How do polarized sunglasses reduce intensity? When unpolarized sunlight enters our eyes through linearly polarized sunglasses the intensity is reduced. How can I understand it from the expression of the intensity? I would prefer some mathematics.
[Physics] How do polarized sunglasses reduce glare
electromagnetic-radiationintensityopticspolarizationvisible-light
Related Solutions
The main idea behind polaroid sunglasses is that reflexion from water, snow and other glary reflectors is mainly polarized in one direction. To understand this, witness the behaviour foretold by the Fresnel Equations (the graph below taken from the Wikipedia "Fresnel Equations" page):
so that you can see for a wide range of scattering angles from these surfaces, the reflected light reaching your eyes is mainly in the $s$-polarized direction (electric field vector orthogonal ("senkrecht" in German) with the plane of polarization), so if you quell this polarization, you get rid of most of the glare from these surfaces.
Why are your lenses twenty degrees off in their polarization axes? I'd say that this is a simple question of production economics. The power through a polaroid varies like $(\sin \theta)^2$, where $\theta$ is the angle between the actual polarizing axes and their ideal directions for quelling a given linear polarization. This functional dependence is very flat for a wide angle range around the null, so, if there is a twenty degree error, the attenuation ratio is still 0.1. So a polarizer that is twenty degrees off is still almost as good as an ideally aligned one for the lower-the-glare-in-human-sight application. Therefore, a manufacturer simply will not go to the extra cost of the quality control needed to align the polarisers more accurately: it really wouldn't make the product any better for the application at hand.
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:
One begins with unpolarised light;
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;
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.
Related Question
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Best Answer
The boring bit is that one polarisation gets through and the other doesn't, so the intensity is cut by half.
But of course suitably opaque glasses could do the same much more cheaply.
The interesting bit comes from reflection of (unpolarised) sunlight off puddles, or swimming pools, or the sea. The reflected light is predominately horizontally polarised. It depends on the angle. At a particular angle, the Brewster angle, when the reflected ray and the refracted ray are at 90 degrees to each other, there is no vertical polarisation in the reflection. So Polaroid sunglasses which accept the vertical and reject the horizontal component cut out half of general light, but much more than half of light reflected off horizontal sheets of water. So you can see what's happening in the pool without being blanked by the glare reflected from the surface.