Light waves and intereference patterns

AI Thread Summary
When shining two flashlights on a wall, no interference pattern is observed because the light waves from the two sources are not coherent and do not interact. For interference to occur, the light waves must be in sync, allowing for constructive or destructive interference. The mathematical representation of this phenomenon involves calculating the probability distribution of the combined wavefunctions, which reveals that without coherence, the interference terms vanish. Using a single light source ensures that the phases are identical, leading to observable interference patterns. Thus, the lack of interference with two flashlights stems from their incoherent nature.
kirsten_2009
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Homework Statement


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You shine two flashlights on a wall. Why don’t you see an interference pattern?

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The Attempt at a Solution



Actually, I thought you would have seen an interference pattern but if you don’t then it’s probably because the waves of light of the two flashlights are not interacting with each other. In other words, they are not adding up as constructive interference or cancelling each other out by destructive interference so they must not be in sync and so the waves of both flashlights make it to the wall with no “interruptions”. If the waves of the two flashlights were exactly in sync then I presume you would see an interference pattern on the wall.
 
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To understand this just find the probability distribution of a linear combination of wavefunctions passing through each slit:
\Psi=R_{1}e^{i\theta_{1}}+R_{2}e^{i\theta_{2}}
and find |\Psi|^{2}

You should end up with R1^2 + R2^2 plus a cross term which exhibits the interference. Then to see why it matters if you have the same laser source or not see how coherent phases affect the cross term. For a single light source plug in \theta_{1}=\theta_{2} since they will have the same phase. If they have different phase, which would happen if you use two different sources, then their difference should vary randomly and so you should take the average value of \theta_{1}-\theta_{2} (which will give you 0 so the interference vanishes)
 
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