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10.6.2: Rendija

  • Page ID
    51207
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    Si una de las dimensiones de la abertura es muy grande comparada con la otra tenemos una rendija. Seguimos teniendo una abertura rectangular; por lo tanto la solución ya la hemos calculado. Sólo debemos introducir el hecho de que en una dirección es mucho mayor que en otra. Es lo mismo que hacer tender una de ellas a \(\infty: \beta \rightarrow \infty\). En ese caso

    • en el eje \(x\) la figura de difracción no cambia.
    • \(\Delta y \rightarrow 0\) : en el eje \(y\) todo se ve comprimido.
    clipboard_e764e4238b21f09722d7dc0ce6d0633dc.png
    Figura \(\PageIndex{1}\): Una rendija: \(b \ll a\)
    clipboard_e810e2871f25fa35a202bac42980c8acc.png
    Figura \(\PageIndex{3}\): Figura de difracción de una rendija vertical. Nótese que la intensidad se distribuye horizontalmente.

    Como se puede ver en la figura \(\PageIndex{2}\), la intensidad se distribuye a lo largo del eje \(x\) según un patrón de difracción, puesto que la abertura es pequeña. Sin embargo, en el eje \(y\) no hay difracción, por lo que vale la óptica geométrica.

    Es un problema unidimensional con

    \[
    \begin{aligned}
    I\left(x^{\prime}\right) &=I(0)\left(\frac{\sin \phi}{\phi}\right)^{2} \\
    &=I(0) \operatorname{sinc}^{2}\left[\left(k \frac{x^{\prime}}{f^{\prime}}-k x\right) \frac{a}{2}\right]
    \end{aligned}
    \]


    10.6.2: Rendija is shared under a CC BY-SA 1.0 license and was authored, remixed, and/or curated by LibreTexts.