196
edits
Changes
Jump to navigation
Jump to search
Line 55:
Line 55: + +
Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 19:29, 2 July 2009
, 19:29, 2 July 2009→Determining Angle of First Diffraction Minimum
In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>. Also, <math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A_0}e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}</math>. So:<br><br>
In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>. Also, <math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A_0}e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}</math>. So:<br><br>
<math>A(r')=\frac{-z'\dot{A_0}}{2\pi c}\int_{z=0} rdrd\phi \frac{e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}</math><br><br>
<math>A(r')=\frac{-z'\dot{A_0}}{2\pi c}\int_{z=0} rdrd\phi \frac{e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}</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}'|}</math><br><br>
Then <math>A(r')=\frac{-z'\hat{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math>