# Multiple slit diffraction

1. Dec 12, 2012

### carlosbgois

Hi there. It's not actually a problem, I was just trying to figure it out, so idk if this is the right section for this post, but anyways: I've managed to derive the expression for the Fraunhofer diffraction in a single slit, such that the distance between a minimum and the central point is given (in an approximation) by $y=\frac{mλD}{a}, m=1, 2, 3, ...$, in which D is the distance from the slit to the screen, and a is the slit gap.

What would happen in the same arrangement, but with a double slit? And with multiple slits? And what if, instead of a slit, I had an obstacle, such that the light would (classicaly) go through the sides, but not over the object?

Many thanks

2. Dec 12, 2012