Optica: Interferencia de haces multiples

INTERFERENCIA DE HACES MULTIPLES

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La interferencia de haces multiples se da cuando un número muy grande de ondas mutuamente coherentes se hacen interferir.

Comenzemos el análisis matemático calculando la diferencia de camino óptico que 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:

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

factorizando

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

obtenemos la relación de Snell

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

de donde se obtiene:

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

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

Ahora definamos las variables.

r es el coeficiente de reflexión de las amplitudes desde el interior del interferometro.

t es el coeficiente de transmisión de las amplitudes para un rayo que viene del exterior.

t' es el coeficiente de transmisión para las amplitudes para un rayo que sale de la cavidad del interferometro.

d es la diferencia de fase entre dos reflexiones consecutivas.

tenemos entonces que para obtener

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\delta =2\pi m$

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

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

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

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

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

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

$I_{rmin}=0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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