Optica: Interferencia de haces multiples

Error al representar (función desconocida «\thetatilde»): {\displaystyle \sin{\thetatilde_{0}}=\frac{BA'}{C'B}=\frac{BA'}{CP}=\frac{BA'}{2d\tan\theta}} la diferencia de camino óptico se puede obtener de la figura (1) donde:

$\sin \theta _{0}={\frac {BA'}{C'B}}={\frac {BA'}{CP}}={\frac {BA'}{2d\tan \theta }}$

$BA'=2d\tan \theta \sin \theta _{0}$ lo cual se pude demostrar que es igual a:

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

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

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

de donde se obtiene:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\delta =2\pi m$

$E_{1t}=E_{0}tt'e^{i\omega t}$

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

$E_{3t}=E_{0}tt'r'^{4}e^{i\left(\omega t-2\delta \right)}$

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

$E_{tT}=E_{0}e^{i\omega t}\left[tt'\sum _{j=1}^{N}r'^{(j-1)}e^{-i\left(j-1\right)\delta }\right]$

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

$I_{rmin}=0$

$\delta =\left(2m+1\right)\pi$

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

$I_{rmax}=I_{i}{\frac {4r^{2}}{(1+r^{2})^{2}}}$

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

${\frac {(1-r^{2})^{2}}{1+r^{4}-2r^{2}\left(1-2\sin ^{2}{\frac {\delta }{2}}\right)}}$

${\frac {\left(1-r^{2}\right)^{2}}{\left(1-r^{2}\right)^{2}+4r^{2}\sin ^{2}{\frac {\delta }{2}}}}$

${\frac {1}{1+\left[{\frac {2r}{1-r^{2}}}\sin {\frac {\delta }{2}}\right]^{2}}}$

$\digamma =\left({\frac {2r}{1-r^{2}}}\right)^{2}$

${\frac {4r}{(1-r)^{2}}}$

${\frac {I_{t}}{I_{i}}}={\frac {1}{1+\digamma \sin ^{2}{\frac {\delta }{2}}}}\equiv A\left(\delta \right)$

${\frac {I_{r}}{I_{t}}}={\frac {\digamma \sin ^{2}{\frac {\delta }{2}}}{1+\digamma \sin ^{2}{\frac {\delta }{2}}}}$

$tt'=T=1-R-A$

$\delta ={\frac {4\pi }{\lambda }}n_{t}d\cos \theta +2\delta _{r}$

${\frac {I_{t}}{I_{i}}}={\frac {T^{2}}{1+R^{2}-2R\cos \delta }}$

${\frac {I_{t}}{I_{i}}}={\frac {T^{2}}{(1-R^{2})}}A(\delta )$

${\frac {I_{t}}{I_{i}}}={\frac {\left(1-R-A\right)^{2}}{\left(1-R\right)^{2}}}A\left(\delta \right)$

$=\left(1-{\frac {A}{1-R}}\right)^{2}A\left(\delta \right)$

$I_{i}\left(1-{\frac {A}{1-R}}\right)^{2}\equiv I_{tmax}$

${\frac {I_{t}}{i_{tmax}}}=A\left(\theta \right)={\frac {1}{1+\digamma \sin ^{2}{\frac {\delta }{2}}}}$

$\delta _{max}=2\pi m$

$\gamma =2{\frac {\delta _{i}}{2}}$

${\frac {1}{2}}={\frac {1}{1+\digamma \sin \left({\frac {{\frac {\delta }{2}}+\delta _{max}}{2}}\right)}}$

$\sin \left({\frac {\delta _{\frac {1}{2}}}{2}}+m\pi \right)=\pm sin\left({\frac {\delta _{\frac {1}{2}}}{2}}\right)$

$\sin ^{2}\left({\frac {\delta _{\frac {1}{2}}}{2}}+m\pi \right)=\sin ^{2}\left({\frac {\delta _{\frac {1}{2}}}{2}}\right)$

$\sin ^{2}\left({\frac {\delta _{\frac {1}{2}}}{2}}\right)={\frac {1}{\digamma }}$

$\delta _{\frac {1}{2}}=2\arcsin \left({\sqrt {\frac {1}{\digamma }}}\right)$

$\gamma =4\arcsin \left({\frac {1}{\sqrt {\digamma }}}\right)$

${\frac {\delta _{\frac {1}{2}}}{2}}={\frac {1}{\sqrt {\digamma }}}$

$\gamma \approxeq \left({\frac {4}{\sqrt {\digamma }}}\right)$

$\Upsilon \equiv {\frac {2\pi }{\gamma }}={\frac {\pi }{2}}{\sqrt {\digamma }}$