I already know that the pattern created by non coherent sources will constantly change ( from the other answers). I want to understand why this is so

My explanation (which I want to confirm): suppose 2 waves with different frequencies meet perfectly in phase at a point p at time t1. The amplitudes add. Now suppose that in the time one of the waves completes one cycle , the other covers half a cycle(different frequency) . This now creates complete destructive interference at the same point. This can happen for different combinations. Thus the pattern is not observable. Am I missing something?


If you have different frequencies then you never quite have a perfect point p on the atomic scale because at p plus a nanometer they are getting out of phase again and the electron in the atom (many nanometers big) has a probability to interact (observation). If you have the same frequencies but they are 180 out of phase then an electron in an atom will have a hard time to interact, however they may get scattered and slightly separated and then the probability of interaction increases again. All these probabilities are governed by quantum mechanics.

  • $\begingroup$ What if we are not interested on these slight shifts? If we are only interested in the macro level observation. $\endgroup$ – Sal_99 Jan 19 '19 at 20:45
  • $\begingroup$ On a macro scale the slight shifts due to scattering etc are what happens most often, hence you observe the photons most often. $\endgroup$ – PhysicsDave Jan 19 '19 at 21:54
  • $\begingroup$ Okay. How do we use this to explain why their is no single observable pattern with non coherent sources ( which my book claims) ? $\endgroup$ – Sal_99 Jan 20 '19 at 7:13
  • $\begingroup$ Check out the link from this site: physics.stackexchange.com/questions/364057/… Also interference is a tricky word, single photon experiments show photons want to travel a certain path of multiples of their wavelength. Superposition is a temporary thing the energy is not cancelled. Dark spots are where there is no energy anyway because of the paths chosen. $\endgroup$ – PhysicsDave Jan 22 '19 at 5:06
  • $\begingroup$ To avoid complication, can you answer the same question but with reference to comparatively simple water waves? $\endgroup$ – Sal_99 Feb 12 '19 at 8:36

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