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Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 16:26, 2 July 2009
, 16:26, 2 July 2009→Determining Angle for First Diffraction Minimum
<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{x}-\mathbf{x}'|}\frac{2\pi}{4} \left[2\delta(|\mathbf{x}-\mathbf{x}'|+c(t-t'))-2\delta(|\mathbf{x}-\mathbf{x}'|-c(t-t'))\right]\Theta</math><br><br>
<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{x}-\mathbf{x}'|}\frac{2\pi}{4} \left[2\delta(|\mathbf{x}-\mathbf{x}'|+c(t-t'))-2\delta(|\mathbf{x}-\mathbf{x}'|-c(t-t'))\right]\Theta</math><br><br>
But the term <math>2\delta(|\mathbf{x}-\mathbf{x}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math><br><br>
But the term <math>2\delta(|\mathbf{x}-\mathbf{x}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math><br><br>
∴<math> G(x,x')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{x}-\mathbf{x}'|-c(t-t'))}{|\mathbf{x}-\mathbf{x}'|}</math><br>