Why do we simply add the intensities for interference from two incoherent sources? What's the proof for that? I know that incoherent means that their phase difference changes with time
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$\begingroup$ You cannot get a stable interference pattern from two incoherent sources. $\endgroup$– R.W. BirdCommented May 6, 2020 at 14:56
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$\begingroup$ Yes so we take average right ? As stated by@Farcher $\endgroup$– SM SheikhCommented May 6, 2020 at 15:05
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$\begingroup$ Yes, as shown by Farcher, when averaged over time, the intensities add. $\endgroup$– R.W. BirdCommented May 6, 2020 at 16:09
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Imagine that you had two sources emitting waves of exactly the same amplitude $A$, same frequency with a constant phase difference.
Where there is superposition of these wave there will be an interference pattern.
At some positions the waves from the two sources will arrive in phase with one another and the resulting amplitude will be $A+A = 2A$ with an intensity proportional to $(2A)^2 = 4A^2$.
At other positions the waves will arrive exactly out of phase with one another and the resultant amplitude will be $A-A =0$.
So the intensity in the region of interference will range from zero to being proportional to $4A^2$.
The average intensity will be proportional to $2A^2$ with is the sum of the intensities due to the individual sources.
So interference does not destroy or create energy rather it channels the flow of different amounts of energy in different directions.
Now with the two same source let the phase of one source relative to the other change with time.
What will be observed?
It will be a moving interference pattern and at any one position the intensity will vary for zero to being proportional to $4A^2$ but on average at that position the intensity will be proportional to $2A^2$, the sum of the intensities of the two individual sources.
You can perhaps now extend this arguement to incoherent sources where there would be superposition and at any instant of time an interference pattern but in the next instant of time the interference pattern would have moved an so at any position the intensity would average out as the sum of the ontensities from the two sources.
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$\begingroup$ 'The average intensity will be proportional to 2A2 with is the sum of the intensities due to the individual sources.' Can you explain, mathematically, how you got this? $\endgroup$– harryCommented Mar 27, 2021 at 1:17
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$\begingroup$ The intensity form one source is $A^2$ so from two incoherent sources it is $2A^2$. For two coherent sources of equal amplitude the intensity ranges from $(A-A)^2=0$ to $(A+A)^2= 4A^2$. $\endgroup$– FarcherCommented Mar 27, 2021 at 7:29
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$\begingroup$ For two coherent sources, yes, that's the range, but how does the net intensity for incoherent sources work out that way? Isn't there a varying phase difference that you have to consider? $\endgroup$– harryCommented Mar 27, 2021 at 7:32
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$\begingroup$ For incoherent sources you are looking at averages over a period of time. With incoherent sources if you took an "instantaneous" picture there would be a variation of the amplitude at different points in space but those would not be the same positions at the next instant of time. In other words the interference pattern produced by incoherent sources is not fixed in space. To illustrate this consider two sound source which differ in frequency by a small amount. As time goes on the interference pattern moves. If you stand in one spot you will hear a succession of maxima and minima called beats. $\endgroup$– FarcherCommented Mar 27, 2021 at 8:19
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$\begingroup$ I got that, but how do you calculate the average of a random variation like that by just... Summing the intensities? $\endgroup$– harryCommented Mar 27, 2021 at 8:36