Light intensity and wave behavior of light

[ViolentCorpse]

Hello PF'ers,

Why is it that two light-bulbs produce twice as much brightness as a single light-bulb would? If we restrict our attention to a single point in space where light from the two sources meet, the waves would cancel half of the time and reinforce half of the time, producing on average no greater brightness than a single bulb working alone, unless the two sources produce all waves exactly in phase all the time, which is highly improbable. The brightness would be constantly fluctuating, but I guess our brains are too slow to observe that (not claiming that this is indeed what actually happens).

I know that coherent light is a prerequisite for interference effects and everyday sources of light aren't coherent, but still light behaves like a wave in most ordinary circumstances so there should be some discernible interference on average at least?

 

[jaydnul]

Photons don't interact with each other. There is a huge, messy network of photons in between you and your computer screen going every which way but you only notice the photons that hit your pupil. A photon's wave function can destructively interfere with itself, but not between two photons.

 

[Orodruin]

The (very) simplified argument: Intensity is proportional to the square of the amplitude. In your case you would have zero amplitude in the destructive spots and twice the amplitude in the constructive. The average intensity would be (0+2^2)/2 = 2 times that of a single source.

Of course, visible light wavelength would be too small to observe this interference even if you had two bulbs sending out coherent light.

 

[Drakkith]

I believe it boils down to the fact that the light emitted from a lightbulb is not coherent, so the interference is completely random, leading to no observable pattern of fringes. In other words, some waves would interfere constructively, some destructively, and most somewhere in between.

 

[ViolentCorpse]

The (very) simplified argument: Intensity is proportional to the square of the amplitude. In your case you would have zero amplitude in the destructive spots and twice the amplitude in the constructive. The average intensity would be (0+2^2)/2 = 2 times that of a single source.

Of course, visible light wavelength would be too small to observe this interference even if you had two bulbs sending out coherent light.

That's a neat, intuitive explanation. Though, I wonder if things wouldn't be the same if there's a single source with no destructive spots at all i.e something like (1+1)^2/2=2?

Thank you so much, Orodruin!