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When unpolarized light is polarized with two polarizers, the intensity becomes $I=I_ocos^2(θ)$ (Malus's law). But when unpolarized light is polarized with only one polarizer, the intensity is reduced to half the intensity of the unpolarized light. Why?

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  • $\begingroup$ Because that's how a polarizer works... Calculate $\theta$. $\endgroup$ – Shahar May 20 '14 at 12:13
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    $\begingroup$ $I=\frac{1}{2\pi}\int_0^{2\pi}I_\text{o}\cos(\theta)^2\,d\theta=I_\text{o}/2.$ $\endgroup$ – DumpsterDoofus May 20 '14 at 12:15
  • $\begingroup$ But now you're taking the integral of the intensity after the second polarizer? $\endgroup$ – Nic May 20 '14 at 12:42
  • $\begingroup$ No, the integral I wrote above assumes only one polarizer. It's integrating over the unpolarized input. $\endgroup$ – DumpsterDoofus May 20 '14 at 17:24
  • $\begingroup$ Where does the cosine term come from? $\endgroup$ – Nic May 20 '14 at 17:31
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Malus's law is about the effect of a polariser on polarised light. You've clearly read a badly written version of it. What your author likely meant to say was:

  1. One begins with unpolarised light;

  2. The first polariser quells the unaligned component of the unpolarised light and outputs polarised light (with half the input's intensity). This polarised output has intensity $I_0$ in your notation;

  3. Of the polarised output from the first polariser, the second polariser lets through a fraction $(\cos\theta)^2$ where $\theta$ is the angle between the axes of the polarisers. So I say again: $I_0$ is the intensity of the polarised input to the second polariser, not the intensity of the unpolarised input to the system of two polarisers. With this proviso, the output intensity is $I_0\,(\cos\theta)^2$.

In Answer to:

But I don't understand why the intensity is lowered to half the input's intensity after the first polariser?

Depolarised light is actually quite a subtle and tricky concept: I discuss ways of dealing with it in my answers here and here. You can think of it intuitively: without a preference for polarisation, perfectly depolarised light must dump half its energy into a polariser: you can take this as a kind of "definition" of depolarised light if you like. The quantum description is much simpler than the classical, so I reproduce it here. We imagine the source producing photons each in pure polarisation states but with random direction. That is, each photon's polarisation axis makes some random, uniformly distributed angle with the polariser's axis. Its pobability of absorption is therefore $(\cos\theta)^2$. So now we simply average this quantity given a uniformly distributed angle:

$$\left<(\cos\theta)^2\right> = \frac{1}{2\pi}\int_0^{2\pi} (\cos\theta)^2\,\mathrm{d}\theta = \frac{1}{2}$$

to find the overall probability of passage through the polariser, and thus the proportion of photons that make it through. For more info on what polarisation actually means for a single photon, see my answer here.

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  • $\begingroup$ But I don't understand why the intensity is lowered to half the input's intensity after the first polariser? $\endgroup$ – Nic May 20 '14 at 12:36
  • $\begingroup$ @Nic See my update at the end of my answer $\endgroup$ – Selene Routley May 20 '14 at 12:58
  • $\begingroup$ I remember having this question during my freshman year, and having to do research in order to obtain a reasonable explanation. I was still unsatisfied with the average value integral answer I found, however, this makes it much more intuitive. Thank you. $\endgroup$ – Gödel Aug 12 '14 at 1:03
  • $\begingroup$ @NaturalPhilosopy I'm not surprised at your having to do research. As I said, depolarised light is quite a subtle and advanced concept: MUCH harder than most instructors make out. The quantum description shows it can be made simple, which is good, but this still isn't the same as "easy", especially on first meeting. $\endgroup$ – Selene Routley Aug 12 '14 at 2:35

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