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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> | + | 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> | + | 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> |
| | + | 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> |
| | + | 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> |