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$$cos^2(x)$ where $I_o$ is the intensity of light after P1. I cannot use this formula directly in this case because of the definition. But, if i understand correctly(which is not likely) P1 simply makes the unpolarised light plane polarised. So suppose now we consider a beam of unpolarised light. This can be thought of as a mixture of many plane polarised lights.consider any one of them and apply malus law. Now $I_o$ is the original intensity and x is the angle of the pass axis w.r.t the chosen plane polarised light. Since this can be chosen absolutely randomly, angle x can vary randomly from 0 to 2pi. We can theorefore find the EXPECTED transmitted intensity to be the average of $cos^2(x)$ as x varies from 0 to 2pi, which returns 1/2.
Is this a correct proof of the question? I am not satisfied myself. Any help with regard to my proof woulf be appreciated. Also, i would like a physical explanation if possible.
Best Answer
Consider the unpolarized light to be a collection $C$ of a large number of plane-polarized waves with polarization distributed evenly over a circle, and each of amplitude $A$ and intensity $I \propto A^2$. Let the transmission axis of the plane-polarizer lie along the $x$-direction. The linear polarizer acts on one of the incident waves $\psi = \vec Ae^{i(kr-\omega t)}$, where $\vec A = \hat x\,A\cos\theta + \hat y\,A\sin\theta$ is the polarization vector of the light, via the transformation $$ \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.