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GR9677 #55
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Alternate Solutions |
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Comments |
tinytoon
2008-11-03 22:05:31 | If you happen to remember that the intensity is proportional to  , then the relation falls out.
Because they are decoupled, as Yosun mentioned, they add independently. The first polarization is in the  -direction, the intensity reduces by  . Therefore you are left with 
It is same for the  -direction and you just add the two in the end, getting:
 |  | tareq
2007-10-05 05:03:27 | The solution of Yosun Chang is wrong ! When the waves recombine, we have first to calculate the amplitude of the resultant wave to account for any interference effect and use it to calculate the intensity. This is done in the first comment above, by Richard.
sawtooth
2007-10-28 17:45:43 |
No interference for vertical polarisations ! If you insist on taking ^2) , the interference contribution is  = 0) .
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| | Richard
2007-09-26 17:38:50 | There is a slightly longer, more mathematically rigorous way to get the same answer:
if the optical path difference is  , the field becomes (letting t and z go to zero)
}=\hat{x}E1-\hat{y}E2) , taking the real part.
The intensity is proportional to the magnitude squared of this vector:  (A) is the answer. | |
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