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

• Pushoam
In summary, 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.
Pushoam

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.

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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.

Pushoam
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.

Pushoam

## 1. What is a wire grid polarizer?

A wire grid polarizer is a thin, flat sheet made of parallel metallic wires that are closely spaced. It is used to polarize light by selectively transmitting or blocking certain orientations of light waves.

## 2. How does a wire grid polarizer work?

A wire grid polarizer works by acting as a filter for light waves. Light is an electromagnetic wave that can oscillate in any direction. As the light passes through the wire grid, only the waves that are aligned parallel to the wires can pass through, while the waves that are perpendicular to the wires are blocked.

## 3. What are the applications of wire grid polarizers?

Wire grid polarizers are commonly used in various optical instruments, such as cameras, microscopes, and polarizing filters for LCD screens. They are also used in telecommunications, astronomy, and scientific research.

## 4. What are the advantages of using wire grid polarizers?

Compared to other types of polarizers, wire grid polarizers offer high transmission efficiency, low absorption, and wide bandwidth. They are also lightweight, compact, and can be easily integrated into optical systems.

## 5. Are there any limitations of wire grid polarizers?

One limitation of wire grid polarizers is that they can only polarize light that is incident at a specific angle. They are also sensitive to temperature variations, which can affect their performance. Additionally, they can be expensive to produce in large sizes.

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