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I read this from Optics (Global Edition) by Eugene Hecht Page 347, Chapter 8: Polarisation. The author says that this is obvious but I seem to be missing something here. Is the incident wave out of phase with the wave propagating in the forward direction?

Thanks for your help!

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  • $\begingroup$ I suspect you are correct, but it would be helpful if you could post a diagram of the system so we know exactly what the configuration is, wavelengths, etc. $\endgroup$
    – Carl Witthoft
    Commented Nov 2, 2020 at 13:43

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Because for a good metal, the current is essentially 90$^\circ$ out of phase with the exciting radiation. This is due to the metal’s conductivity, which is largely imaginary. Then the re-radiated fields have another ~90$^\circ$ phase shift, bringing them (nearly) completely out of phase with the exciting radiation. Transmitted light is cancelled, and the emitted reflection sets up a standing wave on the incident side of the polarizer (with a node at the plane of the wires).

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