# Example: Biconcave lens

Published 2012-11-25 | Author: Alain Matthes

The example is made with the “intersections” library. We fill directly the center part. We need to determine the angle defined by the center I of an arc and a point of the arc (here D).

Originally posted to TeX.SE.

Do you have a question regarding this example, TikZ or LaTeX in general? Just ask in the LaTeX Forum.
Oder frag auf Deutsch auf TeXwelt.de. En français: TeXnique.fr.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections}

\begin{document}
\begin{tikzpicture}
\coordinate (I)  at (1.9,0);
\path [name path=arc1, draw=none](-1.9,-1.5)
\path [name path=arc2, draw=none](1.9,1.5)
\path [name path=rect, draw=none](-0.9,-0.9) rectangle (0.9,0.9);

\path [name intersections={of = arc1 and rect}];
\coordinate (A)  at (intersection-1);
\coordinate (B)  at (intersection-2);

\path [name intersections={of = arc2 and rect}];
\coordinate (C)  at (intersection-1);
\coordinate (D)  at (intersection-2);

\pgfmathanglebetweenpoints{\pgfpointanchor{I}{center}}{%
\pgfpointanchor{D}{center}}
\let\tmpan\pgfmathresult

\fill[red!50] (A)--(D)
arc[start angle=\tmpan, end angle=360-\tmpan,radius=1.5] -- (B)

\draw [brown, ultra thick] (A) -- (D);
\draw [brown, ultra thick] (B) -- (C) ;
\draw [blue, thick] (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick] (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];

\begin{scope}[>=latex]
\draw [->] (-3,0) -- (3,0);
\draw [->] (0,-1.25) -- (0,1.5);
\end{scope}
\end{tikzpicture}
\end{document} 