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# Quantum Electro Dynamics

Posted on 2011-05-05
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I'm reading a popular book on QED and it explains that while light appears to travel in a direct line from the bulb to the mirror and then to the observer, in reality the light is bouncing off the mirror every which way, and it is only the frequency cancellation that makes these other less-direct paths not directly observable.

What about light going from the bulb to the observer without bouncing off of anything? Is that also traveling in many different ways with the less direct paths cancelling out? Or in that case, is the light that you see just following the direct path?

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Question by:MajorBigDeal
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>  Is that also traveling in many different ways with the less direct paths cancelling out?
yes
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Author Comment

ID: 35696239
Would that include a path where the light leaves the bulb from the side facing away from the observer and then makes some corkscrew motions (all without bouncing off of anything) and then turns around to go back towards the observer and then goes into the eye of the observer?
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You can even block a set of less direct paths to keep them from canceling out.
A familiar apparatus for doing this is a diffraction grating.

The Principle is simialr to Huygens'.
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But a diffraction grating requires a reflection so that is off the table.   Let me try to explain the part I am trying to clarify (I appreciate your assistance BTW).

In the mirror example, I get that the light bounces off the mirror all over the place.  If it happens to bounce in the direction of the observer then it will contribute to the final result.   But what if it bounces away from the observer (as most of it must do). Then it will not contribute to the observed result. Is that correct - would you agree with this?

So if we don't allow any reflections,  can any light leaving the bulb from the side opposite the observer, turn around and follow a path to the observer? Or is it only the light that leaves the bulb towards the observer that contributes to the observed result? As you can probably tell, I am very confused.
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ID: 35696565
There are also transmission diffraction gratings.

If it bounces away from the observer it does not contribute to the observed result. (unless there are other observers)
If it destructively interferes it does not contribute to the observed result.

All paths contribute, but the screwier the path, the more it interferes with nearby paths.
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Thanks!
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