Changes

So, <math> \mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}=\frac{e^{ikr'}}{r'^2}\int_0^a rdr\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math><br><br>

So, <math> \mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}=\frac{e^{ikr'}}{r'^2}\int_0^a rdr\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math><br><br>

The integral <math>\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math> is the integral representation of the zero order Bessel function of the first kind with <math> kr\sin{\theta}' \quad</math> as the argument.  This gives us the equation:<br><br>

The integral <math>\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math> is the integral representation of the zero order Bessel function of the first kind with <math> kr\sin{\theta}' \quad</math> as the argument.  This gives us the equation:<br><br>

<math>\mathcal{J}(r')=\frac{e^{ikr'}}{r'^2}\int_0^a rdr 2\pi J_0(kr \sin{\theta}') </math><br><br>

:<math>\mathcal{J}(r')=\frac{e^{ikr'}}{r'^2}\int_0^a rdr 2\pi J_0(kr \sin{\theta}') </math><br><br>

To simplify the math, we make use of the fact that we can represent this Bessel functions as the derivative of a Bessel function of a different order.  In general, the formula to compute this derivative is <br><br> <math>z^{v-k}J_{v-k}(z)=\left(\frac{1}{z}\frac{\part}{\part z}\right)^kz^vJ_v(z)</math><br><br>

To simplify the math, we make use of the fact that we can represent this Bessel functions as the derivative of a Bessel function of a different order.  In general, the formula to compute this derivative is <br><br>  

:<math>z^{v-k}J_{v-k}(z)=\left(\frac{1}{z}\frac{\part}{\part z}\right)^kz^vJ_v(z)</math><br><br>

In this case, we take <math>v=k=1 \quad</math> and <math>z=kr\sin{\theta}' \quad</math>.  So<br><br>

In this case, we take <math>v=k=1 \quad</math> and <math>z=kr\sin{\theta}' \quad</math>.  So<br><br>

<math>J_0(kr\sin{\theta}')=\left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')=\frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')</math><br><br>

:<math>J_0(kr\sin{\theta}')=\left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')</math><br><br>

This gives us the equation<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{r'^2}\int_0^a rdr \frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')</math><br><br>

<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}\left[aJ_1(ka\sin{\theta}')-0J_1(0k\sin{\theta}')\right]=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')</math><br><br> and <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'} 2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')=\frac{z'\dot{A_0}a}{c}\frac{e^{ikr'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>

:<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{r'^2}\int_0^a rdr \left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')</math><br><br>

:<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k^2\sin^2{\theta}'r'^2}\int_0^{ka\sin{\theta'}} dx \frac{d}{dx}xJ_1(x)</math>

<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}\left[aJ_1(ka\sin{\theta}')-0J_1(0k\sin{\theta}')\right]=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')</math><br><br> and <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-i\omega t'} 2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')=\frac{z'\dot{A_0}a}{c}\frac{e^{ikr'-i\omega t'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')</math><br><br>

To find the angle to the diffraction minimum, we must find the zeroes of this amplitude function.  This will occur when <math>J_1(ka\sin{\theta}')=0 \quad</math><br><br>

To find the angle to the diffraction minimum, we must find the zeroes of this amplitude function.  This will occur when <math>J_1(ka\sin{\theta}')=0 \quad</math><br><br>

[[Image:Bessels_J0.svg|thumb|300px|right|Plot of Bessel function of the first kind, J_&alpha;(x), for integer orders &alpha;=0,1,2.]]

[[Image:Bessels_J0.svg|thumb|300px|right|Plot of Bessel function of the first kind, J_&alpha;(x), for integer orders &alpha;=0,1,2.]]

To the right is a graph of three Bessel functions of the first order, specifically <math> J_0(x), J_1(x), and J_2(x) \quad</math>.  As it is shown, the first zero of <math>J_1(x) \quad</math> will <br>

To the right is a graph of three Bessel functions of the first order, specifically <math> J_0(x), J_1(x), and J_2(x) \quad</math>.  As it is shown, the first zero of <math>J_1(x) \quad</math> will <br>

occur at <math>x=0 \quad</math>.  This will correspond to the center of the pattern, at <math>\theta=0 \quad</math>.  Here, we would expect a bright spot, so <math>A(r') \quad</math> should be positive and finite.  At <math>\theta=0 \quad</math> the term <math>\frac{J_1(ka\sin{\theta}')}{\sin{\theta}'}</math> is positive and finite, so this expression gives the correct tamplitude at <math>\theta=0 \quad</math>.  The next zero of <math> J_1(x) \quad</math> corresponds to the first minumum of the diffraction pattern.  In this case, this zero occurs at x=3.832.  So, <math>ka\sin{\theta}'=3.832 \quad</math> . Since<math> k=\frac{2\pi}{\lambda}\quad</math> and <math> a=\frac{D}{2}\quad</math><br><br>

occur at <math>x=0 \quad</math>.  This will correspond to the center of the pattern, at <math>\theta=0 \quad</math>.  Here, we would expect a bright spot, so <math>A(r') \quad</math> should be positive and finite.  At <math>\theta=0 \quad</math> the term <math>\frac{J_1(ka\sin{\theta}')}{\sin{\theta}'}</math> is positive and finite, so this expression gives the correct amplitude at <math>\theta=0 \quad</math>.  The next zero of <math> J_1(x) \quad</math> corresponds to the first minumum of the diffraction pattern.  In this case, this zero occurs at x=3.832.  So, <math>ka\sin{\theta}'=3.832 \quad</math> . Since<math> k=\frac{2\pi}{\lambda}\quad</math> and <math> a=\frac{D}{2}\quad</math><br><br>

  <math>\frac{2\pi D\sin{\theta}'}{2\lambda}=3.832\rightarrow \sin{\theta}'= \frac{1.22\lambda}{D}</math>

  <math>\frac{2\pi D\sin{\theta}'}{2\lambda}=3.832\rightarrow \sin{\theta}'= \frac{1.22\lambda}{D}</math>