Polarization of EM waves is preserved after reflection/refraction -- Why?

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Hello,
here's my question:
during the usual derivation of Fresnel's equations, it is assumed that an incident EM wave (plane monochromatic) is transverse electric or magnatic and that it keeps this nature after reflection and transmission.
How can this be proven?
Thank you!
 

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  • #2
hutchphd
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Because the boundary condition itself is isotropic in the plane. Notice it will change the sense of circular polarization upon reflection.
 
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I'm sorry I have to bring up again this discussion but I don't get why from isotropic boundary conditions it's so easy to conclude TM/TE "conservation" upon reflection/refraction.
Apart from that, here, in Feynman's lectures
https://www.feynmanlectures.caltech.edu/II_33.html
below Fig. 33-6, he says there's a lot of algebra behind it...
Has anyone ever done this calculation? I don't seem to find it anywhere and I'd really like to see it.
Thanks everyone!
 
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hutchphd
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Its done in my 1962 edition of Jackson Classical Electrodynamics. The details are, well, detailed. So long as you understand Brewster's angle and why a circularly polarized wave changes helicity upon reflection you know what you need IMHO.
 
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Delta2
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Its done in my 1962 edition of Jackson Classical Electrodynamics. The details are, well, detailed. So long as you understand Brewster's angle and why a circularly polarized wave changes helicity upon reflection you know what you need IMHO.
1962 edition of Jackson? I thought the first edition was 1973...
 
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Delta2
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NVM my mistake its 1962. All this time i thought i had the first edition but i actually had the second edition (which is from 1975 not 1973) which i bought back at 1999 when I couldn't find the 3rd edition in Greek bookstores.
 
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sophiecentaur
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it keeps this nature after reflection and transmission.
That's not always the case. I only need one example to show this: If a linearly polarised radio wave is intercepted by a dipole that's orientated away from the plane of the E field then there will be a wave, re-radiated from the dipole that's in the plane of the dipiole.
 
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Its done in my 1962 edition of Jackson Classical Electrodynamics. The details are, well, detailed. So long as you understand Brewster's angle and why a circularly polarized wave changes helicity upon reflection you know what you need IMHO.
In chapter 7.5 Jackson assumes it (pag 218, "all the electric fields are shown directed away from the viewer"). He actually discards the first boudary condition, saying it gives nothing...
It is satisfied, and that's good, but, as Feynman said, we need to prove that this is the only solution (together with the other one, the transverse magnetic).
Is it proven somewhere else in the book?
 
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1962 edition of Jackson? I thought the first edition was 1973...
I graduate school I used my 1962 ed from my undergraduate course rather than buy the second edition. The first test, do problems 11.2, 14.7 .....
 
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It's worth noting that isotropy of the boundary conditions in the plane means that the induced surface current will be in the same direction as the incident electric field component in the plane. To change linear polarization, the induced currents would need to have components normal to this direction. This would violate the stated isotropy.
 
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tech99
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If the surface is
It's worth noting that isotropy of the boundary conditions in the plane means that the induced surface current will be in the same direction as the incident electric field component in the plane. To change linear polarization, the induced currents would need to have components normal to this direction. This would violate the stated isotropy.
If the reflecting surface is not a perfect conductor, we may find a pseudo Brewster Angle effect. Waves approaching at this angle, if in the plane of the paper, are weakly reflected. Waves normal to the paper are strongly reflected. But if we have a wave polarised at 45 degrees to the paper, it may be resolved into two waves in the principle planes. These two waves suffer different reflection effects, in amplitude and phase, so when they are re-combined to give the emergent ray we might find it to be elliptically polarised.
 
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