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eliben

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This is my first post to physicsforums. I hope I've picked the right forum for this question, please correct me if not

To answer my question you'll have to have the book "QED, the strange theory of light and matter" in front of your eyes.

I'm now reading Feynman's QED book and ran into something I'm not sure I fully understand towards the end of chapter 2 (page 71). He explains why the probabilities for a photon to reflect and pass through a double-surface glass add up, and I don't think he does it very well.

He says (on page 71, just before figure 45):

But look how nicely things work out: the extra turns made by the stopwatch timing a photon during steps 3 and 5 (on its way to A) are exactly equal to the extra turns it makes timing a photon during steps 5 and 7 (on its way to B). That means when the two reflection arrows are cancelling each other to make a final arrow representing zero reflection, the ar-

rows for transmission are reinforcing each other to make an arrow of length 0.96 + 0.04, or 1--when the probability of reflection is zero, the probability of transmission is 100%

I'm not sure I get

**why**this is true. After all, won't steps 5 + 7 cancel out the transmission just as steps 3 + 5 cancel out the reflection ?

The only thing I can think of that solves it is that reflection in air from glass turns the "clock" by 180 degrees, while the reflection from inside the glass does not. Is this the solution ?