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GR9677 #54
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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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,

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

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

Alternate Solutions
 dc7719572016-10-24 15:04:54 So what I did was take the real parts of the original wave. You see that x will look like cos and y will look like -cos, as many people have stated. So we know that when x=0, y=0. Then you can see that when x=1, y=-1 and vice versa. Plotting just these three points you see that the answer is B.Reply to this comment nerv2012-11-09 14:15:26 Note that $e^{i \pi} = -1$, so $\vec{E} = (E,-E) e^{i[kz-\omega t]}$. But $(E,-E)$ is just a vector $135^o$ to the x-axis, i.e. (B). Reply to this comment flyboy6212010-10-24 17:36:41 I think the easiest way is to draw the axes as described in the problem (looking down the z axis toward the origin, and make a sketch of the first quarter-period or so. Clearly the x component goes like cos(z-t) and the y component goes like -cos(z-t) because of the phase difference of $\pi$. If you sketch that, you can quickly see that the two components are always in phase, and the resultant E-field points in the +x, -y direction, or the -x, +y direction. Thus (B) is correct.Reply to this comment
dc771957
2016-10-24 15:04:54
So what I did was take the real parts of the original wave. You see that x will look like cos and y will look like -cos, as many people have stated. So we know that when x=0, y=0. Then you can see that when x=1, y=-1 and vice versa. Plotting just these three points you see that the answer is B.
nerv
2012-11-09 14:15:26
Note that $e^{i \pi} = -1$, so

$\vec{E} = (E,-E) e^{i[kz-\omega t]}$.

But $(E,-E)$ is just a vector $135^o$ to the x-axis, i.e. (B).
mpdude8
2012-04-20 15:35:37
I just pictures what was going on in my mind. On a piece of paper, you have the combination of something going in the -x direction and the -y direction. The phase shift pi swings the -y direction around to the +y direction. The net wave is a combination of this -- 135 degrees from the x-axis.

The fact that you're viewing from the z-plane just makes the wave look like a line.
someone
2011-11-06 16:02:58
You can do it by retaining the complex exponentials. Let the phase $\Phi = kz - \omega t =0$ then you have $\vec{E} = \hat{x} -\hat{y}$. The polarization is linear, so they must pass thorough the origin together to describe a line of angle $135^\circ$
flyboy621
2010-10-24 17:36:41
I think the easiest way is to draw the axes as described in the problem (looking down the z axis toward the origin, and make a sketch of the first quarter-period or so.

Clearly the x component goes like cos(z-t) and the y component goes like -cos(z-t) because of the phase difference of $\pi$. If you sketch that, you can quickly see that the two components are always in phase, and the resultant E-field points in the +x, -y direction, or the -x, +y direction. Thus (B) is correct.
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.
 dumbguy2007-10-16 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?
 shak2010-10-23 12:48:07 no any solution makes sense for me..may be because i am kinda dubm!:) can anyone clarify...at t=0, E1 is in the x direction , and E2 is in the -y direction , and resultant is in the fourth quarter which is not 135 degree to the +x axis, but -45 degree.. thank you for reply..
 ngendler2015-10-23 03:58:18 The line at -45 degrees is the same line that goes through 135 degrees (135+180 = 360-45)
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.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$