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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$\begingroup$ Related? Fringe width and spacing and number of slits in diffraction experiments $\endgroup$– FarcherCommented Aug 22, 2019 at 9:04
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2$\begingroup$ Possible duplicate of Fringe width and spacing and number of slits in diffraction experiments $\endgroup$– Emilio PisantyCommented Aug 22, 2019 at 11:03
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$\begingroup$ But I am not able to find solution to a numerical problem even after using the formula for resultant amplitude due to three slits. $\endgroup$– user237666Commented Aug 22, 2019 at 11:04
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1 Answer
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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.
