What happens to waves with different polarizations in a wire grid polarizer?

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In a wire grid polarizer, waves with polarization parallel to the wires are primarily absorbed, but some energy can pass through the gaps between the wires, depending on the spacing relative to the wavelength. Conversely, waves with perpendicular polarization experience some reflection, with the amount reflected influenced by the wire's width compared to the wavelength. The discussion highlights that while textbooks may not fully explain these interactions, mathematical relationships can quantify the energy transmission and reflection based on wire dimensions. Misconceptions about electron movement and absorption in the wires are also addressed, emphasizing the need for careful interpretation of sources. Understanding these principles is crucial for accurately analyzing wave behavior in polarizers.
Pushoam
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It is said that the wave whose polarization is parallel to the wire, gets absorbed by the wire.
But, there is empty space between the wires. So, that part of the above - said wave which passes through the empty space will not get absorbed by the wire. So, the transmitted wave must have the wave whose polarization is parallel to the wire.

On the other hand, some part (maybe small) of all of the wave with polarization perpendicular to the wire should get absorbed by the wires.
So, the resultant transmitted wave should have the wave with both kinds of polarization.

The books which I have come across leave the above-mentioned ideas.
Could anyone please clarify it?
 

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You are right that some energy from waves who have E-field parallel to the wires will pass through the gaps. But if we do the math we can prove that the energy that will pass is proportional to the ratio ##\frac{a}{\lambda}## where ##a## is the spacing between the parallel wires and ##\lambda## the wavelength of the waves. So if we arrange for ##a## to be sufficiently small compared to ##\lambda## only a small percentage of energy will pass and the rest will be reflected.

Also the percentage of the energy from waves who have E-field perpendicular to the wire that gets reflected is proportional to ##\frac{b}{\lambda}## where ##b## the width(the vertical dimension) of the parallel wires. So if ##b## is small in comparison to ##\lambda## very small percentage of this energy will be reflected and the rest will pass.
 
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Thank you.
 
Pushoam said:
the wave whose polarization is parallel to the wire, gets absorbed by the wire
It's reflected and not absorbed. To be absorbed, you would need resistance wire of just the correct resistance value.
There is also a nonsense statement in the attached passage which says that "the electricity trans cannot move very far across the width of the wire". At a mean speed of a mm per second (drift velocity of electrons in a metal) and, assuming we have a classical motion for the electrons, how far in either direction would the electrons go in the period of oscillation of the RF wave? In fact, the whole passage is pretty nonsensical and would be better ignored. I wonder what surprises the other 32 pages have for the reader. Treat it all with care and stick to a proper textbook.
 
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