Rendijas

Rendija

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Para un rendija (Figura 1)

$\Psi (P)=c\int \limits _{l_{1}}^{l_{2}}e^{i\kappa \alpha \xi }\,d\xi$

$\Psi (P)=c{\frac {e^{-i\kappa \alpha \xi }}{-i\kappa \alpha }}|_{-a}^{\;a}$

Evaluando.

$\Psi (P)={\frac {c}{-i\kappa \alpha }}(e^{-i\kappa \alpha a}-e^{i\kappa \alpha a})$

multiplicando por 2/2.

$\Psi (P)={\frac {2c}{\kappa \alpha }}({\frac {e^{i\kappa \alpha a}-e^{-i\kappa \alpha a}}{2i}})$

$\Psi (P)={\frac {2c}{\kappa \alpha }}(Sen(\kappa \alpha a))$

$\Psi (P)=2ca{\frac {Sen(\kappa \alpha a)}{\kappa \alpha a}}$

Se nombra.

$\gamma =\kappa \alpha a$

Nuevamente se obtiene.

$\Psi (P)=2ca{\frac {Sen(\gamma )}{\gamma }}$

Elevando al cuadrado. $|\Psi (P)|=4c^{2}a^{2}({\frac {Sen\gamma }{\gamma }})^{2}$

Doble rendija

$\Psi (P)=c\{\int \limits _{-d-a}^{-d+a}e^{-i\kappa \alpha \xi }\,d\xi +\int \limits _{d-a}^{d+a}e^{-i\kappa \alpha \xi }\,d\xi \}$

$\Psi (P)=c\{{\frac {e^{-i\kappa \alpha \xi }}{-i\kappa \alpha \xi }}|_{-d-a}^{-d+a}+{\frac {e^{-i\kappa \alpha \xi }}{-i\kappa \alpha \xi }}|_{d-a}^{d+a}\}$

$\Psi (P)=c{\frac {1}{-i\kappa \alpha }}\{e^{-i\kappa \alpha (-d+a)}-e^{-i\kappa \alpha (-d-a)}+e^{-i\kappa \alpha (d+a)}-e^{-i\kappa \alpha (d-a)}\}$

Mulptiplicando por un factor 2\2.

$\Psi (P)={\frac {2c}{2}}{\frac {1}{-i\kappa \alpha }}\{e^{i\kappa \alpha (d-a)}-e^{-i\kappa \alpha (-d-a)}+e^{i\kappa \alpha (-d-a)}-e^{-i\kappa \alpha (d-a)}\}$

Reacomodando los terminos de la suma de exponenciasles.

$\Psi (P)={\frac {2c}{2}}{\frac {1}{-i\kappa \alpha }}\{(e^{i\kappa \alpha (d-a)}-e^{-i\kappa \alpha (d-a)})-(e^{i\kappa \alpha (d+a)}-e^{-i\kappa \alpha (d+a)})\}$

$\Psi (P)=-2c{\frac {1}{\kappa \alpha }}\{{\frac {(e^{i\kappa \alpha (d-a)}-e^{-i\kappa \alpha (d-a)})}{2i}}-{\frac {(e^{i\kappa \alpha (d+a)}-e^{-i\kappa \alpha (d+a)})}{2i}}\}$

$\Psi (P)=-2c{\frac {1}{\kappa \alpha }}\{Sen(\kappa \alpha (d-a))-Sen(\kappa \alpha (d+a))\}$

Usando la indentidad.

$Sen(a)-Sen(b)=2Cos({\frac {1}{2}}(a+b)Sen{\frac {1}{2}}(a-b))$

Entonces.

$Sen(\kappa \alpha (d-a))-Sen(\kappa \alpha (d+a))=\{2Cos({\frac {1}{2}}(\kappa \alpha d-\kappa \alpha a+\kappa \alpha d+\kappa \alpha a))Sen({\frac {1}{2}}(\kappa \alpha d-\kappa \alpha a-\kappa \alpha d-\kappa \alpha a))\}$

haciendo las sumas y eliminando terminos se obtiene.

$\Psi (P)={\frac {4c}{\kappa \alpha }}\{Sen(\kappa \alpha a)Cos(\kappa \alpha d)\}$

Con un cambio de variable se obtiene.

$\Psi (P)=4ca\{{\frac {Sen(\gamma )}{\gamma }}Cos(\delta )\}$

N Rendijas

Para n dendijas se tiene se tiene la solución construida de la siguiente manera, se puede observar el arreglo de las n rendigas en la figura 1.

$\Psi (p)=c\{\int \limits _{0}^{b}e^{i\kappa \alpha \xi }\,d\xi +\int \limits _{h}^{h+b}e^{i\kappa \alpha \xi }\,d\xi +\int \limits _{2h}^{2h+b}......+\int \limits _{(n-1)h}^{(n-1)h+b}e^{i\kappa \alpha \xi }\,d\xi \}$

Donde cada uno de los sumandos representan la solución a la rejilla individual, para cada una de las integrales se observa .

$\int \limits _{l_{1}}^{l_{2}}e^{i\kappa \alpha \xi }\,d\xi ={\frac {e^{i\kappa \alpha \xi }}{\kappa \alpha \xi }}|_{l_{1}}^{l_{2}}={\frac {1}{\kappa \alpha }}{\frac {e^{i\kappa \alpha l_{2}}-e^{i\kappa \alpha l_{1}}}{i}}$

Resolviendo la primera integral y evaluando en los limites.

$1={\frac {1}{\kappa \alpha }}{\frac {e^{i\kappa \alpha b}-1}{i}}$

$2={\frac {1}{\kappa \alpha }}{\frac {e^{i\kappa \alpha (h+b)}-e^{i\kappa \alpha h}}{i}}$

.

.

.

$n={\frac {1}{\kappa \alpha }}{\frac {e^{i\kappa \alpha (n-1)h+b}-e^{i\kappa \alpha (n-1)h}}{i}}$

La segunda ecuación se puede reescribir de la siguiente manera.

${\frac {1}{\kappa \alpha }}e^{i\kappa \alpha h}{\frac {(e^{i\kappa \alpha b}-1)}{i}}$

Para la eneada ecuación.

${\frac {1}{\kappa \alpha }}e^{i\kappa \alpha (n-1)h}{\frac {(e^{i\kappa \alpha b}-1)}{i}}$

Donde se puede observar un termino comun en todas los terminos, se puede reescribir la ecuacion como sigue.

$\Psi (p)=c\{{\frac {1}{\kappa \alpha }}{\frac {e^{i\kappa \alpha b}-1}{i}}\}\{1+e^{i\kappa \alpha h}+e^{i\kappa \alpha 2h}+...e^{i\kappa \alpha (n-1)h}\}$

Recordando.

$\sigma =1+x+x^{2}+...+x^{n-1}$

$\sigma x=x+x^{2}+x^{3}+...+x^{n}$

Restando las dos ecuaciones anteriores.

$x\sigma -\sigma =x^{n}-1$

$\sigma (x-1)=x^{n}-1$

$\sigma ={\frac {x^{n}-1}{x-1}}$

Si se observa la ecuación se reescribe.

$\sigma ={\frac {x^{n}-1}{x-1}}={\frac {e^{i\kappa \alpha nh}-1}{e^{i\kappa \alpha h}-1}}$

en la ecuacion.

$\Psi (P)=c{\frac {e^{i\kappa \alpha b}-1}{\kappa \alpha }}\{{\frac {e^{i\kappa \alpha nh}-1}{e^{i\kappa \alpha h}-1}}\}$

$\Psi (P)={\frac {2c}{\kappa \alpha }}e^{i\kappa \alpha {\frac {b}{2}}}({\frac {e^{i\kappa \alpha {\frac {b}{2}}}-e^{-i\kappa \alpha {\frac {b}{2}}}}{2i}}){\frac {e^{i\kappa \alpha n{\frac {h}{2}}}({\frac {e^{i\kappa \alpha n{\frac {h}{2}}}-e^{-i\kappa \alpha n{\frac {h}{2}}}}{2i}})}{e^{i\kappa \alpha {\frac {h}{2}}}({\frac {e^{i\kappa \alpha {\frac {h}{2}}}-e^{-i\kappa \alpha {\frac {h}{2}}}}{2i}})}}$

$\Psi (P)={\frac {2c}{\kappa \alpha }}e^{i\kappa \alpha {\frac {b}{2}}}(Sen({\frac {\kappa \alpha b}{2}})){\frac {e^{i\kappa \alpha n{\frac {h}{2}}}(Sen({\frac {\kappa \alpha nh}{2}}))}{e^{i\kappa \alpha {\frac {h}{2}}}(Sen({\frac {\kappa \alpha nh}{2}}))}}$

$\Psi (P)=cbe^{i\kappa \alpha {\frac {b}{2}}}e^{i\kappa \alpha (n-1){\frac {h}{2}}}{\frac {(Sen({\frac {\kappa \alpha b}{2}}))}{\frac {\kappa \alpha b}{2}}}{\frac {Sen(n\kappa \alpha {\frac {h}{2}})}{Sen(\kappa \alpha {\frac {h}{2}})}}$

$\Psi (P)=c\{n\}{\frac {Sen(\beta )}{\beta }}{\frac {Sen(n\beta )}{n\beta }}$

$|\Psi (P)|=I_{0}({\frac {Sen(\beta )}{\beta }})^{2}({\frac {Sen(n\beta )}{n\beta }})^{2}$