86 lines
2.9 KiB
Markdown
86 lines
2.9 KiB
Markdown
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# Exercise 6
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**Generating points according to Fraunhofer diffraction**
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The diffraction of a plane wave thorough a round slit must be simulated by
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generating $N =$ 50'000 points according to the intensity distribution
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$I(\theta)$ on a screen at a great distance $L$ from the slit iself:
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$$
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I(\theta) = \frac{E^2}{2} \left( \frac{2 \pi a^2 \cos{\theta}}{L}
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\frac{J_1(x)}{x} \right)^2 \with x = k a \sin{\theta}
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$$
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where:
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- $E$ is the electric field amplitude, default set $E = \SI{1e4}{V/m}$;
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- $a$ is the radius of the slit aperture, default set $a = \SI{0.01}{m}$;
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- $\theta$ is the angle specified in @fig:fenditure;
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- $J_1$ is the Bessel function of first order;
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- $k$ is the wavenumber, default set $k = \SI{1e-4}{m^{-1}}$;
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- $L$ default set $L = \SI{1}{m}$.
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\begin{figure}
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\hypertarget{fig:fenditure}{%
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\centering
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\begin{tikzpicture}
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\definecolor{cyclamen}{RGB}{146, 24, 43}
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% Walls
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\draw [thick] (-1,3) -- (1,3) -- (1,0.3) -- (1.2,0.3) -- (1.2,3)
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-- (9,3);
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\draw [thick] (-1,-3) -- (1,-3) -- (1,-0.3) -- (1.2,-0.3) -- (1.2,-3)
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-- (9,-3);
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\draw [thick] (10,3) -- (9.8,3) -- (9.8,-3) -- (10,-3);
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% Lines
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\draw [thick, gray] (0.7,0.3) -- (0.5,0.3);
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\draw [thick, gray] (0.7,-0.3) -- (0.5,-0.3);
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\draw [thick, gray] (0.6,0.3) -- (0.6,-0.3);
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\draw [thick, gray] (1.2,0) -- (9.8,0);
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\draw [thick, gray] (1.2,-0.1) -- (1.2,0.1);
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\draw [thick, gray] (9.8,-0.1) -- (9.8,0.1);
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\draw [thick, cyclamen] (1.2,0) -- (9.8,-2);
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\draw [thick, cyclamen] (7,0) to [out=-90, in=50] (6.6,-1.23);
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% Nodes
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\node at (0,0) {$2a$};
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\node at (5.5,0.4) {$L$};
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\node [cyclamen] at (5.5,-0.4) {$\theta$};
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\node [rotate=-90] at (10.2,0) {screen};
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\end{tikzpicture}
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\caption{Fraunhofer diffraction.}\label{fig:fenditure}
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}
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\end{figure}
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Once again, $\theta$, which must be evenly distributed on half sphere, can be
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generated only as a function of a variable $x$ uniformely distributed between
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0 and 1. Therefore:
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\begin{align*}
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\frac{d^2 P}{d\omega^2} = const = \frac{1}{2 \pi}
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&\thus d^2 P = \frac{1}{2 \pi} d\omega^2 =
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\frac{1}{2 \pi} d\phi \sin{\theta} d\theta \\
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&\thus \frac{dP}{d\theta} = \int_0^{2 \pi} d\phi \frac{1}{2 \pi} \sin{\theta}
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= \frac{1}{2 \pi} \sin{\theta} \, 2 \pi = \sin{\theta}
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\end{align*}
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\begin{align*}
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\theta = \theta (x) &\thus
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\frac{dP}{d\theta} = \frac{dP}{dx} \cdot \left| \frac{dx}{d\theta} \right|
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= \left. \frac{dP}{dx} \middle/ \, \left| \frac{d\theta}{dx} \right| \right.
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\\
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&\thus \sin{\theta} = \left. 1 \middle/ \, \left|
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\frac{d\theta}{dx} \right| \right.
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\end{align*}
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Since $\theta \in [0, \pi/2]$, then the absolute value symbol can be omitted:
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\begin{align*}
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\frac{d\theta}{dx} = \frac{1}{\sin{\theta}}
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&\thus d\theta \sin(\theta) = dx
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\\
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&\thus - \cos (\theta') |_{0}^{\theta} = x(\theta) - x(0) = x - 0 = x
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\\
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&\thus - \cos(\theta) + 1 =x
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\\
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&\thus \theta = \text{acos} (1 -x)
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\end{align*}
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