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diff --git a/Master/texmf-dist/doc/generic/pst-diffraction/pst-diffraction-doc.tex b/Master/texmf-dist/doc/generic/pst-diffraction/pst-diffraction-doc.tex new file mode 100644 index 00000000000..f38f06f02e3 --- /dev/null +++ b/Master/texmf-dist/doc/generic/pst-diffraction/pst-diffraction-doc.tex @@ -0,0 +1,451 @@ +%% $Id: pst-diffraction-docE.tex 134 2009-09-27 12:28:50Z herbert $ +\documentclass[11pt,english,BCOR10mm,DIV12,bibliography=totoc,parskip=false,smallheadings, + headexclude,footexclude,oneside]{pst-doc} +\usepackage{pst-grad,pst-diffraction} +\let\pstDiffractionFV\fileversion + +\usepackage{libertinus} +\usepackage{biblatex} +\addbibresource{pst-diffraction-doc.bib} + +\lstset{pos=t,wide=true,language=PSTricks, + morekeywords={psdiffractionRectangle,psdiffractionCircle,psdiffractionCircular},basicstyle=\footnotesize\ttfamily} +\lstdefinestyle{syntax}{backgroundcolor=\color{blue!20},numbers=none,xleftmargin=0pt,xrightmargin=0pt, + frame=single} +\lstdefinestyle{example}{backgroundcolor=\color{red!20},numbers=none,xleftmargin=0pt,xrightmargin=0pt, + frame=single} +\newcommand*\psp{\texttt{pspicture}\xspace} +% +\begin{document} + +\title{\texttt{pst-diffraction}} +\subtitle{Diffraction patterns for diffraction from circular, rectangular and triangular +apertures; v.\pstDiffractionFV} +\author{Manuel Luque \\ Herbert Vo\ss} +\docauthor{Herbert Voß} +\date{\today} +\maketitle + +\tableofcontents + +\clearpage + +\begin{abstract} +\noindent + +\vfill\noindent +Thanks to: Julien Cubizolles, +Doris Wagner, +Timothy Van Zandt, Keno Wehr, +Michael Zedler. +\end{abstract} + +\section{Optical setup} + +\begin{center} +\begin{pspicture}(0,-3)(12,3) +\pnode(0,0){S} \pnode(4,1){L'1} \pnode(4,-1){L'2} \pnode(6,1){E'1} \pnode(6,-1){E'2} +\pnode(6,0.5){E1}\pnode(6,-0.5){E2}\pnode(8.5,1.5){L1}\pnode(8.5,0.5){L2}\pnode(11.5,1.25){P} +% lentille L' +\pscustom[fillstyle=gradient,linecolor=blue,gradend=white]{% + \code{0.5 0.83333 scale} + \psarc(4,0){4.176}{-16.699}{16.699} + \psarc(12,0){4.176}{163.30}{196.699}} +% lentille L +\pscustom[fillstyle=gradient,linecolor=blue,gradend=white]{% + \code{1 1.5 scale} + \psarc(4.5,0){4.176}{-16.699}{16.699} + \psarc(12.5,0){4.176}{163.30}{196.699}} +\pspolygon[linestyle=none,fillstyle=vlines, + hatchcolor=yellow](S)(L'1)(E'1)(E1)(L1)(P)(L2)(E2)(E'2)(L'2) +\uput[90](4,1){$L'$}\uput[90](8.5,2){$L$} +\psdot(S)\uput[180](S){S} +\psline(S)(12,0)\psline[linewidth=2\pslinewidth](6,2)(6,0.5)\psline[linewidth=2\pslinewidth](6,-2)(6,-0.5) +\psline[linestyle=dashed](6,0.5)(6,-0.5)\psline(11.5,-3)(11.5,3)\psline(S)(L'1)(E'1)\psline(S)(L'2)(E'2) +\uput[0](P){P} +\psline(E1)(L1)(P)\psline(E2)(L2)(P)\psline[linestyle=dashed](8.5,0)(P) +%\rput(8.5,0){\psarc{->}(0,0){1.5}{0}{!1.25 3 atan}\uput[0](1.5;15){$\theta$}} +\uput[-90](10,0){$f$}\uput[0](6,2){E}\uput[135](6,0){T}\uput[45](11.5,0){O} +\end{pspicture} +\end{center} + +Monochromatic light rays diverging from the focal point S of a positive lens L' emerge parallel to +the axis and strike the aperture stop E with the aperture T. +The light bends behind the aperture, this bending is called diffraction: +Every point in the opening acts as if it was a point source (Huygens's principle) and the +light waves of all those points overlap and produce an interference pattern (diffraction +pattern) on a screen. When the screen is very far away, the observed patterns are called +Fraunhofer diffraction patterns. In this case one can assume that the rays from the aperture +striking the same point P on the screen are parallel.\\ +In practice one wants to realize a short distance between the aperture stop and the screen. +Hence one sets up a converging lens L after the opening and installs the screen +into the focal plane (containing the points P and O) of this lens. Parallel rays incident on +the lens are then focused at a point P in the focal plane. + +With the following PSTricks-commands we can draw the diffraction patterns for different +geometric forms +of apertures. It is understood that only monochromatic light is used. The aperture stops can +have rectangular, circular or triangular openings. + +The options available are the dimensions of the aperture under consideration and of the particular optical +setting, e.g. the radius in case of an circular opening. Moreover one can choose the wavelength +of the light (the associated color will be given automatically by the package). + +There are three commands, for rectangular, circular and triangular openings respectively: + +\begin{BDef} +\Lcs{psdiffractionRectangle}\OptArgs\\ +\Lcs{psdiffractionCircular}\OptArgs\\ +\Lcs{psdiffractionTriangle}\OptArgs +\end{BDef} + + +\section{The color} +The desired color is defined by specifying the associated wavelength $\lambda$ (in nanometers). +Red for instance one gets by the option \Lkeyword{lambda}=632 because +red light has the wavelength $\lambda_{\textrm{rot}}=632\,\textrm{nm}$. + +The conversion of the wavelength into the associated \texttt{RGB}-value is done by PostScript. +The code is similar to the code of a FORTRAN program which can be found here: \\ +\url{http://www.midnightkite.com/color.html} + +\section{Diffraction from a rectangular aperture} + +\begin{center} +\begin{pspicture}(-2,-1)(2,1.5) +\psframe(-0.5,-1)(0.5,1) +\pcline{<->}(-0.5,1.1)(0.5,1.1) +\Aput{$a$} +\pcline{<->}(0.6,1)(0.6,-1) +\Aput{$h=k\times a$} +\end{pspicture} +\end{center} + +The width of the rectangle with the area $h=k\times a$ is defined by the letter \Lkeyword{a}, +the height by \Lkeyword{k}. +The focal length is specified by \Lkeyword{f}, the desired resolution in pixels [pixel]. +With the option \Lkeyword{contrast} one can improve the visibility of the minor secondary +maxima more. +We get a black and white picture if we use the option \Lkeyword{colorMode}=0, +the option \Lkeyword{colorMode}=1 provides the associated negative pattern. The options +\Lkeyword{colorMode}=2 and \Lkeyword{colorMode}=3 render color pictures in the +\Index{CMYK} and \Index{RGB} color model respectively. + +By default the settings are as follows: + + +\begin{tabular}{@{}lll@{}} +\Lkeyword{a}=0.2e-3 in m; & \Lkeyword{k}=1; & \Lkeyword{f}=5 in m;\\ +\Lkeyword{lambda}=650 in nm; & \Lkeyword{pixel}=0.5; & \Lkeyword{contrast}=38, greates value;\\ +\Lkeyword{colorMode}=3; & \Lkeyword{IIID}=\false. +\end{tabular} + +\bigskip +\noindent +\begin{pspicture}(-3.5,-3.5)(3.5,3.5) +\psdiffractionRectangle[f=2.5] +\end{pspicture} +\hfill +\begin{pspicture}(-1.5,-2.5)(3.5,3.5) +\psdiffractionRectangle[IIID,Alpha=30,f=2.5] +\end{pspicture} + +\begin{lstlisting}[style=example] +\begin{pspicture}(-3.5,-3.5)(3.5,3.5) +\psdiffractionRectangle[f=2.5] +\end{pspicture} +\hfill +\begin{pspicture}(-1.5,-2.5)(3.5,3.5) +\psdiffractionRectangle[IIID,Alpha=30,f=2.5] +\end{pspicture} +\end{lstlisting} + + + +\noindent\begin{pspicture}(-2,-4)(2,4) +\psdiffractionRectangle[a=0.5e-3,k=0.5,f=4,pixel=0.5,colorMode=0] +\end{pspicture} +\hfill +\begin{pspicture}(0,-3)(4,4) +\psdiffractionRectangle[IIID,a=0.5e-3,k=0.5,f=4,pixel=0.5,colorMode=0] +\end{pspicture} + + +\begin{lstlisting}[style=example] +\begin{pspicture}(-2,-4)(2,4) +\psdiffractionRectangle[a=0.5e-3,k=0.5,f=4,pixel=0.5,colorMode=0] +\end{pspicture} +\hfill +\begin{pspicture}(0,-3)(4,4) +\psdiffractionRectangle[IIID,a=0.5e-3,k=0.5,f=4,pixel=0.5,colorMode=0] +\end{pspicture} +\end{lstlisting} + + + +\noindent +\begin{pspicture}(-2.5,-2.5)(3.5,3) +\psdiffractionRectangle[a=0.5e-3,k=2,f=10,lambda=515,colorMode=1] +\end{pspicture} +\hfill +\begin{pspicture}(-1.5,-2)(3.5,3) +\psdiffractionRectangle[IIID,Alpha=20,a=0.5e-3,k=2,f=10,lambda=515,colorMode=1] +\end{pspicture} + + +\begin{lstlisting}[style=example] +\begin{pspicture}(-2.5,-2.5)(3.5,3) +\psdiffractionRectangle[a=0.5e-3,k=2,f=10,lambda=515,colorMode=1] +\end{pspicture} +\hfill +\begin{pspicture}(-1.5,-2)(3.5,3) +\psdiffractionRectangle[IIID,Alpha=20,a=0.5e-3,k=2,f=10,lambda=515,colorMode=1] +\end{pspicture} +\end{lstlisting} + + +\noindent +\begin{pspicture}(-3.5,-1)(3.5,1) +\psdiffractionRectangle[a=0.5e-3,k=20,f=10,pixel=0.5,lambda=450] +\end{pspicture} +\hfill +\begin{pspicture}(-3.5,-1)(3.5,4) +\psdiffractionRectangle[IIID,Alpha=10,a=0.5e-3,k=20,f=10,pixel=0.5,lambda=450] +\end{pspicture} + +\begin{lstlisting}[style=example] +\begin{pspicture}(-3.5,-1)(3.5,1) +\psdiffractionRectangle[a=0.5e-3,k=20,f=10,pixel=0.5,lambda=450] +\end{pspicture} +\hfill +\begin{pspicture}(-3.5,-1)(3.5,4) +\psdiffractionRectangle[IIID,Alpha=10,a=0.5e-3,k=20,f=10,pixel=0.5,lambda=450] +\end{pspicture} +\end{lstlisting} + +\section[Diffraction from two rectangular apertures]{Diffraction from two rectangular apertures% +\protect\footnote{This simulation was provided by Julien Cubizolles.}} +It is also possible to render the diffraction pattern of two congruent rectangles +(placed parallel such that their base is located on the $x$-axis) +by using the option \Lkeyword{twoSlit}. +By default this option is deactivated. +The distance of the two rectangles is specified by the option $s$. +The default for $s$ is $12e^{-3}\,\mathrm{m}$. + + +\begin{center} +\noindent +\begin{pspicture}(-4,-1)(4,1) +\psdiffractionRectangle[a=0.5e-3,k=10,f=10,pixel=0.5,lambda=650,twoSlit,s=2e-3] +\end{pspicture} +\end{center} + +\begin{lstlisting}[style=example] +\begin{pspicture}(-4,-1)(4,1) +\psdiffractionRectangle[a=0.5e-3,k=10,f=10,pixel=0.5,lambda=650,twoSlit,s=2e-3] +\end{pspicture} +\end{lstlisting} + +\begin{center} +\begin{pspicture}(-2,-1)(4,4) +\psdiffractionRectangle[IIID,Alpha=20,a=0.5e-3,k=10,f=10,pixel=0.5,lambda=650,twoSlit,s=2e-3] +\end{pspicture} +\end{center} + +\begin{lstlisting}[pos=t,style=example,wide=false] +\begin{pspicture}(-2,-1)(4,4) +\psdiffractionRectangle[IIID,Alpha=20,a=0.5e-3,k=10,f=10,pixel=0.5,lambda=650,twoSlit,s=2e-3] +\end{pspicture} +\end{lstlisting} + + + + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Diffraction from a circular aperture} +The radius of the circular opening can be chosen via the letter \Lkeyword{r}, e.g. +\Lkeyword{r}=1e-3. The default is $r=1$ mm. In the first quadrant +PSTricks displays the graph of the intensity distribution (the maximum in the center will be +cropped if its height exceeds the margin of the environment \Lenv{pspicture*}). + +\hspace*{-1cm}% +\begin{LTXexample}[pos=t,style=example,wide=false] +\begin{pspicture}(-3.5,-3.5)(3.5,3.5) +\psdiffractionCircular[r=0.5e-3,f=10,pixel=0.5,lambda=520] +\end{pspicture} +% +\begin{pspicture}(-3.5,-1.5)(3.5,3.5) +\psdiffractionCircular[IIID,r=0.5e-3,f=10,pixel=0.5,lambda=520] +\end{pspicture} +\end{LTXexample} + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Diffraction from two circular apertures} +Only the case of equal radii is provided, this common radius can be defined like in the +previous section via \Lkeyword{r}=\dots. +Furthermore one has to give the half distance of the circles measured from their centers by +\Lkeyword{d}=\dots, e.g. \Lkeyword{d}=3e-3. Also the option +\Lkeyword{twoHole} has to be used.\\ +The rendering process could take some time in this case\dots + + +\begin{pspicture}(-3,-3.5)(3.5,3.5) +\psdiffractionCircular[r=0.5e-3,f=10,d=3e-3,lambda=515,twoHole] +\end{pspicture} +% +\begin{pspicture}(-3.5,-1.5)(3.5,3.5) +\psdiffractionCircular[IIID,r=0.5e-3,f=10,d=3e-3,lambda=515,twoHole] +\end{pspicture} + + +\begin{lstlisting}[style=example] +\begin{pspicture}(-3,-3.5)(3.5,3.5) +\psdiffractionCircular[r=0.5e-3,f=10,d=3e-3,lambda=515,twoHole] +\end{pspicture} +% +\begin{pspicture}(-3.5,-1.5)(3.5,3.5) +\psdiffractionCircular[IIID,r=0.5e-3,f=10,d=3e-3,lambda=515,twoHole] +\end{pspicture} +\end{lstlisting} + + +\hspace*{-1cm}% +\begin{pspicture}(-3,-3)(3.5,4) +\psdiffractionCircular[r=0.5e-3,f=10,d=2e-3,lambda=700,twoHole,colorMode=0] +\end{pspicture} +% +\begin{pspicture}(-3.5,-2)(3.5,3.5) +\psdiffractionCircular[IIID,r=0.5e-3,f=10,d=2e-3,lambda=700,twoHole,colorMode=0] +\end{pspicture} + +\begin{lstlisting}[style=example] +\begin{pspicture}(-3.5,-3)(3.5,4) +\psdiffractionCircular[r=0.5e-3,f=10,d=2e-3,lambda=700,twoHole,colorMode=0] +\end{pspicture} +% +\begin{pspicture}(-3.5,-2)(3.5,3.5) +\psdiffractionCircular[IIID,r=0.5e-3,f=10,d=2e-3,lambda=700,twoHole,colorMode=0] +\end{pspicture} +\end{lstlisting} + +Not in every case bands occur in the central circle. The number $N$ of those inner +bands is given by $N=2.44\frac{d}{r}$. Thus this effect is not observable until $N\geq2$ +or $d=\frac{2r}{1.22}$ (see +\url{http://www.unice.fr/DeptPhys/optique/diff/trouscirc/diffrac.html}). + +\hspace*{-1cm}% +\begin{pspicture}(-3,-3.5)(3,3.5) +\psdiffractionCircular[r=0.5e-3,f=10,d=4.1e-4,lambda=632,twoHole] +\end{pspicture} +% +\begin{pspicture}(-3.5,-1.5)(3.5,3) +\psdiffractionCircular[IIID,r=0.5e-3,f=10,d=4.1e-4,lambda=632,twoHole] +\end{pspicture} + + +\begin{lstlisting}[style=example] +\begin{pspicture}(-3,-3.5)(3,3.5) +\psdiffractionCircular[r=0.5e-3,f=10,d=4.1e-4,lambda=632,twoHole] +\end{pspicture} +% +\begin{pspicture}(-3.5,-1.5)(3.5,3) +\psdiffractionCircular[IIID,r=0.5e-3,f=10,d=4.1e-4,lambda=632,twoHole] +\end{pspicture} +\end{lstlisting} + + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Diffraction from a triangular aperture} + +Only the case of an equilateral triangle is provided, whose height \Lkeyword{h} has to be +defined as an option. As is generally known, $h$ can be computed from the length $s$ of +its side by $h=\frac{\sqrt{3}}{2}s$. A black and white picture can be obtained by using the +option \Lkeyword{colorMode}=0. + + + +\begin{center} +\begin{pspicture}(-1,-1)(1,1) +\pspolygon*(0,0)(1;150)(1;210) +\pcline{|-|}(-0.732,-1)(0,-1) +\Aput{$h$} +\end{pspicture} +\end{center} + +\makebox[\linewidth]{% +\begin{pspicture}(-3,-3)(3,2.5) +\psdiffractionTriangle[f=10,h=1e-3,lambda=515,contrast=38] +\end{pspicture} +\quad +\begin{pspicture}(-3,-3)(3,2.5) +\psdiffractionTriangle[f=10,h=1e-3,colorMode=1,contrast=38,lambda=515] +\end{pspicture} +\quad +\begin{pspicture}(-3,-3)(3,2.5) +\psdiffractionTriangle[f=10,h=1e-3,colorMode=0,contrast=38,lambda=515] +\end{pspicture}} + + +\begin{lstlisting}[style=example] +\begin{pspicture}(-3,-3)(3,2.5) +\psdiffractionTriangle[f=10,h=1e-3,lambda=515,contrast=38] +\end{pspicture} +\quad +\begin{pspicture}(-3,-3)(3,2.5) +\psdiffractionTriangle[f=10,h=1e-3,colorMode=1,contrast=38,lambda=515] +\end{pspicture} +\quad +\begin{pspicture}(-3,-3)(3,2.5) +\psdiffractionTriangle[f=10,h=1e-3,colorMode=0,contrast=38,lambda=515] +\end{pspicture} +\end{lstlisting} + + +\makebox[\linewidth]{% +\begin{pspicture}(-3,-2)(3,3.5) +\psdiffractionTriangle[IIID,f=10,h=1e-3,lambda=515,contrast=38] +\end{pspicture} +\quad +\begin{pspicture}(-3,-2)(3,3.5) +\psdiffractionTriangle[IIID,f=10,h=1e-3,colorMode=1,contrast=38,lambda=515] +\end{pspicture} +\quad +\begin{pspicture}(-3,-2)(3,3.5) +\psdiffractionTriangle[IIID,f=10,h=1e-3,colorMode=0,contrast=38,lambda=515] +\end{pspicture}} + +\begin{lstlisting}[style=example] +\begin{pspicture}(-3,-2)(3,3.5) +\psdiffractionTriangle[IIID,f=10,h=1e-3,lambda=515,contrast=38] +\end{pspicture} +\quad +\begin{pspicture}(-3,-2)(3,3.5) +\psdiffractionTriangle[IIID,f=10,h=1e-3,colorMode=1,contrast=38,lambda=515] +\end{pspicture} +\quad +\begin{pspicture}(-3,-2)(3,3.5) +\psdiffractionTriangle[IIID,f=10,h=1e-3,colorMode=0,contrast=38,lambda=515] +\end{pspicture} +\end{lstlisting} + + + +\clearpage +\section{List of all optional arguments for \texttt{pst-diff}} +\Loption{pst-diff} is the short form for the keywords in the package \LPack{pst-diffraction}. + +\xkvview{family=pst-diff,columns={key,type,default}} + + + +\raggedright +\nocite{*} +\printbibliography +%\bibliography{pst-diffraction-doc} + +\printindex + +\end{document} |