Homework Help: Integral simplification using Bessel functions

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1. May 31, 2017

roam

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?

Any suggestions would be greatly appreciated.

2. May 31, 2017

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.

3. May 31, 2017

roam

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
4. May 31, 2017

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...

5. May 31, 2017

@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
6. Jun 1, 2017

@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 $(d/2)J_1(\frac{d}{2})$.

7. Jun 1, 2017

roam

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 $( \frac{\pi d^2}{4 \lambda z})^2 [ \frac{2 J_1(\frac{\pi r d}{\lambda z})}{(\frac{\pi r d}{\lambda z})}]^2$?

8. Jun 1, 2017

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
9. Jun 2, 2017

@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.