Polarization of white light beam

hamdal

I want to find out that what happens if we polarize a white light which has all of the wavelenghts of 400-700 (nm).Does it devide into colors?What happens to it?

Khashishi

Nothing much happens to it, except it is polarized. You lose half the light. A lot of sunglasses are polarized.

AbsoluteZer0

I believe you've mistaken Scattering with Polarization. Scattering is the splitting of white light into its component frequencies (red, orange, yellow, green, blue, indigo, and violet.)

Polarization is, in a simple sense, the limiting of photon vibrations to one vibrational plane. Photons can vibrate in a horizontal plane as well as a vertical plane. In order to understand this better, take the "picket fence analogy:"



If the picket fence is lined horizontally, then only light vibrating in the horizontal plane will be able to cross through. Polarization decreases the amount of light that goes through. The frequency of the light that is polarized will remain the same, the only thing that will change is the alignment of vibration.

http://en.wikipedia.org/wiki/Scattering

sophiecentaur

Unfortunately, the 'picket fence' analogy can be very misleading. It has to be presented very well if it is not to be taken wrongly. It always has to be stressed that the slots are not just 'selective' (only letting certain waves through) but they take one vector component of all waves that arrive. If the fence is merely 'selective' then it will only let an infinitesimally small fraction of the energy through as opposed to half the energy which is what we find with real polarisers.
The pictures in the above post are not good, in this respect; the bottom picture really does suggest that none of the slightly tilted wave would get through.

Also, there is no point in talking of "photons vibrating" in a particular direction. If this were a proper model then a polariser would be taking a fraction of the energy of any photon that is not perfectly aligned with the polariser's plane. We all know that photons have an energy value that is determined by the frequency so how could we only have a fraction of this energy? Why oh why do people seem to think that photons help in these arguments? It just shoots any explanation in the foot. (Even that picket fence argument manages to avoid photons - that is one small thing in its favour)

AbsoluteZer0

So the picket fence analogies are entirely useless?

phinds

I don't think the centaur said useless, the message was "misleading --- handle with care", which is a good message

EDIT: I think that as an explicitly non-technical ANALOGY, it is a helpful explanation for non-technical people.

nasu

Are you sure you mean Scattering?
Anyway, is quite misleading to equate Scattering with "splitting into components".
Scattering may result in some spectral separation but so can refraction or interference, for example.

hamdal

No no no...I definetely didn't mean scattering,diffraction or any other things...I read that the colors appearing on a CD case is because the light polarizes and then Birefringence happens and some how we see colors...I'm trying to see why Birefringence and polarization gives us colors!
and by the way...when scattering, refraction or defraction happens the frequency stays the same, it's the wavelenghts that change!

acording to the Snell's law we have:
(V_1)/(V_2) =(λ_2)/(λ_1)
and we have:
V=fλ
so if:
(V_2)(λ_2)=(V_1)(λ_1)
then:
f_2=f_1

sophiecentaur

Possibly but you have to expect probelms down the line. You could ask what do they want to know for if the analogy is either a poor one or needs loads of qualifications. When asked about polarisation, I always refer to TV antenna on peoples' roofs. They all recognise one and they also seem to grasp very quickly how the signal out varies as the orientation of the antenna is moved through a complete turn. It's a short step to move into light polarisation. No fudges involved there, I think and nothing they are going to need to 'un-learn' later. They even get the fact that exact cross polarisation cancellation is very hard to achieve and you don't need to mention the word 'vector'.

sophiecentaur

I'm not sure what you read but the mechanism for that sort of thing is complicated and involves more than just birefringence. It involves the phenomenon of interference and the differing path lengths for the two rays through the medium. Certainly not the best thing to cut your teeth on in this topic. (Notice that the colours you get in examples like this are not the same as the 'rainbow'. They are more like oil film colours which are more 'garish' and unnatural looking.)

btw, your final line
is not a 'result' ("then"); it's a starting boundary condition for phase continuity at the interface.

hamdal

So you mean it can be diffraction?can it be Birefringence and diffraction at the same time?I need a theory for this thing!for experiments I have a few ideas but for the theory I have almost nothing!

sophiecentaur

I think you will need to say in more detail what you have read. It doesn't quite make sense to me. Birefringence will not, in itself, produce separation of wavelengths. You will get dispersion at an interface when the angle of incidence is not zero and, with a birefringent material, I guess you can expect the two rays to be dispersed differently and at different angles but, for parallel sided blocks, the 'prism' effect is minimal.
As I said earlier. This is a complicated scenario and you would need to describe more precisely what you want to know about it. If you want to understand how colours can be formed under different circumstances then it would be better for you to look at all possible mechanisms (dispersion, interference in thin films, diffraction by regular gratings etc.) and then see which one best describes what you have been seeing. This calls for some private study, I think, rather than 'question and answer' as, at the moment, your questions are not well defined enough to answer.

hamdal

dispersion, interference in thin films and diffraction by regular gratings!got it...thank you for your help!

sophiecentaur

Ah. What you needed was the relevant buzzwords. I see now. ;-)