Light intensity and polarization

[gespex]

I have read about light and polarization. One thing I don't understand though... If you would look through a circular polarization filter, you notice that the light is less intense.
But if you look at photons that are unpolarized; they become polarized when they go through the filter. I have read that the x and y components are separated (as x^2 + y^2 = 1).

So I believe the amplitude of the wave of polarized light becomes smaller, and this causes the decreased intensity, is this correct? But if that's the case, how come this photon still has exactly the same amount of energy?

Thanks in advance

 

[Bill_K]

gespex, In answer to your last question, the energy of a photon depends only on its frequency, not the amplitude. The amplitude tells you how many photons you have.

Actually for situations like you're describing, I'm puzzled that anyone would want to talk about photons. As far as I know, Maxwell's equations have not yet been repealed! And it is often much easier to talk about electromagnetic waves than photons. Photons are only appropriate in other situations - at very high energy or very low intensity.

Regarding your first question, filters by definition remove part of the input, so naturally the output is less. Your idea that a photon would "become polarized" when passing through the filter is not true - part of the wave (i.e. some of the photons) is/are simply not transmitted.

 

[gespex]

gespex, In answer to your last question, the energy of a photon depends only on its frequency, not the amplitude. The amplitude tells you how many photons you have.

Actually for situations like you're describing, I'm puzzled that anyone would want to talk about photons. As far as I know, Maxwell's equations have not yet been repealed! And it is often much easier to talk about electromagnetic waves than photons. Photons are only appropriate in other situations - at very high energy or very low intensity.

Regarding your first question, filters by definition remove part of the input, so naturally the output is less. Your idea that a photon would "become polarized" when passing through the filter is not true - part of the wave (i.e. some of the photons) is/are simply not transmitted.

Thanks for your answer. And that's understandable. But that doesn't answer my main question: what makes the light's intensity less at the other side of a polarization filter? Does that mean energy and light intensity, in this case, are completely unrelated?
Because from what I understand, a circular polarization filter removes one of the wave components but not the other. Meaning that nearly all the waves are let through, with the only exception being waves that are polarized perpendicular to the passed through polarization. The only difference would be the amplitude of the wave, if I can believe the random webpage I read.
This means that nearly all energy would pass through, right? And yet, the intensity of the light on the other side is visibly less.
So, where is the flaw in my logic, or are intensity and energy in light simply unrelated?

 

[James R]

And that's understandable. But that doesn't answer my main question: what makes the light's intensity less at the other side of a polarization filter? Does that mean energy and light intensity, in this case, are completely unrelated?

You need to distinguish between the energy of one photon and the energy of a group of photons passing through the filter. If an individual photon happens to be passed through the filter, then its energy will be unchanged. If it isn't passed through, then the photon is simply not there on the other side.

Overall, your light wave consists of billions and trillions of individual photons. Some pass through; others do not. What you call the "light intensity" is a number that is proportional to the number of photons present. The energy is of course related to the number of photons that pass through the filter - just add up the individual energies of all the photons that make it through.