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There is doubt regarding concept of polarization in my mind regarding a YouTube video https://youtu.be/ji3o61_ORT4?si=j4zBy5dLg1AhJpNu&t=6

in this video Scott Milam shows 2 polarizer perpendicularly aligned with each other but still we can see some of the intensity of light passing through it ,it must be zero according to Malus law but still we are able to see some of the light intensity coming through it? can somebody plz explain what may be the reason for the following.

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    $\begingroup$ looks like cheaper polarizers .... no polarizer is 100.000% $\endgroup$
    – PhysicsDave
    Commented Oct 28 at 15:33

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The only reason I can think is the light is not in a pure state of polarization, i.e., there is some degree of depolarization check it here. Depolarized light is basically the incoherent mixture of two or more states of polarizations. Such incoherent mixtures does not present interference effects, since we don't have correlation between the component waves in the mixture.

Depolarized light is the most common way we find light in nature. We can produce it through polarized light using depolarizers, but usually it is not intentionally created by thermal fluctuations, inhomogeneities and impurities in the light source, etc. Perfect Polarized light is a idealized case, where such imperfections are negligible. Of course it's possible to produce sources of high polarized light, but it seems not the case in this video

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    $\begingroup$ Can you explain this line "Such incoherent mixtures does not present interference effects, since we don't have correlation between the component waves in the mixture." present in your statement in a more simpler way @Ruffolo. $\endgroup$
    – Assassins Hunter
    Commented Oct 29 at 8:13
  • $\begingroup$ A depolarized light can be thought as a mixture of waves. Usually, when we have two or more waves in phase, we have interference effect, where they will sum up it's amplitudes. When it happens, we say the waves are in a coherent mixture. Now we have a situation where waves are not in phase. In such cases, the mixture does not show interference effect. The resulting intensity is just the sum of intensities. We call it incoherent mixture $\endgroup$
    – Ruffolo
    Commented Oct 29 at 9:18
  • $\begingroup$ When we turn on two light bulbs at home, we don't observe interference between their lights. There are a mixture of both waves, but it is a incoherent mixture. It happens because they are not in phase. Now, sometimes it's said that there are no correlation between the waves. This term comes from quantum optics as I now, but you should check. $\endgroup$
    – Ruffolo
    Commented Oct 29 at 9:21
  • $\begingroup$ The 2 bulbs are incoherent because their frequency is not exactly equal or their phase difference is time dependent, it not because of they are not in phase but rather their phase difference is time dependent and that is why they are incoherent. @Ruffolo. $\endgroup$
    – Assassins Hunter
    Commented Oct 29 at 10:11
  • $\begingroup$ Please rectify me @Ruffolo if my above statement is wrong with proper explaination. $\endgroup$
    – Assassins Hunter
    Commented Oct 29 at 11:08
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It's probably a polaroid (https://en.wikipedia.org/wiki/Polaroid_(polarizer)). Cheep and easy, but not 100%. Maybe a company data sheet has a 2 x 2 matrix describing it.

Wiki (https://en.wikipedia.org/wiki/Polarizer) says silver nano-particle embedded in thin glass plates can get something looking like:

$$ T = \left[\begin{array}{cc} 0.987 & 0 \\ 0 & 10^{-5} \end{array}\right]^{\frac 1 2} $$

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  • $\begingroup$ Can you please tell what does "T" matrix actually represents? @JEB. $\endgroup$
    – Assassins Hunter
    Commented Oct 29 at 8:16
  • $\begingroup$ transmission of [algined, anti-aligned] light, though my it needs a square since it operates on amplitudes...imma fix that now. $T, R, A$ are pretty common for transmission, reflection and absorption when talking about polarizations, and a 2x2 operates on a 2x1 matrix of polarization states, though there are many choices for basis. Here, it's linear in the OP, so it's linear here. $\endgroup$
    – JEB
    Commented Nov 11 at 15:34

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