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.

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Biconcave lens

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  \coordinate (I)  at (1.9,0);     
  \path [name path=arc1, draw=none](-1.9,-1.5) 
     arc[start angle=-90, end angle=90,radius=1.5];
  \path [name path=arc2, draw=none](1.9,1.5)   
     arc[start angle=90, end angle=270,radius=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);


  \fill[red!50] (A)--(D)  
        arc[start angle=\tmpan, end angle=360-\tmpan,radius=1.5] -- (B)
        arc[start angle=\tmpan-180, end angle=180-\tmpan,radius=1.5] ; 

  \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];

    \draw [->] (-3,0) -- (3,0);
    \draw [->] (0,-1.25) -- (0,1.5);


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