# Interference using three slits

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?

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.