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In Fig. 33–6(a) we see that the field of amplitude b is radiated by the motion of charges inside the glass which are responding to a field a inside the glass, and that therefore b is proportional to a. We might suppose that since our two figures are exactly the same, except for the direction of polarization, the ratio B/A would be the same as the ratio b/a. This is not quite true, however, because in Fig. 33–6(b) the polarization directions are not all parallel to each other, as they are in Fig. 33–6(a). It is only the component of A which is perpendicular to B, Acos(i+r), which is effective in producing B. Now we use a trick. We know that in both (a) and (b) of Fig. 33–6 the electric field in the glass must produce oscillations of the charges, which generate a field of amplitude −1, polarized parallel to the incident beam, and moving in the direction of the dotted line. But we see from part (b) of the figure that only the component of A that is normal to the dashed line has the right polarization to produce this field, whereas in Fig. 33–6(a) the full amplitude a is effective, since the polarization of wave a is parallel to the polarization of the wave of amplitude −1. Therefore we can write

Acos(i−r)/a=−1/−1,(33.2)

QUESTION: How is it that the component of A perpendicular to B is Acos(i+r)?

And how is Acos(i-r) the component of A that is normal to the dashed line?

Satheesh Paul Antonysamy
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  • Extend B backwards, into the glass. Note that the angle between A and -B is i+r. – andars May 05 '17 at 18:30
  • But that's the component of A that is (anti-) parallel to B, not perpendicular to B. There might be some confusion, because A and B are supposed to be amplitudes, so Feynman might have meant them to be the little dashed vectors that are perpendicular to what the figure shows as A and B. It's conceivable that there was a breakdown between transcribing the text and drawing the figure, but unless I'm misunderstanding, there does seem to be a problem with the passage (but probably not with the results). – NickD May 05 '17 at 19:58

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Here is Feynman's original blackboard figure:

enter image description here

He didn't say (for example), "It is only the component of A which is perpendicular to B, A cos(i+r), which is effective in producing B." What he actually said was, "... in this case I have the factors of polarization; it's only that component of this polarization which is in this direction which is effective in that case. And therefore it isn't just B/A, but it has an extra factor which is the cosine of the angle between here and here. That is, if I were to project this direction here, and ask for the component in this direction, which is the only component that's effective to produce capital 'B', it would be less than it would be in this case by the cosine of that angle, so you see where that comes from." The transcribed text and figure were edited by Leighton.

Mike Gottlieb

Editor, The Feynman Lectures on Physics

www.feynmanlectures.info www.feynmanlectures.caltech.edu

Michael A. Gottlieb
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    It's pretty nice to be able to go back and check the original material! Thanks. – NickD May 06 '17 at 22:57
  • This has been corrected. We changed, "It is only the component of A which is perpendicular to B, Acos(i+r), which is effective in producing B" to "It is only the component of the electric field in the glass which is perpendicular to B, Acos(i+r), which is effective in producing B." And we changed "...only the component of A that is normal to the dashed line has the right polarization to produce this field,..." to "...only the component of the electric field in the glass that is normal to the dashed line has the right polarization to produce this field,...". – Michael A. Gottlieb May 18 '17 at 11:03
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    The corrections are now in place in the online edition of FLP and will appear in future printings of the books. The corrections will be added to our posted FLP errata and Satheesh Paul will be added to our list of FLP-NM errata contributors. – Michael A. Gottlieb May 18 '17 at 11:03
  • Thank you, Sir. It is truly an honor to have my name associated with FLP. – Satheesh Paul Antonysamy Jan 17 '19 at 15:53