Exercise 6

Generating points according to Fraunhofer diffraction

The diffraction of a plane wave thorough a round slit must be simulated by generating N = 50'000 points according to the intensity distribution I(\theta) on a screen at a great distance L from the slit iself:


  I(\theta) = \frac{E^2}{2} \left( \frac{2 \pi a^2 \cos{\theta}}{L}
  \frac{J_1(x)}{x} \right)^2 \with x = k a \sin{\theta}

where:

E is the electric field amplitude, default set E = \SI{1e4}{V/m};
a is the radius of the slit aperture, default set a = \SI{0.01}{m};
\theta is the angle specified in @fig:fenditure;
J_1 is the Bessel function of first order;
k is the wavenumber, default set k = \SI{1e-4}{m^{-1}};
L default set L = \SI{1}{m}.

\begin{figure} \hypertarget{fig:fenditure}{% \centering \begin{tikzpicture} \definecolor{cyclamen}{RGB}{146, 24, 43} % Walls \draw [thick] (-1,3) -- (1,3) -- (1,0.3) -- (1.2,0.3) -- (1.2,3) -- (9,3); \draw [thick] (-1,-3) -- (1,-3) -- (1,-0.3) -- (1.2,-0.3) -- (1.2,-3) -- (9,-3); \draw [thick] (10,3) -- (9.8,3) -- (9.8,-3) -- (10,-3); % Lines \draw [thick, gray] (0.7,0.3) -- (0.5,0.3); \draw [thick, gray] (0.7,-0.3) -- (0.5,-0.3); \draw [thick, gray] (0.6,0.3) -- (0.6,-0.3); \draw [thick, gray] (1.2,0) -- (9.8,0); \draw [thick, gray] (1.2,-0.1) -- (1.2,0.1); \draw [thick, gray] (9.8,-0.1) -- (9.8,0.1); \draw [thick, cyclamen] (1.2,0) -- (9.8,-2); \draw [thick, cyclamen] (7,0) to [out=-90, in=50] (6.6,-1.23); % Nodes \node at (0,0) {$2a$}; \node at (5.5,0.4) {$L$}; \node [cyclamen] at (5.5,-0.4) {$\theta$}; \node [rotate=-90] at (10.2,0) {screen}; \end{tikzpicture} \caption{Fraunhofer diffraction.}\label{fig:fenditure} } \end{figure}

Once again, \theta, which must be evenly distributed on half sphere, can be generated only as a function of a variable x uniformely distributed between 0 and 1. Therefore:

\begin{align*} \frac{d^2 P}{d\omega^2} = const = \frac{1}{2 \pi} &\thus d^2 P = \frac{1}{2 \pi} d\omega^2 = \frac{1}{2 \pi} d\phi \sin{\theta} d\theta \ &\thus \frac{dP}{d\theta} = \int_0^{2 \pi} d\phi \frac{1}{2 \pi} \sin{\theta} = \frac{1}{2 \pi} \sin{\theta} , 2 \pi = \sin{\theta} \end{align*}

\begin{align*} \theta = \theta (x) &\thus \frac{dP}{d\theta} = \frac{dP}{dx} \cdot \left| \frac{dx}{d\theta} \right| = \left. \frac{dP}{dx} \middle/ , \left| \frac{d\theta}{dx} \right| \right. \ &\thus \sin{\theta} = \left. 1 \middle/ , \left| \frac{d\theta}{dx} \right| \right. \end{align*}

Since \theta \in [0, \pi/2], then the absolute value symbol can be omitted:

\begin{align*} \frac{d\theta}{dx} = \frac{1}{\sin{\theta}} &\thus d\theta \sin(\theta) = dx \ &\thus - \cos (\theta') |_{0}^{\theta} = x(\theta) - x(0) = x - 0 = x \ &\thus - \cos(\theta) + 1 =x \ &\thus \theta = \text{acos} (1 -x) \end{align*}

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Exercise 6

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