Create biconcave lens

Question

I have been trying for hours.. can anyone help me? I can't fill with color the "shape" that is the lens.

\documentclass[border=12pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{intersections}

\begin{document}

\begin{tikzpicture}[scale=1]

\begin{scope}[>=latex]
\draw [->] (-3,0) -- (3,0);
\draw [->] (0,-1.25) -- (0,1.5);
\end{scope}
\definecolor{lensBlue}{RGB}{217,232,250}



\path [name path=arc1](-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\path [name path=arc2](1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];
\path [name path=rect](-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);

\draw (A) -- (D);
\draw (B) -- (C) ;

\draw (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];


\end{tikzpicture}

\end{document}

Answer 1

Here are two possible choices:

In the first case I used fill=white to white out the pink fill in in arc sections. But if you want that filled in you can do as in the second case. The drawing of the aixs was moved to the end so that it is on top. You could also achieve the same effect with the backgrounds library using on background layer.

Code:

\documentclass[border=3pt]{standalone}

\usepackage{tikz} %pgf-tikz pakcage
\usetikzlibrary{intersections}

\definecolor{lensBlue}{RGB}{217,232,250}

\begin{document}
\begin{tikzpicture}

\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);

\draw [brown, ultra thick] (A) -- (D);
\draw [brown, ultra thick] (B) -- (C) ;

\fill [red!50] (A) -- (D) -- (C) -- (B) -- cycle;

\draw [blue, thick, fill=white] (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick, fill=white] (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];

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

\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);

\draw [brown, ultra thick] (A) -- (D);
\draw [brown, ultra thick] (B) -- (C) ;

\fill [red!50] (A) -- (D) -- (C) -- (B) -- cycle;

\draw [blue, thick, fill=yellow!50] (-1.9,-1.5) arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick, fill=yellow!50] (1.9,1.5) arc[start angle=90, end angle=270,radius=1.5];

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

\end{document}

Answer 2

Two methods too

A) Without library intersections

We draw and fill a rectangle and then we add the two arcs.

\documentclass[border=3pt]{standalone}
\usepackage{tikz} 

\begin{document}
\begin{tikzpicture}

\filldraw [red!50,draw=brown, ultra thick] (-1,-0.9) rectangle (1,0.9);

\draw [blue, thick, fill=white] (-1.9,-1.5) 
      arc[start angle=-90, end angle=90,radius=1.5];
\draw [blue, thick, fill=white] (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}

B) With the library intersections but without filling outside

We fill directly the center part. We need to do this to determine the angle defined by the center I of an arc and a point of the arc (here D).

\documentclass[border=3pt]{standalone}
\usepackage{tikz} 
\usetikzlibrary{intersections}

\begin{document}
\begin{tikzpicture}
 \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);

\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)
        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];

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

Answer 3

User interfaces:

\const{Major}{3}% semi major
\const{Minor}{2}% semi minor
\const{Xo}{3.25}% distance from origin to ellipse center
\const{ThetaD}{50}

\documentclass[pstricks,border=12pt,nomessages]{standalone}
\usepackage{pstricks-add,fp}

\def\const#1#2{%
    \expandafter\FPeval\csname#1\endcsname{round(#2:17)}%only 17 digits after . is allowed
    \pstVerb{/#1 \csname#1\endcsname\space def}%
}

% user defined data
\const{Major}{3}% semi major
\const{Minor}{1.5}% semi minor
\const{Xo}{3.25}% distance from origin to ellipse center
\const{ThetaD}{60}

% internal used data
\const{AlphaD}{180-ThetaD}
\const{Alpha}{pi-ThetaD*pi/180}% supplementary angle relation
\const{DeltaAx}{Major*cos(Alpha)}
\const{DeltaAy}{Minor*sin(Alpha)}
\const{DeltaBx}{-Major}
\const{DeltaBy}{0}


\psset
{
    unit=1.5cm,
    linecolor=red,
}

\def\quadrant#1{%!
        \psscalebox{#1}{
        \pnode(!Xo 0){O}
        \pnode[!DeltaAx DeltaAy](O){A}
        \pnode[!DeltaBx DeltaBy](O){B}
        \pnode(0,0){Origin}
        \pscustom*[linecolor=yellow,origin={O}]
        {
            \psline(Origin)(Origin|A)(A)
            \psellipticarc(O)(!Major Minor){(A)}{(B)}
        }
        \psellipticarc[linestyle=dashed,linecolor=gray,origin={O},dimen=middle](O)(!Major Minor){90}{(A)}
        \psline(Origin|A)(A)
        \psellipticarc[origin={O},dimen=middle](O)(!Major Minor){(A)}{(B)}        
       }}

\begin{document}

\begin{pspicture}[showgrid=false](-\Xo,-\Minor)(\Xo,\Minor)
    \psforeach{\i}{1 1,1 -1,-1 -1,-1 1}{\quadrant{\i}}
    \psaxes[labels=none,ticks=none,linecolor=gray]{->}(0,0)(-\Xo,-\Minor)(\Xo,\Minor)
\end{pspicture} 

\end{document}