2D Fraunhoffer diffraction

  • Thread starter Tibo123
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Hello,

My name is Thibaut, I am a student in physics and I currently struggling with something that some of you might find obvious.

I did an experiment with a laser 650nm passing through a 2D squared grid. The light interferes and diffracts and a photomultiplier collect only the central spot of the diffraction pattern 1m away from the grid. We progressively deposit on the bars of the grids a material that blokes the light and makes the holes of the grids look gradually smaller. We see a decrease in the light intensity collected at the photomultiplier. The intensity of the light decreases with x^4, where x is the size of the hole. I would like to be able to show numerically that trends.

I have tried a few things already but without any success. I am currently making the 2D FT of the pattern of the grid but while the holes shrink, I only observe a x^2 trend. I wonder if it is possible to include in my calculation the beam but I don't know enough about FT to tell.

I am opened to your suggesting,

Best Regards,

Thibaut
 

Answers and Replies

  • #2
blue_leaf77
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Suppose ##(x,y)## are the coordinates in the object plane, ##(u,v)## coordinates in the photomultiplier plane, and ##E_0## the amplitude of the incident plane wave. The field in the multiplier plane is then given by
$$
\tilde{E}(u,v) \propto \int_{-x/2}^{x/2}\int_{-x/2}^{x/2} E_0 e^{ikux/L}e^{ikvy/L} dx dy
$$
Since you are only interested in the central maximum, we can set ##u=v=0##. Thus
$$
\tilde{E}(0,0) \propto \int_{-x/2}^{x/2}\int_{-x/2}^{x/2} E_0 dx dy = E_0 x^2
$$
The photomultiplier measures the intensity, therefore the intensity of the central maximum will be ##I(0,0) \propto |\tilde{E}(0,0)|^2##.
 
  • #3
Andy Resnick
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Hello,

My name is Thibaut, I am a student in physics and I currently struggling with something that some of you might find obvious.<snip>

Thibaut

Another approach to blue_leaf77:

The grid transmits the incident field proportionally to the area of the holes (x2), and since the intensity is proportional to the field, the received intensity is proportional to x4.
 
  • #4
blue_leaf77
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the intensity is proportional to the field
In an effort to avoid confusion for the OP, I think you forgot to say "(modulus) square of the field".
 
  • #5
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Thank you guys. I will try again what you suggested blue but I fail to do so in the past.

I don't understand the argument of Andy. What do you mean when you say "since the intensity is proportional to the field, the received intensity is proportional to x4"?

Regards,

Thibaut
 
  • #6
Andy Resnick
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In an effort to avoid confusion for the OP, I think you forgot to say "(modulus) square of the field".

Oops- my error. Point is, the result can be obtained radiometrically as well.
 
  • #7
Andy Resnick
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Thank you guys. I will try again what you suggested blue but I fail to do so in the past.

I don't understand the argument of Andy. What do you mean when you say "since the intensity is proportional to the field, the received intensity is proportional to x4"?

Regards,

Thibaut

Right- I meant to say the intensity is proportional to the square of the field. Sorry for the confusion.
 
  • #8
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I see what you mean now. It does make my point indeed. Thanks guys.

I am looking for a way to express the intensity of the light as a function of x and y for a light passing through a squared grid. I find one equation for "diffraction by many slits". But I can't find any in the literature for a network a squared frames. If you know that this equation exists somewhere in a book, could you please give me the references? I would like to simulate it in python and see the intensity profile as I have seen it with my eyes on a white sheet of paper I use to put before the photomultiplier just to have a look.

Regards,

Thibaut
 
  • #9
blue_leaf77
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So, in your application, the aperture plane contains a group of small rectangular apertures of area ##xy## spaced uniformly from each other hence forming a regular array, am I wrong? But how far does this array extend? If it extends to infinity, the simple far-field diffraction relation will not hold. Anyway, if it's of finite extent, the entire grid may be mathematically modeled as a convolution between an array of delta functions
$$
d(x,y) = \sum_{mn} \delta(x-pm,y-qn)
$$
where ##p## and ##q## are distances between adjacent rectangle in ##x## and ##y## directions, and the rectangular function ##f(x,y) = \textrm{rect}(x/A,y/B)## where ##A## and ##B## are the size of individual rectangle in ##x## and ##y## directions, respectively. If far-field condition is satisfied, the diffraction field will be proportional to the Fourier transform of ##f \ast d##, which then leads to the familiar array theorem in the Fourier plane.
 
Last edited:
  • #10
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It's a grid so I should have called it an ordered array of squared apertures. I would say that my laser beam covers 6 to 8 apertures on one side, so 36 to 64 holes in total. So it's not infinite.
Thanks.
 
  • #11
blue_leaf77
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It seems like you will not be dealing with uniform illumination, and it's not yet clear about the phase of the impinging laser. If the grid is placed in the focus of a laser, the wavefront at the grid plane can be approximated as being plane, therefore you only need to take care of the amplitude distribution. If the amplitude distribution is denoted by ##A(x,y)##, the input field will just be the product between this distribution function and the convolution between ##f## and ##g## mentioned in my previous comment.
 
  • #12
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Thanks, I have the array theorem under my eyes from the book "Optics" of Hecht. I think I will try to simulate the result of this convolution with python and I will show you what I obtained.
 

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