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1.4.1: Polarización lineal

  • Page ID
    51098
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    Un caso particular de elipse es una recta que pasa por el origen (Figura \(\PageIndex{1}\)). Esto significa que el campo está vibrando en una misma dirección del espacio. Esta situación degenerada se produce cuando
    \[
    \delta_{y}-\delta_{x}=\left\{\begin{array}{l} \notag
    0 \\
    \pi
    \end{array}\right.
    \]
    \({ }^{6}\) Hemos descompuesto el vector amplitud compleja en módulo y argumento y expresado su fase como \(-\delta_{x}\) \(\mathrm{o}-\delta_{y}\). El haber elegido fases \(\delta_{x}\) y \(\delta_{y}\) no cambiaría nada.

    clipboard_ec35e04cef832ec0164cd442411037154.png
    Figura \(\PageIndex{1}\): Una elipse puede degenerar en una recta.

    o bien cuando \(\left|E_{0 x}\right|=0\) o \(\left|E_{0 y}\right|=0\). Entonces diremos que la polarización es lineal y que la luz es linealmente polarizada. Podemos usar la notación
    \[
    \begin{aligned}
    \mathbf{E}_{0} &=\left(\begin{array}{c}
    E_{0 x} \\
    E_{0 y}
    \end{array}\right)=\left(\begin{array}{c}
    \left|E_{0 x}\right| e^{-i \delta_{x}} \\
    \left|E_{0 y}\right| e^{-i \delta_{y}}
    \end{array}\right) \\
    &=e^{-i \delta_{y}}\left(\begin{array}{c}
    \left|E_{0 x}\right| e^{-i\left(\delta_{x}-\delta_{y}\right)} \\
    \left|E_{0 y}\right|
    \end{array}\right)
    \end{aligned}
    \]
    aprovechamos la condición de diferencia de fase 0 o \(\pi\) para poner

    \[
    \mathbf{E}_{0}=e^{-i \delta_{y}}\left(\begin{array}{c}
    \pm\left|E_{0 x}\right| \\
    \left|E_{0 y}\right|
    \end{array}\right) \propto \text { vector real } \notag
    \]
    es decir, siempre que haya proporcionalidad a un vector real (aunque el coeficiente sea complejo) se dirá que la luz está linealmente polarizada.


    1.4.1: Polarización lineal is shared under a CC BY-SA 1.0 license and was authored, remixed, and/or curated by Alvaro Tejero Cantero.