# Why does diffraction depend on wavelength? [closed]

We've all heard of the diffraction of radio waves over a mountain and the diffraction of water waves through a gap, but why does this effect depend on wavelength?

I'm looking for as simple answer as possible - if it's a bit hand-wavey that might be OK.

Another thing that troubles me is when I read the effect of diffraction depends on the size of the wavelength compared to the gap/ obstacle. But, how do you define the size of the obstacle if it's an edge, e.g. the tip of a mountain, the edge of a wall, the edge of a razor blade?

## closed as too broad by John Rennie, Kyle Kanos, Brandon Enright, Alexander, Qmechanic♦Apr 13 '14 at 18:54

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• That's an awfully broad question. You're basically asking us to explain diffraction to you. I would start by browsing the Hyperphysics pages on diffraction and come back to us with specific questions. – John Rennie Apr 12 '14 at 18:29
• @JohnRennie I'm not interested in the complex interference patterns that occur, just the basic principle as to why it should depend on wavelength. I can't think of a simple, intuitive reason. – User 17670 Apr 12 '14 at 18:37
• @JohnRennie Could you answer my second question - about defining size? – User 17670 Apr 12 '14 at 18:38
• @JohnRennie Thanks for providing the link. I've now read all of the pages therein, and I can't find the answer to my question. Fancy giving it a bash? – User 17670 Apr 12 '14 at 18:52