How Can One Derive the Diffraction Pattern Formula from a 1D Aperture?

Click For Summary
SUMMARY

The discussion focuses on deriving the diffraction pattern formula from a one-dimensional aperture using the Huygens-Fresnel principle. The aperture is defined by an opacity function, f(u), which can represent various configurations such as single or multiple slits. The amplitude of the wave passing through the aperture is expressed as A f(u) cos(2π/λ√((x-u)²+L²)) du, leading to a diffraction integral that requires squaring for intensity calculation. The participants emphasize the need to normalize the integral by the total length of the aperture and suggest using numerical integration tools like Mathematica or MATLAB for specific cases.

PREREQUISITES
  • Understanding of the Huygens-Fresnel principle
  • Familiarity with diffraction patterns and wave optics
  • Basic knowledge of integral calculus
  • Experience with numerical integration software such as Mathematica or MATLAB
NEXT STEPS
  • Explore the application of the Huygens-Fresnel principle in wave optics
  • Learn about numerical integration techniques in Mathematica
  • Study the Fourier transform and its relation to diffraction patterns
  • Investigate the effects of varying the opacity function f(u) on diffraction outcomes
USEFUL FOR

Physics students, optical engineers, and researchers in wave optics seeking to understand diffraction patterns and their mathematical derivations.

struggling_student
Messages
9
Reaction score
1
I've been trying to derive a formula for diffraction pattern formed by casting a plane-wave through a generic 1D aperture onto a screen distanced ##L## from the aperture. The aperture is described by an opacity function ##f:\mathbb{R} \rightarrow [0,1]## so it can be a single slit, multiple slits, shaded glass with varying opacity. By the Huygens-Fresnel principle every point on that aperture is a spherical wave and we weigh them by infinitesimal ##du## so that it can be integrated.

Let ##u## be the position on the aperture relative to some chosen axis which also goes through the screen. Let the position on the screen relative to that axis be ##x##. The opacity function is a function of ##u##, i.e. ##f=f(u)##.

The wave that goes through point ##u## on reaching the screen has amplitude

$$A f(u) \cos\left(\frac{2\pi}{\lambda}\sqrt{(x-u)^2+L^2 }\right) du,$$

and the resulting diffraction will be

$$A \int_{\mathbb{R}} f(u) \cos\left(\frac{2\pi}{\lambda}\sqrt{(x-u)^2+L^2 }\right) du.$$

It's a function of ##x## and we would square it to get intensity. I'm not sure how to proceed or what I did wrong. This approach is the only approach I am interested in. I'm trying to obtaining something similar to Fourier transform. What's missing here?

IMG_20210807_0001.jpg
 
  • Love
Likes   Reactions: Delta2
Science news on Phys.org
I would like delete this post I am convinced nobody cares. How does one delete?
 
I care but I am not so good in this type of problems that's why I hesitate to type my thoughts. However I liked the generalization of this problem (function ##f(u)##) and the way you use Huygens-Fresnel principle to integrate over all possible sources. The only problem I see with that integral expression is that :
  • you have to divide by the total length of the aperture, pretty much the same way you divide by ##b-a## when you calculate the average value of a function ##f(x)## over the interval ##[a,b]##, $$\mu=\frac{1}{b-a}\int_a^b f(x)dx$$
  • your integral though it doesn't seem very complicated, yet it doesn't seem to have a closed analytical antiderivative, at least wolfram alpha can't find it (I tried it with ##f(u)=1## the constant function).
Have you try to do numerical integration using some math software (mathematica, MATLAB e.t.c) for the case f(u) is the function that corresponds to a single slit to see what diffraction pattern you get?

What do you mean when you say you want to get something similar to Fourier transform? Fourier transform of ##f(u)##?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
50
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K