A straight edge diffraction pattern is not the same as a slit or multiple slit diffraction pattern. The spacing between fringes is not equal but gradually gets smaller the farther out you go as in the image below. I have come up with a simple way to calculate the exact locations of these dark and bright fringes and was wanting to compare my calculations to others. I am hoping someone has an equation or solution to this problem (one that I can understand). I have Googled this many times but the only equations I am finding seem way to complex for what is needed. For example if we have a monochromatic light source of $500nm$ wavelength directed toward a single sharp edge and on to an observation screen one meter away,

  • what would the distances from the shadows edge to the 1st bright spot or any spot as far out as you want to calculate?

enter image description here

  • $\begingroup$ Use the Fresnel approximation. First Fourier transform, then advance the phases by the right amounts, then Fourier transform back. $\endgroup$ – Andrew Steane Aug 16 at 18:33
  • $\begingroup$ @AndrewSteane Thanks but that’s still confusing. I have found a much simpler way. See page 10 of my paper “Single Edge Certainty” at billalsept.com You will find Y=the square root of lamda x L x (m+3/4) $\endgroup$ – Bill Alsept Aug 16 at 19:52

The trick is that because the "slit" is infinitely wide, you shouldn't work in the far-field approximation (Fraunhofer diffraction integral which leads to Fourier optics), but with distances computed to the square order (Fresnel diffraction integral). The results are appropriately named Fresnel integrals (this time this is a name of a special analytical function that is well documented). So, to get the maxima and minima, you need to find zeroes of the derivatives of these functions (which is fortunately much easier than actually computing the function). What you get is, that the squares of maximum positions are equally spaced.

The principle behind the diffraction integrals is simply adding contributions with phase delays proportional to the distance from that point. Google will tell you everything you need to know about the Huygens principle and the derivation of the result for straight edge diffraction (example - warning, they use imaginary units).

Maybe a paper referring to measurements (follow the references to learn more).

And a formula for max/min:

$$x_{max/min}(m)=\sqrt{\frac{\lambda L (L-y)}{y}(m+\tfrac{3}{4})}$$ where even $m$ are maxima and odd $m$ are minima. $y$ is the distance of the light source to the edge and $L$ is the distance of light source to the screen. For infinite $y$ and $L$ (sunlight, for instance), only the $L-y=D$ distance from the edge to the screen matters, and you are left with $$x_{max/min}(m)=\sqrt{\lambda D (m+\tfrac{3}{4})}$$ Now take $\lambda=500\,\rm{nm}$ and $D=1\,\rm m$. You get the max/min pattern: 0.61mm, 0.9mm, 1.17mm, 1.37mm,...

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  • $\begingroup$ So how would you calculate the 1st bright fringe in the example I set up? I cant find anything on Google that sets up a single edge experiment and then gives you the derivations. There is nothing to compare to?? So again the light source is 500nm wavelength and the screen is 1 meter (or 1,000,000,000nm) beyond the single edge. How would you calculate the distance from shadow edge to center of 1st bright spot? $\endgroup$ – Bill Alsept Jan 28 '16 at 9:20
  • $\begingroup$ Actually I kinda found something to compare my answers with and that was images with graphs like the one above. In that case (L) the distance from edge to screen was the distance from the moons limb to Earth. $\endgroup$ – Bill Alsept Jan 28 '16 at 9:28
  • $\begingroup$ I still cannot find any examples of how to calculate the 1st bright spot for the set up I described above (500nm wavelength and 1 meter distance). Or any example of derivations already made from another set up. All I need to know is what measurement others are calculating for a single edge fringe pattern so I can compare. If anyone knows how to do it please help. Thanks $\endgroup$ – Bill Alsept Jan 29 '16 at 0:06
  • $\begingroup$ Maybe someone has suggestions for a better way to search for this answer. Thanks $\endgroup$ – Bill Alsept Jan 29 '16 at 6:08
  • $\begingroup$ Expanded my answer. $\endgroup$ – orion Jan 29 '16 at 15:50

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