Elementary derivation for the minima of the interference pattern in the case of a thin wire without referring to the Babinet principle

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?

• You can consider it like young double slit experiment, with the two diagonaly opposite side, b, as width of slit and do calculations.. – Anubhav Goel Jan 7 at 20:15
• 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? – Julia Jan 8 at 12:05
• You dont need to consider that,, all of that aint causing any diffraction, only borders cause diffraction. – Anubhav Goel Jan 8 at 12:45

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..😁..