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Lets say there is a Young double slit interference apparatus, but with three slits placed at $y= - d$, $0$, $d$, and where the screen is at $X = D$ parallel to the $y$ axis. Can there be any areas on the screen where the intensity is at a minimum? If yes, then which points will be those, and how can I mathematically find those points?

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If $D$ is sufficiently large you can suppose the first and second have the phase differencw $\phi$, and the first and the second have phase difference $2\phi$. Then you have $$\sin(\omega t)+\sin(\omega t+\phi)+\sin(\omega t+2\phi)=\sin(\omega t)+\sin(\omega t) \cos\phi)+\cos(\omega t) \sin(\phi)+\sin(\omega t)+\cos(2\phi)+\cos(\omega t)*\sin(2\phi)$$ so $$1+\cos(\phi)+\cos(2\phi)=0$$ and $$\sin(\phi)+\sin(2\phi)=0$$ Let Wolfram do the calculating.

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