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?

  • $\begingroup$ You can consider it like young double slit experiment, with the two diagonaly opposite side, b, as width of slit and do calculations.. $\endgroup$ Jan 7, 2019 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, 2019 at 12:05
  • $\begingroup$ You dont need to consider that,, all of that aint causing any diffraction, only borders cause diffraction. $\endgroup$ Jan 8, 2019 at 12:45

1 Answer 1


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..😁..


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.