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In a circle there's infinite amount of degrees (eg. 0 deg, 0.00000000000...1 deg etc.) In a ground school we are thought that there's 360 degrees in a circle.

A landscape behind my window is incoherent light source, so it randomly emits photons with all polarization directions.

When I put a polarizer between landscape and my eye... i can still see the everything. But how is that possible if the polarizer transmits only $1/\infty$ of all photons (since there's infinite amount of directions of polarization)?

Even if we assume that there's just 360 degrees in circle... The landscape behind my window is not 360 times darker when I observe it through filter (eg. polarization glasses).

Why won't polarizer dim the light severely?

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    $\begingroup$ You seem to be thinking that a polarizing filter is some sort of notch or delta-function filter which only transmits photons with polarizations precisely aligned to the filter and and that all other photons are rejected, but that's not how it works. Look up Malus's Law for polarization filters. $\endgroup$
    – user93237
    Apr 17, 2018 at 17:05
  • $\begingroup$ In addition to @SamuelWeir: the idea that our units (degrees) would have something to the transmission (1/360) is not good physics reasoning. $\endgroup$
    – JEB
    Apr 18, 2018 at 16:32

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A polarizer does not work by transmitting one very specific polarization angle and blocking all other - even only very slightly different - polarizations completely.

Instead a polarizer obeys Malus' law: If the angle between the axis of polarization of the incident light and the polarizer's polarization angle is $\theta$, then the fraction of transmitted intensity to incident intensity is $\cos^2(\theta)$. Or, equivalently speaking in terms of photons, the probability that a photon with definite polarization at such an angle is transmitted and not absorbed is $\cos^2(\theta)$.

The reason for this is that any incident polarization can always be seen as a linear superposition of the polarizer's polarization and the one orthogonal to it. The orthogonal polarization is completely blocked and the parallel polarization is completely transmitted - this is the definition of what a perfect polarizer does. Now, an incident electric field vector $\vec E$ at angle $\theta$ is $$ \vec E_0 = \cos(\theta) E_{||} + \sin(\theta)E_{\perp}$$ in terms of the field in the parallel direction $E_{||}$ and the field in the orthogonal direction $E_\perp$. After the polarizer, all that is left is $\vec E_1 = \cos(\theta) E_{||}$. Therefore, the ratio of the intensities is $$ \frac{I_1}{I_0} \propto \frac{(\vec E_1)^2}{(\vec E_0)^2} = \cos^2(\theta),$$ which shows Malus' law.

If the polarization angle of the the incident radiation is uniformly randomly distributed among the possible angles, then the average intensity will be $$ \frac{1}{2\pi}\int^{2\pi}_0\cos^2(\theta)\mathrm{d}\theta = \frac{1}{2},$$ i.e. "totally unpolarized" photons have, on average, an equal chance to be absorbed by or transmitted through the polarizer.

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Polarizers don’t just filter photons, they also change the polarization of the photons that make it through. If you send 1000 incoherent photons through a polarizer 500 on average will make it through and all of them will become parallel with the Polarizer. Thats why polarization filters do not dim (coherent) light completely. The closer a photon is polarized parallel with the slit the better chance it will make it through. A photon that is almost perpendicular to the slit can still make it through but has less chance. It will then become polarized as it goes through.

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    $\begingroup$ This answer kind of misses the point by just asserting that exactly half of "incoherent" photons will make it through. This is: 1. Not true because their passing or not-passing is probabilistic, so it will rarely be exactly half. 2. Only true in the sense of probabilities if you assume their polarization is uniformly random among all polarizations, and then it follows from the more specific claim that the probability for a single photon at polarization $\theta$ to pass through the polarizer is $\cos^2(\theta)$. 3. Fails to point out that there is also a classical view on this phenomenon. $\endgroup$
    – ACuriousMind
    Apr 17, 2018 at 21:27
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    $\begingroup$ @ACuriousMind I never said exactly, why would I? (1) they’re passing through or not is a matter of fact and not probabilities. We know this because they did. (2) incoherent just means a low degree of coherence. On the average 50% will make it through. It would never even be close to Deming completely. How and why would you argue against that? Oh and welcome back, I haven’t seen your name for a while. $\endgroup$ Apr 17, 2018 at 21:41
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If you use a polarizing filter for ultraviolet light you could see that visible light will be dimmed more as if you use a polarising filter for visible light. The ratio of reflected and absorbed light to the light which is going through the filter depends from the slits width.

If 50% of monochromatic light goes through the filter, this means that for some orientation of the filter the light with the polarisation direction from 0° to 90° and 180° to 270° goes through this filter. Behind the filter all light is polarised in the same direction. From this you can conclude that the lights electric and magnetic field components gets rotated and aligned.

To prove the last conclusion one has to put two filters behind one another. If one filter has the orientation to the other filter of 90°, no light is going through. But, now place a third filter between the others and orient this filter in the direction of 45°. Will you see light going through? If yes, does this prove that light is rotated by a well designed filter?

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