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Integral simplification using Bessel functions

  1. May 31, 2017 #1
    1. The problem statement, all variables and given/known data

    I need to simplify the following integral

    $$f(r, \theta, z) =\frac{1}{j\lambda z} e^{jkr^2/2z} \int^{d/2}_0 \int^{2\pi}_0 \exp \left( -\frac{j2\pi r_0 r}{z\lambda} \cos \theta_0 \right) r_0 \ d\theta_0 dr_0 \tag{1}$$

    Using the following integrals:

    $$\int^{2\pi}_0 \cos (z \cos \theta) d \theta = 2 \pi J_0 (z) \tag{i}$$
    $$\int^{2\pi}_0 \sin (z \cos \theta) d \theta =0 \tag{ii}$$
    $$\int^{z}_0 z J_0 (u) du=zJ_1 (z) \tag{iii}$$

    P. S. This integral is meant to give the radial intensity profile of the diffraction pattern of the circular aperture (##|f(r, \theta, z)|^2 ##). I am required to express this as a simple function of a ##J_1(.)^2## function. Here ##J_n (z)## denotes the Bessel function of order ##n##.

    2. Relevant equations


    3. The attempt at a solution

    Leaving aside the constant ##A=\frac{1}{j\lambda z} e^{jkr^2/2z}## in the front, the integral becomes:

    $$\int^{d/2}_0 r_0 \int^{2\pi}_0 \exp \left( -\frac{j2\pi r_0 r}{z\lambda} \cos \theta_0 \right) \ d\theta_0 dr_0 $$

    $$=\int^{d/2}_0 r_0 \int^{2\pi}_0 \cos \left( -\frac{j2\pi r_0 r}{z\lambda} \cos \theta_0 \right) + j \sin \left( -\frac{j2\pi r_0 r}{z\lambda} \cos \theta_0 \right) \ d\theta_0 dr_0 $$

    Using (i) and (ii):

    $$=\int^{d/2}_0 r_0 2 \pi J_0 \left( \frac{j2\pi r_0 r}{z\lambda} \right) dr_0 $$

    Here I had to use a substitution ##u=j2\pi r r_0/z \lambda##, and ##dr_0 = \frac{z \lambda}{j2\pi r} du##:

    $$\left( \frac{z\lambda}{j 2 \pi r} \right)^2 2 \pi \int^{\frac{z \lambda d}{4 j \pi r}}_0 u J_0 (u) du$$

    Using (iii) this becomes:

    $$\therefore f(r, \theta, z) = A. \frac{d}{r} \left( \frac{z \lambda}{2 \pi r} \right)^2 J_1 \left( \frac{z \lambda d}{4 j \pi r} \right)$$

    However, in textbooks, this intensity distribution (known as 'Airy pattern') is given by:

    $$|f(r, \theta, z)|^2 = \left( \frac{\pi d^2}{4 \lambda^2} \right)^2 \left[ \frac{2 J_1 \left( \frac{d \pi r}{z \lambda} \right)}{\frac{\pi d r}{\lambda z}} \right].$$

    So, why is my answer so different? What is the mistake? :confused:

    Any suggestions would be greatly appreciated.
     
  2. jcsd
  3. May 31, 2017 #2

    Charles Link

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    I think the error is in the limit on your integral. When ## r_o=d/2 ##, ## u=... ##. ## \\ ## Also another error before that: ## e^{jx}=cos(x)+jsin(x) ##. There is no ## j ## in the cos(x) or ## sin(x) ## and no j in the ## J_o(x). ## It is not ## J_o(jx) ##, etc.
     
  4. May 31, 2017 #3
    Thank you so much for pointing out the mistake in ##dr_0##. The j in cos and sin was a typo.

    So, using ##u=\frac{2\pi r}{z \lambda}r_0## and ##dr_0 = \frac{z \lambda}{2\pi r} du## we find:

    $$f(r, \theta, z) = A \int^{d/2}_0 \left( \frac{z \lambda}{2 \pi r} \right)^2 2 \pi u J_0 (u) du$$

    $$= A \left( \frac{z \lambda}{2 \pi r} \right)^2 \pi d \ J_1 \left( \frac{d}{2} \right)$$

    But according to textbooks, squaring this should give the intensity as:

    $$I(r) = \left( \frac{\pi d^2}{4 \lambda^2} \right)^2 \left[ \frac{2 J_1 \left( \frac{d \pi r}{z \lambda} \right)}{\frac{d \pi r}{z \lambda}} \right]^2$$

    I don't think I've made any other mistakes. So why is my answer so different? Have they used some properties of the Bessel functions to further manipulate this?
     
    Last edited: May 31, 2017
  5. May 31, 2017 #4

    Charles Link

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    Setting ## r_o=d/2 ## gives you for the upper limit on the integral as ## u=\pi r d/(z \lambda ) ## when ## r_o=d/2 ##. ## \\ ## Editing...okay, I see what you did=let me look it over a little more...
     
  6. May 31, 2017 #5

    Charles Link

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    @roam What you have written from the textbooks is not completely correct. I think you are a whole lot closer. Upon doing some algebra and taking ## |f|^2 ##, I get for your result ## (\frac{\pi d}{\lambda z})^2 J_1^2 (\frac{\pi r d}{\lambda z}) ##. ## \\ ## A google of the topic, (along with a change from ## \lambda^2 ## to ## \lambda z ## in the first term of your "textbook" solution) gives the result ## ( \frac{\pi d^2}{4 \lambda z})^2 [ \frac{2 J_1(\frac{\pi r d}{\lambda z})}{(\frac{\pi r d}{\lambda z})}]^2 ##. If this is the case, the difference in the two expressions is something like a factor of ## 4r^2 ##. ## \\ ## (Editing note: I see you correctly squared the ## J_1 ## term in the textbook solution you gave in post #3 for the Airy disc. In the OP, the exponent of 2 is absent. ) ## \\ ## Additional note is the on-axis intensity in one google article was simply given as ## I_o ##, and what I used is more in agreement with a little logic along with a result from a recent Insights article which I authored: The Diffraction Limited Spot Size with Perfect Focusing In this case, you have a screen at distance ## z ## rather than a lens that will give you the far field at distance ## f ## as in my article, but notice ## \frac{\pi d^2}{4 } ## is the area of the aperture, and the on-axis intensity is proportional to the square of this area. (See my article). Also, the on-axis intensity is inversely proportional to the square of the wavelength (and not the fourth power) (see my article again), and the on-axis intensity will fall off as ## \frac{1}{z^2} ##. Thereby, it makes sense that the term is ## \lambda z ## rather than ## \lambda^2 ## inside the parentheses in the denominator of the first term. ## \\ ## Editing: Upon repeating the algebra, etc., I get complete agreement between your result and the textbook result that I gave. (I resolved the ## 4 r^2 ## factor.)
     
    Last edited: Jun 1, 2017
  7. Jun 1, 2017 #6

    Charles Link

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    @roam I spotted a mistake you made near the last line of post #3: ## r_o=d/2 ##, but the ## zJ_1(z) ## needs to be ## (\frac{\pi r d}{\lambda z}) J_1(\frac{\pi r d}{\lambda z}) ## and not ## (d/2)J_1(\frac{d}{2}) ##.
     
  8. Jun 1, 2017 #7
    Thank you for your correction!

    So we have:

    $$I(r) = \left[ 2 \pi \left( \frac{z\lambda}{2 \pi r} \right)^2 \frac{\pi r d}{\lambda z} J_1 \left( \frac{\pi r d}{\lambda z} \right) \right]^2$$

    which simplifies to:

    $$I(r) = \left( \frac{z \lambda d}{2 r} \right)^2 \left[ J_1 \left( \frac{\pi r d}{\lambda z} \right) \right]^2$$

    But here I am not getting the inverse square law you've mentioned.

    How did you manipulate the last expression to get ##( \frac{\pi d^2}{4 \lambda z})^2 [ \frac{2 J_1(\frac{\pi r d}{\lambda z})}{(\frac{\pi r d}{\lambda z})}]^2##? :confused:
     
  9. Jun 1, 2017 #8

    Charles Link

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    You have a term ## |A|=\frac{1}{\lambda z} ## that you haven't included. You need to include that term. Multiplying and dividing by ## \frac{\pi rd}{\lambda z} ## will put an exponent of 2 on the term ## (\frac{\pi r d }{\lambda z}) ## in the middle, and putting a 2 on the ## J_1 ## term will cause a division by 2 on the first term. I think that should complete it.
     
    Last edited: Jun 1, 2017
  10. Jun 2, 2017 #9

    Charles Link

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    @roam I just wanted to do a follow-up on this calculation and comment that most of the features of the Airy disc (such as the width of the pattern and the on-axis intensity) are more readily computed by using a square (or rectangular (dimensions axb)) aperture to create the diffraction pattern, and looking simply at the far-field pattern. Assuming ## I(\theta, \phi)=I_o i_a(\theta) i_b(\phi) ## where ## \theta ## is azimuthal angle and ## \phi ## is elevation angle, and ## i_a(\theta) ## defines the normalized diffraction pattern in a single dimension for slit width ## a ## . (Angles are assumed small enough that the coordinate system remains well-defined. This is the method I used in my Insights article.). ## \\ ## The on-axis intensity/irradiance (watts/m^2) can be found squaring the result for the electric field amplitude ## E(0,0,z) ## from the complete diffraction integral by setting the phase term equal to zero in the integral. The result is ## I_o=E_i^2 \frac{A^2}{\lambda^2 z^2} ##, where there may be an additional constant (depending on the units) that relates power per unit area ## I ## to electric field squared ## E^2 ##. (## E_i ## is the incident electric field amplitude). This result in the far-field holds regardless of whether the aperture is circular or rectangular .## \\ ## In order to have energy conservation, the effective area of the diffraction pattern in the far-field ## A_{effective} ##, must be ## A_{effective}=\frac{\lambda^2 z^2}{A} ##. (Note that total power ## P=I_o A_{effective}=E_i^2 A ## ). This is consistent with the parameter ## \frac{\pi rd}{\lambda z } ## inside of ## J_1 ##. The effective width of this function will be (aside from a constant) such that ## \Delta r=\frac{\lambda z}{d} ##, so that the effective area of the pattern will indeed be on the order of ## A_{effective}=\frac{\lambda^2 z^2}{A} ##. (This result follows quickly from considering the width of the pattern for a rectangular aperture.) ## \\ ## Your post was actually the first time I have worked through the Airy disc calculation in its entirety, and I was glad that you had already posted the major part of the calculation. :) :) ## \\ ## Editing: There is actually one additional result that should follow from the results above: ## \int\limits_{0}^{+ \infty} \frac{J_1^2(x)}{x^2} x \, dx = ##Constant, that will correctly normalize the result that we obtained. And here is a "link": Equation (76) says the Constant is 1/2. http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html ## \\ ## This result is consistent with ##2 \pi \int\limits_{0}^{+\infty} (\frac{ 2 J_1(\frac{\pi rd}{\lambda z})}{\frac{\pi rd }{\lambda z}})^2 r \, dr=A_{effective}=\frac{\lambda^2 z^2}{A} ## where ## A=\frac{\pi d^2}{4} ## where the ## 2 \pi ## in front comes from a ## d \phi ## integration in polar coordinates.
     
    Last edited: Jun 2, 2017
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