction is minus instead?

Similarly for [itex]\alpha[/itex] = [itex]-\pi[/itex]/2 We'd have;

[itex]E_{x}[/itex] = [itex]E_{R}cos(kz-wt)[/itex]

[itex]E_{y}[/itex] = [itex]E_{R}sin(kz-wt)[/itex]

Is this correct?

[itex]\alpha[/itex] = [itex]\pi[/itex]/2 should correspond to left circular polarization, and [itex]\alpha[/itex] = [itex]-\pi[/itex]/2 to right.

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# Homework Help: Circular Polization type in an EM wave.

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