A polarized filter is exposed to a unpolarized light source. The output of the filter should be of lower intensity, hence lower energy.
Should not the filtered light be of a lower frequency to account for the lower energy ( E = h v) ?
Answer
Light is quantized so the energy E=nhν, where n is the number of photons in the optical pulse. (If you don't like to think in terms of pulses, then P=nhν denotes the optical power of the beam, with n here being the number of photons per unit time). The polarizing filter absorbs or reflects some of the photons in the light beam, which is why the energy/power measured after the polarizer is reduced. But the frequency remains intact.
You may now like to think of what would happen if a single photon (i.e. n=1) was subjected to this experiment. The answer is that either a photon with energy equal to that of the photon emitted by the source will be found, or nothing will be measured.
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