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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

Qmechanic
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Bill Alsept
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  • Use the Fresnel approximation. First Fourier transform, then advance the phases by the right amounts, then Fourier transform back. – Andrew Steane Aug 16 '20 at 18:33
  • @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) – Bill Alsept Aug 16 '20 at 19:52
  • Bill, you might find the Barnett & Harris paper in J. Optical Soc. Amer. 52 (6) 637-643 (1962) a useful source for the data you wanted. Your 1st order fringe value of m + 0.75 = 0 + 0.75, for an infinitely large source to edge distance, does not seem to hold for laboratory-sized edge to source distances. – Tony Hitchins Aug 18 '23 at 20:01
  • @Tony Hitchins my paper was updated some time ago. You can find the new derivation starting on page 13 of "Single Edge Certainty" at billalsept.com. The actual shape of the edges will affect the quality and shape of the pattern. My calculations are based on a perfect situation. Have you read the the whole paper or the section on Occultation? Thanks for the link, this looks interesting. – Bill Alsept Aug 19 '23 at 08:54

2 Answers2

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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,...

Glorfindel
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orion
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  • 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? – Bill Alsept Jan 28 '16 at 09:20
  • 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. – Bill Alsept Jan 28 '16 at 09:28
  • 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 – Bill Alsept Jan 29 '16 at 00:06
  • Maybe someone has suggestions for a better way to search for this answer. Thanks – Bill Alsept Jan 29 '16 at 06:08
  • Expanded my answer. – orion Jan 29 '16 at 15:50
  • Thank you! that's what I'm looking for. I will see how it compares. – Bill Alsept Jan 29 '16 at 15:56
  • Your calculation was the same as mine which was 612372.4644 nm for the first max. Sorry to be a pest but could you please go through your last equation because it is different than mine. We are solving for y so how do you get y into your equation? On the last equation you have m on both sides? Could you explain m in that situation. Thanks – Bill Alsept Jan 29 '16 at 18:08
  • I just took a quick limit $L,y\to\infty$. When $D=L-y$ remains constant, but you send the light source infinitely far, the difference between the distances becomes negligible and $L/y\approx 1$. – orion Jan 29 '16 at 18:17
  • So in your last equation m equals what? – Bill Alsept Jan 29 '16 at 18:21
  • Depends on what you're computing. Put in $0$ to get the first maximum, $1$ for the first minimum, $2$ for the second maximum, and so on. – orion Jan 29 '16 at 18:30
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The discussion so far has stated that the x value of the first order fringe is 6.1E-04 when L-D is infinite, D =1 m, lambda is 5E-07 and 0.75 is added to m. The subsidiary question is whether there is any independent supporting data for that answer. Using the equation in Barnett and Harris for the Fresnel variable, v, yields a value of 1.22 corresponding to x = 6.1 E-04. The U. of Rochester thesis of Harris, JS, provides a comparable set-up: lambda = 5.461 E-07, D = 1.905 m and L-D is undefined. The experimental values for the first order fringe were x = 9.504 E-04 and v = 1.22. Using the x = square root(lambda.D.(m+0.75) equation yields x = 9.509E-04 from which v is calculated as 1.22. Thus (m+0.75) is valid for the first bright fringe when L-D is infinite. When L-D is not infinite, as in the Barnett and Harris experiment (J. Optical Soc. Amer. 52 (6) 637-643 (1962)), (m+0.75) does not equal 0.75 for the first bright fringe.

Tony
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