7.4 KiB
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 setE = \SI{1e4}{V/m}
;a
is the radius of the slit aperture, default seta = \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 setk = \SI{1e-4}{m^{-1}}
;L
default setL = \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, the try and catch method described in @sec:3 was implemented and
the same procedure about the generation of \theta
was employed. This time,
though, \theta
must be evenly distributed on half sphere:
\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*}
If \theta
is chosen to grew together with x
, then the absolute value 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*}
The sample was stored and plotted in a histogram with a customizable number $n$
of bins default set n = 150
. In \textcolor{red}{fig} an example is shown.
\textcolor{red}{missing plot.}
Gaussian noise convolution
The sample must then be smeared with a gaussian noise with the aim to recover
the original sample afterwards, implementing a deconvolution routine.
For this purpose, a 'kernel' histogram with a odd number m
of bins and the
same bin width of the previous one, but a smaller number of them (m < n
), was
filled with m
points according to a gaussian distribution with mean \mu
,
corresponding to the central bin, and variance \sigma
.
Then, the original histogram was convolved with the kernel in order to obtain
the smeared signal. The procedure is summed up in \textcolor{red}{fig}.
\textcolor{red}{missing plots.}
The third histogram was obtained by keeping the same edges of the original signal and a number of bins n +m -1 (?). The convolution was obtained by permorming the dot product between the invere kernel and the clean signal for each relative position of the two histograms. For a better understaing, see \textcolor{red}{fig}.
\begin{figure} \hypertarget{fig:dot_conv}{% \centering \begin{tikzpicture} \definecolor{cyclamen}{RGB}{146, 24, 43} % original histogram \draw [thick, cyclamen, fill=cyclamen!15!white] (0.0,0) rectangle (0.5,2.5); \draw [thick, cyclamen, fill=cyclamen!15!white] (0.5,0) rectangle (1.0,2.8); \draw [thick, cyclamen, fill=cyclamen!15!white] (1.0,0) rectangle (1.5,2.3); \draw [thick, cyclamen, fill=cyclamen!15!white] (1.5,0) rectangle (2.0,1.8); \draw [thick, cyclamen, fill=cyclamen!15!white] (2.0,0) rectangle (2.5,1.4); \draw [thick, cyclamen, fill=cyclamen!15!white] (2.5,0) rectangle (3.0,1.0); \draw [thick, cyclamen, fill=cyclamen!15!white] (3.0,0) rectangle (3.5,1.0); \draw [thick, cyclamen, fill=cyclamen!15!white] (3.5,0) rectangle (4.0,0.6); \draw [thick, cyclamen, fill=cyclamen!15!white] (4.0,0) rectangle (4.5,0.4); \draw [thick, cyclamen, fill=cyclamen!15!white] (4.5,0) rectangle (5.0,0.2); \draw [thick, cyclamen, fill=cyclamen!15!white] (5.0,0) rectangle (5.5,0.2); \draw [thick, cyclamen] (6.0,0) -- (6.0,0.2); \draw [thick, cyclamen] (6.5,0) -- (6.5,0.2); \draw [thick, <->] (0,3.3) -- (0,0) -- (7,0); % kernel histogram \draw [thick, cyclamen, fill=cyclamen!15!white] (1.0,-1) rectangle (1.5,-1.2); \draw [thick, cyclamen, fill=cyclamen!15!white] (1.5,-1) rectangle (2.0,-1.4); \draw [thick, cyclamen, fill=cyclamen!15!white] (2.0,-1) rectangle (2.5,-1.8); \draw [thick, cyclamen, fill=cyclamen!15!white] (2.5,-1) rectangle (3.0,-1.4); \draw [thick, cyclamen, fill=cyclamen!15!white] (3.0,-1) rectangle (3.5,-1.2); \draw [thick, <->] (1,-2) -- (1,-1) -- (4,-1); % arrows \draw [thick, <->] (1.25,-0.2) -- (1.25,-0.8); \draw [thick, <->] (1.75,-0.2) -- (1.75,-0.8); \draw [thick, <->] (2.25,-0.2) -- (2.25,-0.8); \draw [thick, <->] (2.75,-0.2) -- (2.75,-0.8); \draw [thick, <->] (3.25,-0.2) -- (3.25,-0.8); \draw [thick, ->] (2.25,-2.0) -- (2.25,-4.2); % smeared histogram \begin{scope}[shift={(0,-1)}] \draw [thick, cyclamen, fill=cyclamen!15!white] (-1.0,-4.5) rectangle (-0.5,-4.3); \draw [thick, cyclamen, fill=cyclamen!15!white] (-0.5,-4.5) rectangle ( 0.0,-4.2); \draw [thick, cyclamen, fill=cyclamen!15!white] ( 0.0,-4.5) rectangle ( 0.5,-2.0); \draw [thick, cyclamen, fill=cyclamen!15!white] ( 0.5,-4.5) rectangle ( 1.0,-1.6); \draw [thick, cyclamen, fill=cyclamen!15!white] ( 1.0,-4.5) rectangle ( 1.5,-2.3); \draw [thick, cyclamen, fill=cyclamen!15!white] ( 1.5,-4.5) rectangle ( 2.0,-2.9); \draw [thick, cyclamen, fill=cyclamen!15!white] ( 2.0,-4.5) rectangle ( 2.5,-3.4); \draw [thick, cyclamen] (3.0,-4.5) -- (3.0,-4.3); \draw [thick, cyclamen] (3.5,-4.5) -- (3.5,-4.3); \draw [thick, cyclamen] (4.0,-4.5) -- (4.0,-4.3); \draw [thick, cyclamen] (4.5,-4.5) -- (4.5,-4.3); \draw [thick, cyclamen] (5.0,-4.5) -- (5.0,-4.3); \draw [thick, cyclamen] (5.5,-4.5) -- (5.5,-4.3); \draw [thick, cyclamen] (6.0,-4.5) -- (6.0,-4.3); \draw [thick, cyclamen] (6.5,-4.5) -- (6.5,-4.3); \draw [thick, cyclamen] (7.0,-4.5) -- (7.0,-4.3); \draw [thick, cyclamen] (7.5,-4.5) -- (7.5,-4.3); \draw [thick, <->] (-1,-2.5) -- (-1,-4.5) -- (8,-4.5); \end{scope} \end{tikzpicture} \caption{Dot product as a step of the convolution between the original signal (above) and the kernel (below). The final result is the lower fledging histogram.}\label{fig:dot_conv} } \end{figure}