# QED: reflection of light in glass

eliben
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):

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 ?

Renge Ishyo
Hi there eliben and welcome. In this example, the two vectors for the light are pointing in opposing directions (see figure 44 and follow the direction of the light as it travels from the source to either the detector at A or through the glass to B...you will see that 3 to 5 goes up towards A and 5 to 7 goes down towards B). One pathway leads towards transmission (5 to 7) and one leads towards reflection (3 to 5). If they are both opposing tendencies and equal in magnitude they cancel, which means the light follows the pathway determined before entering the glass (which if you cover up the glass completely with your finger in figure 44, you can see that the light points from the source directly towards B and not at all towards A).

Some notes to help you with in this book: The "clock" that Feynman talks about is really the frequency of the light wave (cycles per second), and the explanations he gives about vectors cancelling each other out so that light is either seen or not seen at a detector is no different than the interference patterns you see in a typical double slit experiment.

Hope it helps.

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eliben
Hi there eliben and welcome. In this example, the two vectors for the light are pointing in opposing directions (see figure 44 and follow the direction of the light as it travels from the source to either the detector at A or through the glass to B...you will see that 3 to 5 goes up towards A and 5 to 7 goes down towards B). One pathway leads towards transmission (5 to 7) and one leads towards reflection (3 to 5). If they are both opposing tendencies and equal in magnitude they cancel, which means the light follows the pathway determined before entering the glass (which if you cover up the glass completely with your finger in figure 44, you can see that the light points from the source directly towards B and not at all towards A).

I don't understand, sorry. The path the light makes inside the glass (3-5-7) doesn't cancel out, because there are two shrinks by 0.2 caused by it (at points 4 and 6).
What does cancel out is that the path inside the glass on the way to A (3-5) sometimes cancels out the direct reflection off the front surface (this is the case in which there's no reflection at all).
Rather, the part I was asking about was why when the reflection path cancels out, the transmission path is at its maximum.

Mentor
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 ?
Yes, that's exactly the solution. The reflection at the air/glass interface gives you the 180 degree phase shift.

Feynman is comparing reflection to transmission. First compare reflection from the front surface (Fig 40) with reflection at the back surface (Fig 42). The first has a phase shift of 180 degrees (at step 2); the second has a phase shift due to steps 3 + 5 (without the phase shift from step 2 or 4).

Now compare transmission by two surfaces (Fig 43) with transmission with an extra "bounce" (Fig 44). The first has a phase shift due to step 3; the second has a phase shift due to steps 3 + 5 + 7. The difference between the two is just from steps 5 + 7.

But the phase shift due to 3 + 5 equals the phase shift due to 5 + 7. So when 3 + 5 equals 180 degrees, reflection is a maximum (don't forget the phase shift at step 2). But then transmission is a minimum, since there's no additional phase shift.

(As much as I love this book, I must admit that it requires careful reading. Feynman could have been clearer.)

eliben
Yes, that's exactly the solution. The reflection at the air/glass interface gives you the 180 degree phase shift.

Feynman is comparing reflection to transmission. First compare reflection from the front surface (Fig 40) with reflection at the back surface (Fig 42). The first has a phase shift of 180 degrees (at step 2); the second has a phase shift due to steps 3 + 5 (without the phase shift from step 2 or 4).

Now compare transmission by two surfaces (Fig 43) with transmission with an extra "bounce" (Fig 44). The first has a phase shift due to step 3; the second has a phase shift due to steps 3 + 5 + 7. The difference between the two is just from steps 5 + 7.

But the phase shift due to 3 + 5 equals the phase shift due to 5 + 7. So when 3 + 5 equals 180 degrees, reflection is a maximum (don't forget the phase shift at step 2). But then transmission is a minimum, since there's no additional phase shift.

(As much as I love this book, I must admit that it requires careful reading. Feynman could have been clearer.)

Thanks a lot !

eliben
Later, at page 105 (and figure 68) Feynman says that for class, the turn is 90 degrees, and not 180 as in Figure 41. What accounts for the difference ?

Mentor
Later, at page 105 (and figure 68) Feynman says that for class, the turn is 90 degrees, and not 180 as in Figure 41. What accounts for the difference ?
Realize that these refer to different things. The 90 degrees is the "turn" (phase shift) associated with a photon scattering from an electron in glass, and that in this more detailed analysis of transmission and reflection, one considers all parts of the glass, not just the front and back surfaces. Feynman relates this to the earlier (and simpler) analysis in the paragraph at the bottom of page 106 beginning: "There is a mathematical trick we can use to get the same answer..."

As to why it's 90 degrees, as opposed to any other value, I do not know. (I suspect it has something to do with the spin of the electron.) Perhaps someone with a better understanding of QED can shed some light on it.