Infinite light polarizers in an array

[sqljunkey]

I was wondering if I placed an infinite number of light polarizers in an array, each rotating an infinitesimal amount from the next, would I be able to get 100% of the photons shun thru them on the other side?

 

[Keith_McClary]

Do you mean like this:
http://www.ilectureonline.com/lectures/subject/PHYSICS/6/64/912

 

[sqljunkey]

Yes thanks.

 

[sqljunkey]

Hi,
Does the intensity of the beam of light matter? What is happening when the polarizer fails to polarize the beam of light?

 

[Keith_McClary]

Hi,
Does the intensity of the beam of light matter? What is happening when the polarizer fails to polarize the beam of light?

The intensity doesn't matter.
In the video, the first polarizer polarizes the beam.

 

[Nugatory]

I was wondering if I placed an infinite number of light polarizers in an array, each rotating an infinitesimal amount from the next, would I be able to get 100% of the photons shun thru them on the other side?

No. At every step, the photon Is either transmitted with the new polarization or absorbed - and the probability of absorption is always non-zero.

You can use Malus's law to calculate the exact probability at each step, and if you know how to evaluate limits you can work out what gets through an infinite number of arbitrarily small steps.

 

[Vanadium 50]

No. At every step, the photon Is either transmitted with the new polarization or absorbed - and the probability of absorption is always non-zero.

I don't think this is right.

Let me work in amplitudes, not intensities. I will show that with an infinite number of polarizers I can flip a polarization from horizontal to vertical with no loss in amplitude. If that is true, I can make an arbitrary rotation with an infinite number of polarizers with no loss in intensity.

Start with n polarizers:

A = A \cos^n (\frac{\pi}{2n})

Take the small angle approximation:
A^\prime \approx A (1 - \frac{\pi^2}{4n^2})^n

Use the approximation (1+\epsilon)^n \approx 1 + n\epsilon

A^\prime \approx A (1 - \frac{\pi^2}{4n})

Which gives, in the large n limit A^\prime = A. If the amplitude is unchanged, so is the intensity.

 

[sophiecentaur]

No. At every step, the photon Is either transmitted with the new polarization or absorbed - and the probability of absorption is always non-zero.

You can use Malus's law to calculate the exact probability at each step, and if you know how to evaluate limits you can work out what gets through an infinite number of arbitrarily small steps.

Malus' Law gives you the amplitude, which is proportional to the root of the photon flux. There is no point in going into photons from Malus' Law.
As @Vanadium 50 says, if you take the limit, you get the directly opposite answer. Maths can be weird. However, I guess the answer to a practical situation would be that each polariser would absorb some co-polar Energy so the answer would actually approach zero.