Optica: Interferencia de haces multiples

$\sin {\widetilde {\theta _{0}}}={\frac {BA'}{C'B}}={\frac {BA'}{CP}}={\frac {BA'}{2d\tan \theta }}$ Error al representar (función desconocida «\thetatilde»): {\displaystyle \sin\thetatilde{\theta_{0}}=\frac{BA'}{C'B}=\frac{BA'}{CP}=\frac{BA'}{2d\tan\theta}} $BA'=2d\tan \theta \sin {\widetilde {\theta _{0}}}$

$\Delta \Lambda ={\frac {2d}{\cos \theta }}nt-2d\tan \theta \sin \theta _{0}ni$

$={\frac {2d}{\cos \theta }}nt\left[1-\sin \theta \sin \theta _{0}{\frac {ni}{nt}}\right]$

$={\frac {2d}{\cos \theta }}nt\left[1-\sin ^{2}\theta \right]$

$\Delta \Lambda =2dnt\cos \theta$

$\delta ={\frac {2\pi }{\lambda _{0}}}\Delta \Lambda ={\frac {4\pi }{\lambda _{0}}}ntd\cos \theta$

$E_{1r}=E_{0}re^{i\omega t}$

$E_{2r}=E_{0}tt'r'e^{i\left(\omega t-\delta \right)}$

$E_{3r}=E_{0}tt'r'^{3}e^{i\left(\omega t-2\delta \right)}$

$E_{Nr}=E_{0}tt'r'^{(2N-3)}e^{i\left(\omega t-\left(N-1\right)\delta \right)}$

$E_{rT}=\sum E_{ir}=E_{0}re^{i\omega t}+\sum _{j=2}^{N}E_{0}tr'^{(2j-3)}t'e^{i\left(\omega t-\left(j-1\right)\delta \right)}$

$=E_{0}e^{i\omega t}\left[r+\left\{\sum _{j=2}^{N}\left(r'^{2}e^{i\delta }\right)^{j-2}\right\}r'tt'e^{-i\delta }\right]$

$E_{r}=E_{0}e^{i\omega t}\left[r+{\frac {r'tt'e^{-i\delta }}{1-r'^{2}e^{-i\delta }}}\right]$

$E_{r}=E_{0}e^{i\omega t}\left[{\frac {r\left(1-e^{-i\delta }\right)}{1-r^{2}e^{-i\delta }}}\right]$

$tt'+r^{2}=1$

$I_{R}=E_{r}E_{r}^{*}=E_{0}^{2}\left[{\frac {r\left(1-e^{i\delta }\right)}{1-r^{2}e^{i\delta }}}\right]\left[{\frac {r\left(1-e^{i\delta }\right)}{1-r^{2}e^{i\delta }}}\right]$

$I_{r}=E_{0}^{2}{\frac {2r^{2}\left(1-\cos \theta \right)}{\left(1+r^{4}\right)-2r^{2}\cos \delta }}$

$E_{t}=E_{0}e^{i\omega t}{\frac {tt'}{1-r^{2}e^{-i\delta }}}$

$I_{t}=I_{i}\left(tt'\right)^{2}{\frac {1}{1-r^{2}e^{-i\delta }}}{\frac {1}{1-r^{2}e^{i\delta }}}$

$I_{t}=I_{i}\left(tt'\right)^{2}{\frac {1}{\left(1+r^{4}\right)-2r^{2}\cos \delta }}$

$I_{t}=I_{i}{\frac {\left(1-r^{2}\right)^{2}}{\left(1+r^{4}\right)-2r^{2}\cos \delta }}$

$I_{r}+I_{t}={\frac {\left(1-r^{2}\right)^{2}+2r^{2}\left(1-\cos \delta \right)}{\left(1+r^{4}\right)-2r^{2}\cos \delta }}I_{i}$

$={\frac {1+r^{4}-2r^{2}-2r^{2}\cos \delta +2r^{2}}{\left(1+r^{4}\right)-2r^{2}\cos \delta }}I_{i}=I_{i}$

$I_{i}=I_{i}=I_{r}+I_{t}$

$\delta =2\pi m$