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 $E = n h \nu$, where $n$ is the number of photons in the optical pulse. (If you don't like to think in terms of pulses, then $P = n h \nu$ 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.
No comments:
Post a Comment