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Silversonic
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My bad for the spelling fail in the title.
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
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]
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.
Homework Statement
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
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]
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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