A light bulb puts out a stream of photons with various wave amplitudes. If we add more light bulbs in the room, how can we just assume that the light intensity increases to greater levels; that would be the wave amplitude increases? Wouldn't there be a wide range of constructive and destructive interference in the waves? So isn't it just as possible that all of the interference is destructive? Obviously this is incorrect, but how do we understand and calculate the constructive or destructive interference in the real world, where we have a number of sources?


2 Answers 2


Normal light sources emit photons with different frequencies/different wave lengths. Therefore, the inference would be restricted to the tiny fraction of the photons that happen to have the same frequency. Also the interference fringe pattern (i.e. the locations of bright and dark spots) depends on the frequency of the interfering photons and thus the patterns produced by photons of different frequencies are different with no net cumulative effect to be observed.

It is also important to realize that the interference does not change the overall intensity of the light. So even if we have two light sources with the same frequency (like in the double slit experiment) and we do observe an interference pattern, a reduction in intensity in one location is always accompanied by an increase an another. Otherwise, energy is not conserved.


The problem is saved the fact that power goes as the square of the field. If the 1st source has a field $ \vec E_1 $, then the average intensity is

$$ I_1 = \frac 1 2 c\epsilon_0||\vec E_1||^2$$

If you add another random source, $\vec E_2$, then:

$$ I = \frac 1 2 c\epsilon_0||\vec E_1 + \vec E_2 ||^2$$ $$ I = \frac 1 2 c\epsilon_0\big(||\vec E_1||^2 + ||\vec E_2 ||^2 + 2\vec E_1\cdot \vec E_2 \big)$$ $$ I = I_1 + I_2 + c\epsilon_0\vec E_1\cdot \vec E_2 $$

So there is an interference term, but when you time-average it over all field directions, it is zero.


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