GR9677 #54



Alternate Solutions 
dc771957 20161024 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 20121109 14:15:26  Note that , so
.
But is just a vector to the xaxis, i.e. (B).
  flyboy621 20101024 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 quarterperiod or so.
Clearly the x component goes like cos(zt) and the y component goes like cos(zt) because of the phase difference of . If you sketch that, you can quickly see that the two components are always in phase, and the resultant Efield points in the +x, y direction, or the x, +y direction. Thus (B) is correct.  

Comments 
dc771957 20161024 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 20121109 14:15:26  Note that , so
.
But is just a vector to the xaxis, i.e. (B).
  mpdude8 20120420 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 xaxis.
The fact that you're viewing from the zplane just makes the wave look like a line.   someone 20111106 16:02:58  You can do it by retaining the complex exponentials. Let the phase then you have . The polarization is linear, so they must pass thorough the origin together to describe a line of angle   flyboy621 20101024 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 quarterperiod or so.
Clearly the x component goes like cos(zt) and the y component goes like cos(zt) because of the phase difference of . If you sketch that, you can quickly see that the two components are always in phase, and the resultant Efield points in the +x, y direction, or the x, +y direction. Thus (B) is correct.   georgi 20070830 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 20071016 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 20101023 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 20151023 03:58:18 
The line at 45 degrees is the same line that goes through 135 degrees (135+180 = 36045)

  poongryudo 20051208 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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