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For the single slit there is a well known elementary derivation along the following line:

  1. If the path difference $\delta$ between the elementary waves coming from the borders of the slit is one wave length $\lambda$ you imagine say 100 elementary waves evenly distributed starting from the slit. Then the path difference between the 1. and 51. wave is $\lambda/2$ so they interfere destructively, the same path difference occurs between the 2. and 52. wave, the 3. and 53, and so on. So you get a interference minimum for $\delta = \lambda$.
  2. If $\delta = k\lambda$ ($k \in \{1,2,3,\dots\}$) you just devide the ray bundle into $k$ bundles, where in each bundle the path difference between the "boundary rays" is $\lambda$ and you can argue as in 1. So for each bundle you get complete destructive interference.
  3. From geometry it follows then that you get minima for $$ \sin(\alpha) = \frac{k\lambda}{b} $$ where $b$ is the width of the slit.

From Babinets principle it follows that the same formula should apply for the interference pattern of a opaque thin wire of thickness $b$.

Is there any direct elementary derivation (like that one for the single slit as sketched out above) for the case of the thin wire?

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  • $\begingroup$ You can consider it like young double slit experiment, with the two diagonaly opposite side, b, as width of slit and do calculations.. $\endgroup$ – Anubhav Goel Jan 7 at 20:15
  • $\begingroup$ But in an idealized double slit (on that level) one would assume that each slit is pointlike. How to deal with the inifinite slit widths then? $\endgroup$ – Julia Jan 8 at 12:05
  • $\begingroup$ You dont need to consider that,, all of that aint causing any diffraction, only borders cause diffraction. $\endgroup$ – Anubhav Goel Jan 8 at 12:45
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For interference( just like thin wire) Minima is at $y =m\lambda D/d$

For diffraction , Maxima is at $y= m\lambda D/d$

Add the two and you, get maxima + minima = all light is passed as if no object is there.

Apply this at each fringe..

As for intensity, If you look at centre, since in diffraction intensity I was divided in each fringe, centre had less light, calculate that, and in double slit intereference, light was added at centre ,which was otherwise dark. Both these contribute to make a centre of normal intensity as if there was no slit or wire.

Intensity at distance from object wont matter as they wont show any diffraction..

Note: I cant find formula for intensities, so, I cant help with math part..😁..

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