Amplitude Normalization of Electromagnetic Waves

[asksage]

I've two electromagnetic waves (light) with amplitudes 1x (normal) and 2x (double) amplitude. And I want to pass these two waves through a "normalizer" expecting 1x amplitude for both waves.

Question is: Is such a "normalizer" possible and/or exists. I'm not looking for any electronic solution. Ideally, it should be some sort of organic polymer and/or photonic band-gap material. Something like a polarizer for example.

Thanks!

 

[sophiecentaur]

You are asking for something quite sophisticated, if it is to work very well. Photochromic sunglasses do the sort of thing you are after but they work on UV, I think. Anything passive, like the sunglasses, will not have an ideal normalising behaviour but could start to 'saturate' at a certain level so that doubling the input level might produce a lot less than double the output level. This is more of a 'compression' than a 'limiting' (to use Audio terminology).
You could do it as well as you chose to if you could employ feedback and an active system.

 

[asksage]

Thank you "sophiecentaur" for your reply.

What if we we restrict our "normalization filter" to output only fix amplitude of the wave i.e. no matter what the amplitude of the input wave is e.g. 1x, 2x, 3x, 4x etc. It outputs only a fixed pre-defined amplitude.

Shouldn't it be easier now? Now, can we make it a passive filter/system without any feedback? So that no matter what the input amplitude is >= x, we get x amplitude on the output.

Thanks once again!

 

[buttkhan]

Ok, I've a question.. What would happen if an electromagnetic wave with a peak-to-peak amplitude of say 400nm is incident on a hole with diameter less than 400nm e.g. say half i.e. 200nm?

Will it pass through? If yes, will the amplitude of the wave on the other side of the hole will still be 400nm?

 

[sophiecentaur]

When you say "peak to peak amplitude" are you referring to the Wavelength or are you using using inappropriate units for the Amplitude of the wave (which should be in V/m)? The wavelength would not be altered by going through an aperture.

@askage: If you need to do this with a Radio Frequency signal then is is straightforward to do this with circuit components on a transmission line. A pair of diodes can easily limit the volts to, say 0.5V pk to pk but I can't think of a substance that would achieve this with higher, optical, frequency signals. I think that the situation changes when you can no longer deal with the variations of the fields using circuit elements / electrons flowing in wires.

A "filter" system would normally be linear - i.e. output proportional to input. All you could hope for (I think) would be some substance that might warm up as the energy through it increases and that this increase in temperature could increase the absorption of the radiation. This would produce, as I said previously, a compression of the input power range, rather than a hard limit to the transmitted power.

Edit. Have you a particular application in mind?

 

[buttkhan]

When you say "peak to peak amplitude" are you referring to the Wavelength or are you using using inappropriate units for the Amplitude of the wave (which should be in V/m)? The wavelength would not be altered by going through an aperture.

I meant Amplitude not Wavelength and sorry for using inappropriate units. So, will amplitude be altered?

 

[sophiecentaur]

OK
I think you may need to modify your view of what goes on here.
Say there is a beam of light, incident on a hole. The aperture will restrict the amount of energy getting through - more or less according to the fraction of the area of the beam that coincides with the hole. But the real issue here is the effect of Diffraction. This is because of the wavelike nature of light which produces interference between all the different parts of the wave front.
The effect of an 'edge' on a plane wave that hits it is to produce a spreading out of the light away from the original direction of the beam (in both directions). If there is a big hole, the majority of the the light goes straight through but, as the hole gets smaller (approaching the wavelength of the light) the spread out energy is a bigger proportion of what gets through because the edges constitute a major part of the 'hole'. The pattern emerging from a tiny pinhole will be a set of concentric rings whilst the pattern emerging from a doorway will be a silhouette of the doorway.
Google "diffraction" to find out all about diffraction / interference and you will see that it is the aperture (hole) size, relative to the wavelength that determines what happens - whether you just get a shadow or a 'fringed' pattern. The actual amplitude of the wave makes no difference.