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Polarizing Sunglasses

How much light to polarizing sunglasses block? To answer this, a couple of things need to be clarified. First, what is meant by ‘how much light’, and second, how polarization works.

To answer the first question, it seems best to measure the power that the sunglasses absorb, i.e. the energy-per-time. Light travels in waves of electromagnetic fields. The electric field has an amplitude, , and the energy delivered by the wave per unit time is proportional to .

Next, let’s tackle polarization. Mathematically, the electric field is represented by a vector. When the wave travels through a polarizer, the polarizer leaves intact only one component of the vector — the component that’s in the same direction as the polarizer. So if the electric field of amplitude E has an angle relative to the polarizer, the wave’s amplitude once it travels past the polarizer is reduced to . The vector also appears to rotate, but that doesn’t matter for our purposes.

What is the power carried by the wave once it gets past the polarizer? That’s simply the square of its new amplitude: . The polarizing sunglasses are of course a polarizer, so they have this effect on an incoming wave.

There’s just one obstacle left to overcome before making the calculation: All the discussion above waves assumes that the waves are already polarized in some direction before reaching the polarizer. But light hitting your glasses is generally unpolarized, so we have to account for that. Unpolarized light can be thought of as a sum of polarized waves in different angles, each with the same amplitude. To get the power absorbed by the glasses, we need to average over the possible angles:

Question: Why is the integration carried out between 0 and pi rather than 0 and 2*pi?

We conclude that using polarization, sunglasses reduce light power by one half.. Note that there are probably other mechanisms by which such glasses absorb light — this calculation only addresses polarization.