Timeline for Why must slit be of order $\lambda$ for radial diffraction?
Current License: CC BY-SA 4.0
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| Sep 24 at 12:58 | comment | added | Solomon Slow | If you arrange a continuum of coherent, point sources along an infinite straight line, the sum of their wave functions gives parallel wave fronts moving perpendicular to the line. Now, cut off half of the line. Make it extend in one direction from an end point. The source at that end point has no neighbors on the one side to which its wave function can be added. Waves from that source radiate out in all directions, not just perpendicular to the line. That's what diffraction is. It's the edge of a slit cutting off half of a line of sources. physics.stackexchange.com/q/554972/74763 | |
| Sep 24 at 0:19 | answer | added | anon | timeline score: 0 | |
| Sep 23 at 20:42 | comment | added | Aidan | Also, if we take the Huygen-Fresnel principle, then every point on the wavefront acts like a source, so why would the edges of the slit acting as sources produce this radial pattern? | |
| Sep 23 at 20:37 | history | edited | Aidan | CC BY-SA 4.0 |
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| Sep 23 at 20:36 | comment | added | Aidan | Thanks for the comment Solomon, maybe I'll reword it. | |
| Sep 23 at 20:28 | comment | added | Solomon Slow | Your illustrations aren't quite right because diffraction isn't something that a slit does, it's something that the edges of a slit do. Each edge is a separate source of diffracted light. When the edges are very close to each other (when the slit is narrow) then (A) you get a macro-scale interference pattern from the diffracted light, and (B) very little un-diffracted light gets through the slit. When the edges are far apart, then the interference fringes are much smaller, and they may be overwhelmed by the much brighter, undiffracted light that passes through the wide opening. | |
| Sep 23 at 20:00 | history | asked | Aidan | CC BY-SA 4.0 |