Why do wire-grid polarizers have a separation less than the wavelength?

[montser]

I've read about wire-grid polarizers from wikipedia. The article says that waves perpendicular to the wires cannot induce electron movement and hence they pass through with minimal energy loss. On the other hand, waves parallel to the wires are reflected by the wires because electron movement is possible. What I don't understand is two things:

1. why the electron movement doesn't also induce a transmitted wave?

2. why must the wires need to have a separation less than the wavelength?

Thanks

 

[sophiecentaur]

You need 2. in order for 1. to happen.
Each wire has an 'effective cross section' so that high enough currents will be induced into it for the grid to act like a sheet.

Why just reflection? is because there is 180degree phase shift and the re-radiated wave is of equal amplitude to the incident wave. That means that forward waves will cancel . The energy has to go somewhere, so it goes back the way it came.

 

[montser]

You need 2. in order for 1. to happen.
Each wire has an 'effective cross section' so that high enough currents will be induced into it for the grid to act like a sheet.

Why just reflection? is because there is 180degree phase shift and the re-radiated wave is of equal amplitude to the incident wave. That means that forward waves will cancel . The energy has to go somewhere, so it goes back the way it came.

what phase shift?between what?and why there is a phase shift?

Thanks for you replay

 

[sophiecentaur]

It's all about so-called Boundary conditions. The E field must be zero at the surface of a perfect conductor (a 'short circuit' therefore no volts) so the wave that is produced by the currents induced in the surface must produce a wave that is in anti-phase with the incident wave. The sum of these waves, in the forward direction is zero.

Having a grid is almost as good as having a continuous conducting surface (for that particular polarisation) as long as the wires are spaced close enough together.

 

[montser]

It's all about so-called Boundary conditions. The E field must be zero at the surface of a perfect conductor (a 'short circuit' therefore no volts) so the wave that is produced by the currents induced in the surface must produce a wave that is in anti-phase with the incident wave. The sum of these waves, in the forward direction is zero.

Having a grid is almost as good as having a continuous conducting surface (for that particular polarisation) as long as the wires are spaced close enough together.

First, thanks for your replay. Now I understand why there is a phase shift. But what I still do not understand is why this re-radiated wave do not cancel also the incident beam? in other word, what breaks the symmetry between the incident and the forward waves?

again thanks

 

[Born2bwire]

First, thanks for your replay. Now I understand why there is a phase shift. But what I still do not understand is why this re-radiated wave do not cancel also the incident beam? in other word, what breaks the symmetry between the incident and the forward waves?

again thanks

With the scattered wave, there is no phase shift between the waves on either side of the grid. Basically the grid behaves like a continuous sheet as described previously and it acts as a new plane wave source. So the waves on either side of the grid only differ in the direction of travel. So on the transmitted side, this means that the spatial phase progression follows that of the incident wave since they both travel in the same direction. But on the reflected side, the scattered wave is traveling in the opposite direction (if we assume normal incidence) as the incident wave. Thus, the phase progression in space is different between the two. For example, the incident wave may be something like

$$\mathbf{E}_{inc} \sim \cos (kx-\omega t)$$

While the reflected wave is,

$$\mathbf{E}_{ref} \sim \cos (-kx-\omega t) = \cos (kx+\omega t)$$

So you can see that for a given time t that the two will interfere constructively (take the case of t = n*2\pi for an easy comparison).

 

[sophiecentaur]

For normal incidence, there will be a standing wave formed from the incident and reflected wave as the two waves interfere constructively and destructively at different points in space.