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One of my school exam questions asked to show(mathematically of physically) that the intensity of unpolarised light passing through a polarizer is halved. I found this rather difficult to prove, but i made an attempt anyway.

MY ATTEMPT: Malus law: if we have 2 polarizers P1 and P2 inclined at angle x with each other, the intensity passing through P2 is $I_o$$cos^2(x)$ where $I_o$ is the intensity of light after P1. I cannot use this formula directly in this case because of the definition. But, if i understand correctly(which is not likely) P1 simply makes the unpolarised light plane polarised. So suppose now we consider a beam of unpolarised light. This can be thought of as a mixture of many plane polarised lights.consider any one of them and apply malus law. Now $I_o$ is the original intensity and x is the angle of the pass axis w.r.t the chosen plane polarised light. Since this can be chosen absolutely randomly, angle x can vary randomly from 0 to 2pi. We can theorefore find the EXPECTED transmitted intensity to be the average of $cos^2(x)$ as x varies from 0 to 2pi, which returns 1/2. Is this a correct proof of the question? I am not satisfied myself. Any help with regard to my proof woulf be appreciated. Also, i would like a physical explanation if possible.

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  • $\begingroup$ Yes, it is correct. The average (expected) transmission when you have random polarizations equally distributed in $\theta$ is $1/2$. $\endgroup$ – Sean E. Lake Sep 2 '16 at 6:39
  • $\begingroup$ Can you please suggest a physical reason? $\endgroup$ – Lelouch Sep 2 '16 at 6:42
  • $\begingroup$ A physical reason for what? For an unpolarized light beam being the same as light with random polarization that changes constantly? I don't personally know why that is the case, and tend to take it as the definition of unpolarized. For what it's worth, you can have partially polarized light in which the probabilities of the two polarizations are not equal (for example, see light reflected near, but not quite at, Brewster's angle hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polar.html ). $\endgroup$ – Sean E. Lake Sep 2 '16 at 6:49
  • $\begingroup$ Also, my assumption that a beam can be treated as a collection of plane polarised lights, and treating eavh individually assume that there is not interaction whatsoever between these plane polarised lights in the beam. Is this assumption true? $\endgroup$ – Lelouch Sep 2 '16 at 6:54
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Consider the unpolarized light to be a collection $C$ of a large number of plane-polarized waves with polarization distributed evenly over a circle, and each of amplitude $A$ and intensity $I \propto A^2$. Let the transmission axis of the plane-polarizer lie along the $x$-direction. The linear polarizer acts on one of the incident waves $\psi = \vec Ae^{i(kr-\omega t)}$, where $\vec A = \hat x\,A\cos\theta + \hat y\,A\sin\theta$ is the polarization vector of the light, via the transformation $$ \psi\underbrace{\mapsto}_{\text{polarizer}} \psi_x = \hat x\,A\cos\theta\,e^{i(kr-\omega t)}. $$ The amplitude of $\psi_x$ is $A\cos\theta$ and the intensity of this wave is $I_x \propto A^2\cos^2\theta$, yet we must integrate over the range of angles $\theta$ to account for the intensity of the collection $I_C$: $$ I_C \propto A^2\int_0^{2\pi}d\theta\,\cos^2\theta = A^2/2 = I/2. $$ So the intensity of the polarized light is $I_C = I/2,$ half the intensity of the unpolarized light.

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