GREPhysics.NET: GR9677 #54

GR9677 #54

Problem

GREPhysics.NET Official Solution

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Optics $\Rightarrow$ }Field Trajectory

The problem gives equi-amplitude, thus the field becomes $\vec{E}=Ee^{i(kz-\omega t)}\hat{x} + E e^{i(kz-\omega t + \pi)}\hat{y}$ . Taking the real part, (applying Euler's Theorem, to wit: $e^{i\theta}=\cos\theta + i\sin\theta$ ) one has $\vec{E}=E\cos(kz-\omega t)\hat{x} + E \cos(kz-\omega t + \pi)\hat{y}$ . Apply the trig identity $\cos(\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin\alpha\sin\beta$ to make the field argument equi-phase, $\vec{E}=E\cos(kz-\omega t)\hat{x} - E \cos(kz-\omega t)\hat{y}$ .

Looking down from the z-axis, one has $z=0\Rightarrow$ $\vec{E}=E\cos(\omega t)\hat{x} - E \cos(\omega t)\hat{y}$ .

Make a table of a few values of t and E,

$\begin{eqnarray} \omega t & \Rightarrow & (\cos(\omega t),-\cos(\omega t))\\ 0 & \Rightarrow & (1,-1)\\ \pi/6 &\Rightarrow& \frac{1}{2}(\sqrt{3},-\sqrt{3})\\ \pi/4 &\Rightarrow& \frac{1}{2}(\sqrt{2},-\sqrt{2}), <br /> \end{eqnarray}$

and one deduces that the points plot out a diagonal line at $135^\circ$ to the x-axis, as in choice (B).

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Comments

georgi
2007-08-30 22:16:29

there is a really really simple way to do this. note you get a circle iff the phase diff is a multiple of pi/2. note that if phase diff is zero you get a line in the 1st quadrant and thus for pi shift you must get the only other solutions that gives a line which is B the 135 degree line.

dumbguy
2007-10-16 19:22:21

For the circle, I'm guessing that what if it is out of phase by -pi/2 it will be counter clockwise and pi/e clockwise? am I correct or is it the other way around?

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poong-ryu-do
2005-12-08 21:32:30

There's a way eaiesr way to do this.

1. see that if waves are in phase a 45 degree straight line is formed y=mx;E1=E2 so x=y; m=1;theta= 45

2. note that wave 1 wave is shifted by pi, hence y=-x;theta =-45or 90+45=135.

gawd havnt got all day to do this! athough its nice to have a proof though.

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