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Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 18:57, 2 July 2009
, 18:57, 2 July 2009→Determining Angle for First Diffraction Minimum
If<math>A(\mathbf{x},t) \quad</math> is independent of <math>\mathbf{x} \quad</math>, then:<br><br>
If<math>A(\mathbf{x},t) \quad</math> is independent of <math>\mathbf{x} \quad</math>, then:<br><br>
<math>A(x')=\frac{-\part}{\part z'}\int_{z'=0}^\infin dS A\left(\mathbf{0},t-\frac{S}{c}\right)=A\left(\mathbf{\emptyset},t'-\frac{z'}{c}\right)</math><br><br>
<math>A(x')=\frac{-\part}{\part z'}\int_{z'=0}^\infin dS A\left(\mathbf{0},t-\frac{S}{c}\right)=A\left(\mathbf{\emptyset},t'-\frac{z'}{c}\right)</math><br><br>
This gives us uniform translations of waves at velocity c. More generally: <br><br>
This gives us uniform translation of waves at velocity c. More generally: <br><br>
<math>A(x')=\frac{-1}{2\pi}\int_{z=0} d^2x\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{x}, t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|}\right)</math><br><br>
<math>A(x')=\frac{-1}{2\pi}\int_{z=0} d^2x\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{x}, t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|}\right)</math><br><br>
<math>=\frac{-1}{2\pi}\int_{z=0} d^2x\left(\frac{A\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|^3}(-z')+\frac{\dot{A}\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|c}\frac{-z'}{|\mathbf{x}-\mathbf{x}'|}\right)</math><br><br>
<math>=\frac{-1}{2\pi}\int_{z=0} d^2x\left(\frac{A\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|^3}(-z')+\frac{\dot{A}\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|c}\frac{-z'}{|\mathbf{x}-\mathbf{x}'|}\right)</math><br><br>
<math>A(x')=\frac{-1}{2\pi}\int_{z=0} d^2x\left(\frac{A\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|^3}(-z')+\frac{1}{c}\frac{\dot{A}\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|^2}(z')\right)</math><br><br>
<math>A(x')=\frac{-1}{2\pi}\int_{z=0} d^2x\left(\frac{A\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|^3}(-z')+\frac{1}{c}\frac{\dot{A}\left(\mathbf{x},t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|^2}(z')\right)</math>