Usuario:Carlos A.Z.

Hola esta es mi pagina soy alumno de optica

Alumno: Carlos Acosta Zepeda

Materia: Optica

Correo: aberk07@hotmail.com

--CAZ 20:50 3 jul 2008 (CDT)

Esta es la ultima hoja de soluciones Gaussianas corregida y con los calculos intermedios desarrollados.

Para la representación polar de ${\frac {1}{z-{\tilde {z}}}}={\frac {1}{R}}+{\frac {2\imath }{kw^{2}}}$ tenemos

$|z|={\sqrt {{\frac {1}{R^{2}}}+{\frac {4}{k^{2}w^{4}}}}}$ y el argumento $\theta =\arctan({\frac {2R}{kw^{2}}})$ entonces.

${\frac {1}{z-{\tilde {z}}}}={\sqrt {{\frac {1}{R^{2}}}+{\frac {4}{k^{2}w^{4}}}}}\exp[\imath \arctan({\frac {2R}{kw^{2}}})]$

Tomando definiciones vemos que

${\frac {R}{w^{2}}}={\frac {z-z_{0}+{\frac {z^{2}}{z-z_{0}}}}{w_{0}^{2}[{\frac {(z-z_{0})^{2}}{z_{R}^{2}}}+1]}}={\frac {z_{R}^{2}}{w_{0}^{2}(z-z_{0})}}={\frac {k}{2}}{\frac {z_{R}}{z-z_{0}}}$

y

${\frac {1}{R^{2}}}+{\frac {4}{k^{2}w^{4}}}={\frac {(z-z_{0})^{2}}{(z-z_{0})^{2}+z_{R}^{2}}}+{\frac {4z_{R}^{4}}{k^{2}w_{0}^{4}[(z-z_{0})^{2}+z_{R}^{2}]^{2}}}$

Definiendo $B^{2}=(z-z_{0})^{2}+z_{R}^{2}$ y mediante $k^{2}=({\frac {2z_{R}}{w_{0}^{2}}})^{2}$

tenemos

${\frac {1}{R^{2}}}+{\frac {4}{k^{2}w^{4}}}={\frac {1}{B^{2}}}(z-z_{0})^{2}+z_{R}={\frac {1}{B^{2}}}$

o

${\frac {1}{R^{2}}}+{\frac {4}{k^{2}w^{4}}}=({\frac {w_{0}}{z_{R}w}})^{2}$

de manera que

${\sqrt {{\frac {1}{R^{2}}}+{\frac {4}{k^{2}w^{4}}}}}\exp[\imath \arctan({\frac {2R}{kw^{2}}})]={\frac {w_{0}}{z_{R}w}}\exp[\imath \arctan({\frac {2R}{kw^{2}}})]={\frac {w_{0}}{z_{R}w}}\exp[\imath \arctan({\frac {z_{R}}{z-z_{0}}})]$

mientras que la fase es.

$\imath k{\frac {(x-x_{1})^{2}+(y-y_{1})^{2}}{2(z-z_{1})}}={\frac {\imath k}{2}}((x-x_{1})^{2}+(y-y_{1})^{2})({\frac {1}{R}}+{\frac {2\imath }{kw^{2}}})=\imath k{\frac {(x-x_{1})^{2}+(y-y_{1})^{2}}{2R}}-{\frac {(x-x_{1})^{2}+(y-y_{1})^{2}}{w^{2}}}$

La amplitud compleja es entonces

${\tilde {u}}=A_{0}{\frac {w_{0}}{z_{R}w}}\exp[\imath \arctan({\frac {2R}{kw^{2}}})]\exp[\imath k{\frac {(x-x_{1})^{2}+(y-y_{1})^{2}}{2R}}]\exp[-{\frac {(x-x_{1})^{2}+(y-y_{1})^{2}}{w^{2}}}]$

Tomando $\psi ={\tilde {u}}\exp[\imath kz]$

La solución de la ecuación diferencial en la aproximación paraxial es una Gaussiana dada por

$\psi =A_{0}{\frac {w_{0}}{z_{R}w}}\exp[\imath \arctan({\frac {2R}{kw^{2}}})]\exp[\imath k{\frac {(x-x_{1})^{2}+(y-y_{1})^{2}}{2R}}]\exp[-{\frac {(x-x_{1})^{2}+(y-y_{1})^{2}}{w^{2}}}]\exp[\imath kz]$

--CAZ 00:48 4 jul 2008 (CDT)

La máxima curvatura en la onda Gaussiana

El radio de curvatura es

$R(z)=(z-z_{0})+{\frac {z_{R}^{2}}{(z-z_{0})}}$

Derivando e igualando a cero tenemos:

${\frac {dR}{dz}}=1-{\frac {z_{R}^{2}}{(z-z_{0})^{2}}}=0$

despejando z obtenemos

$z=z_{0}\pm z_{R}$

--CAZ 01:17 4 jul 2008 (CDT)