GREPhysics.NET
GR | # Login | Register
   
  GR9677 #54
Problem
GREPhysics.NET Official Solution    Alternate Solutions
This problem is still being typed.
Optics}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,


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

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
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.Alternate Solution - Unverified
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).
Alternate Solution - Unverified
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.
Alternate Solution - Unverified
Comments
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.Alternate Solution - Unverified
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).
Alternate Solution - Unverified
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.
NEC
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^\circNEC
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.
Alternate Solution - Unverified
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?
shak
2010-10-23 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..
ngendler
2015-10-23 03:58:18
The line at -45 degrees is the same line that goes through 135 degrees (135+180 = 360-45)
NEC
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.

NEC

Post A Comment!
You are replying to:
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.

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

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