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== Special Case ==
 
== Special Case ==
 
Picture an opaque screen with a circular aperture of radius a.<br><br>  
 
Picture an opaque screen with a circular aperture of radius a.<br><br>  
Let<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}'|}</math><br><br>
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Let<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}'|^2}</math><br><br>
 
Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math><br><br>
 
Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math><br><br>
 
But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>
 
But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>
 
::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br>
 
::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br>
 
::<math>=r'-\frac{2r\rho'\cos\phi}{2r'}\frac{\rho'}{r'}\sin\theta'</math><br><br>
 
::<math>=r'-\frac{2r\rho'\cos\phi}{2r'}\frac{\rho'}{r'}\sin\theta'</math><br><br>
so that <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math>
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so that <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math><br><br>
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In this particular case, we are dealing with far-field effects only, so <math>\frac{2r\sin\theta'\cos\phi}{r'}\rightarrow 0 </math> and <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx 1</math><br><br>
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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>
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