Hello, 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): 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 ?