carlosbgois
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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 [itex]y=\frac{mλD}{a}, m=1, 2, 3, ...[/itex], 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
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