55

I was trying to draw something like that

enter image description here

Searching through the net to see if it already exists in tikz I came accross in texample.net the following image which looks a lot like what I want to draw.

enter image description here

I also found in our site a way to draw a circularly polarized light. I tried to combine a bit the two codes but I don't seem to be able to achieve something useful.

"My code" is

\documentclass[a4paper,11pt]{article}

\usepackage{kerkis}
\usepackage{tikz}

\usetikzlibrary{%
    calc,%
    fadings,%
    shadings%
}
\usetikzlibrary{arrows,snakes,shapes}
\usetikzlibrary{backgrounds}
\usetikzlibrary{
    shapes.geometric,
    decorations.pathreplacing
}
\begin{document}

\begin{tikzpicture}[x={(0.866cm,-0.5cm)}, y={(0.866cm,0.5cm)}, z={(0cm,1cm)}, scale=1.0,
    %Option for nice arrows
    >=stealth, %
    inner sep=0pt, outer sep=2pt,%
    axis/.style={thick,->},
    wave/.style={thick,color=#1,smooth},
    polaroid/.style={fill=black!60!white, opacity=0.3},
]
    % Colors
    \colorlet{darkgreen}{green!50!black}
    \colorlet{lightgreen}{green!80!black}
    \colorlet{darkred}{red!50!black}
    \colorlet{lightred}{red!80!black}

    % Frame
    \coordinate (O) at (0, 0, 0);
    \draw[axis] (O) -- +(14, 0,   0) node [right] {x};
    \draw[axis] (O) -- +(0,  2.5, 0) node [right] {y};
    \draw[axis] (O) -- +(0,  0,   2) node [above] {z};

    \draw[thick,dashed] (-2,0,0) -- (O);

    % monochromatic incident light with electric field
    \draw[wave=blue, opacity=0.7, variable=\x, samples at={-2,-1.75,...,0}]
        plot (\x, { cos(1.0*\x r)*sin(2.0*\x r)}, { sin(1.0*\x r)*sin(2.0*\x r)})
        plot (\x, {-cos(1.0*\x r)*sin(2.0*\x r)}, {-sin(1.0*\x r)*sin(2.0*\x r)});

    \foreach \x in{-2,-1.75,...,0}{
        \draw[color=blue, opacity=0.7,->]
            (\x,0,0) -- (\x, { cos(1.0*\x r)*sin(2.0*\x r)}, { sin(1.0*\x r)*sin(2.0*\x r)})
            (\x,0,0) -- (\x, {-cos(1.0*\x r)*sin(2.0*\x r)}, {-sin(1.0*\x r)*sin(2.0*\x r)});
    }

    \filldraw[polaroid] (0,-2,-1.5) -- (0,-2,1.5) -- (0,2,1.5) -- (0,2,-1.5) -- (0,-2,-1.5)
        node[below, sloped, near end]{Polaroid};%

    %Direction of polarization
    \draw[thick,<->] (0,-1.75,-1) -- (0,-0.75,-1);

    % Electric field vectors
    \draw[wave=blue, variable=\x,samples at={0,0.25,...,6}]
        plot (\x,{sin(2*\x r)},0)node[anchor=north]{$\vec{E}$};

    %Polarized light between polaroid and thin section
    \foreach \x in{0, 0.25,...,6}
        \draw[color=blue,->] (\x,0,0) -- (\x,{sin(2*\x r)},0);

    \draw (3,1,1) node [text width=2.5cm, text centered]{Polarized light};

    %Crystal thin section
    \begin{scope}[thick]
        \draw (6,-2,-1.5) -- (6,-2,1.5) node [above, sloped, midway]{Crystal section}
                -- (6, 2, 1.5) -- (6, 2, -1.5) -- cycle % First face
            (6,  -2, -1.5) -- (6.2, -2,-1.5)
            (6,   2, -1.5) -- (6.2,  2,-1.5)
            (6,  -2,  1.5) -- (6.2, -2, 1.5)
            (6,   2,  1.5) -- (6.2,  2, 1.5)
            (6.2,-2, -1.5) -- (6.2, -2, 1.5) -- (6.2, 2, 1.5) 
                -- (6.2, 2, -1.5) -- cycle; % Second face

        %Optical indices
        \draw[darkred, ->]       (6.1, 0, 0) -- (6.1, 0.26,  0.966) node [right] {$n_{g}'$}; % index 1
        \draw[darkred, dashed]   (6.1, 0, 0) -- (6.1,-0.26, -0.966); % index 1
        \draw[darkgreen, ->]     (6.1, 0, 0) -- (6.1, 0.644,-0.173) node [right] {$n_{p}'$}; % index 2
        \draw[darkgreen, dashed] (6.1, 0, 0) -- (6.1,-0.644, 0.173); % index 2
    \end{scope}

    %Second polarization
    \draw[polaroid]   (12, -2,  -1.5) -- (12, -2,   1.5)  %Polarizing filter
        node [above, sloped,midway] {Polaroid} -- (12, 2, 1.5) -- (12, 2, -1.5) -- cycle;
    \draw[thick, <->] (12, -1.5,-0.5) -- (12, -1.5, 0.5); %Polarization direction

\tikzset{%
    xyz path/.style args={\x=#1; \y=#2; \z=#3; (#4)}{
        insert path={
            \foreach \step [evaluate={\x=#1; \y=#2; \z=#3;}] in {#4}{   
                -- (\x, \y, \z) } 
        }
    },
    cosine path/.style args={#1:#2}{
        xyz path={\x=cos(\step); \y=0; \z=\step/360; (#1, 5, ..., #2)},
        insert path={ coordinate (cosine path end) }
    },
    sine path/.style args={#1:#2}{
        xyz path={\x=0; \y=sin(\step); \z=\step/360; (#1, 5, ..., #2)},
        insert path={ coordinate (sine path end) }
    },
    spiral path/.style args={#1:#2}{
        xyz path={\x=cos(\step); \y=sin(\step); \z=\step/360; (#1, 5, ..., #2)},
        insert path={ coordinate (spiral path end) }
    },
    marker/.style={
        insert path={
            node [fill, circle,  inner sep=0pt, minimum size=#1] {}
        }
    }
}

\def\lastangle{135}
\def\cycles{5}

\foreach \cycle in {0,...,\cycles}{
    \tikzset{shift={(0, 0, \cycle)}}
    \ifnum\cycle=\cycles
        \let\endangle=\lastangle
    \else
        \def\endangle{360}
    \fi

    \draw [blue, very thick] (1, 0, 0) [spiral path={0:\endangle}];
}

\end{tikzpicture}
\end{document}

My output is

enter image description here

How to move the circular spiral to the crystal section?

Improve this question
2
  • Off-topic: what's with the English-to-Greek transliterations? Surely the Greek word for "polarized" isn't exactly that? Let alone "light".
    – Ryan Reich
    Commented May 12, 2013 at 17:53
  • 6
    @RyanReich: You are right!!! I am writing in Greek as a main language(no XeLaTeX), so while I was struggling with the code I didn't care about what you noticed! Just for the record the greek word for polarized is πολωμένο and it's read poloMEno and the greek word for light is φως and it's read fos!
    – Thanos
    Commented May 12, 2013 at 18:03

2 Answers 2

Reset to default
52

To be honest, I don't understand the spiral code to well, at lest not why switching the roles of x and z leads to strange results. So I would recommend using Tikz's plot operation:

Code

\documentclass[a4paper,11pt]{article}

\usepackage{kerkis}
\usepackage{tikz}

\usetikzlibrary{%
    calc,%
    fadings,%
    shadings%
}
\usetikzlibrary{arrows,snakes,shapes}
\usetikzlibrary{backgrounds}
\usetikzlibrary{
    shapes.geometric,
    decorations.pathreplacing
}
\begin{document}

\begin{tikzpicture}[x={(-30:1cm)}, y={(30:1cm)}, z={(90:1cm)}, scale=1.0,
    %Option for nice arrows
    >=stealth, %
    inner sep=0pt, outer sep=2pt,%
    axis/.style={thick,->},
    wave/.style={thick,color=#1,smooth},
    polaroid/.style={fill=black!60!white, opacity=0.3},
]
    % Colors
    \colorlet{darkgreen}{green!50!black}
    \colorlet{lightgreen}{green!80!black}
    \colorlet{darkred}{red!50!black}
    \colorlet{lightred}{red!80!black}

    % Frame
    \coordinate (O) at (0, 0, 0);
    \draw[axis] (O) -- +(14, 0,   0) node [right] {x};
    \draw[axis] (O) -- +(0,  2.5, 0) node [right] {y};
    \draw[axis] (O) -- +(0,  0,   2) node [above] {z};

    \draw[thick,dashed] (-2,0,0) -- (O);

    % monochromatic incident light with electric field
    \draw[wave=blue, opacity=0.7, variable=\x, samples at={-2,-1.75,...,0}]
        plot (\x, { cos(1.0*\x r)*sin(2.0*\x r)}, { sin(1.0*\x r)*sin(2.0*\x r)})
        plot (\x, {-cos(1.0*\x r)*sin(2.0*\x r)}, {-sin(1.0*\x r)*sin(2.0*\x r)});

    \foreach \x in{-2,-1.75,...,0}{
        \draw[color=blue, opacity=0.7,->]
            (\x,0,0) -- (\x, { cos(1.0*\x r)*sin(2.0*\x r)}, { sin(1.0*\x r)*sin(2.0*\x r)})
            (\x,0,0) -- (\x, {-cos(1.0*\x r)*sin(2.0*\x r)}, {-sin(1.0*\x r)*sin(2.0*\x r)});
    }

    \filldraw[polaroid] (0,-2,-1.5) -- (0,-2,1.5) -- (0,2,1.5) -- (0,2,-1.5) -- (0,-2,-1.5)
        node[below, sloped, near end]{Polaroid};%

    %Direction of polarization
    \draw[thick,<->] (0,-1.75,-1) -- (0,-0.75,-1);

    % Electric field vectors
    \draw[wave=blue, variable=\x,samples at={0,0.25,...,6}]
        plot (\x,{sin(2*\x r)},0)node[anchor=north]{$\vec{E}$};

    %Polarized light between polaroid and thin section
    \foreach \x in{0, 0.25,...,6}
        \draw[color=blue,->] (\x,0,0) -- (\x,{sin(2*\x r)},0);

    \draw (3,1,1) node [text width=2.5cm, text centered]{Polarized light};

    %Crystal thin section
    \begin{scope}[thick]
        \draw (6,-2,-1.5) -- (6,-2,1.5) node [above, sloped, midway]{Crystal section}
                -- (6, 2, 1.5) -- (6, 2, -1.5) -- cycle % First face
            (6,  -2, -1.5) -- (6.2, -2,-1.5)
            (6,   2, -1.5) -- (6.2,  2,-1.5)
            (6,  -2,  1.5) -- (6.2, -2, 1.5)
            (6,   2,  1.5) -- (6.2,  2, 1.5)
            (6.2,-2, -1.5) -- (6.2, -2, 1.5) -- (6.2, 2, 1.5) 
                -- (6.2, 2, -1.5) -- cycle; % Second face

        %Optical indices
        \draw[darkred, ->]       (6.1, 0, 0) -- (6.1, 0.26,  0.966) node [right] {$n_{g}'$}; % index 1
        \draw[darkred, dashed]   (6.1, 0, 0) -- (6.1,-0.26, -0.966); % index 1
        \draw[darkgreen, ->]     (6.1, 0, 0) -- (6.1, 0.644,-0.173) node [right] {$n_{p}'$}; % index 2
        \draw[darkgreen, dashed] (6.1, 0, 0) -- (6.1,-0.644, 0.173); % index 2
    \end{scope}

    %Second polarization
    \draw[polaroid]   (12, -2,  -1.5) -- (12, -2,   1.5)  %Polarizing filter
        node [above, sloped,midway] {Polaroid} -- (12, 2, 1.5) -- (12, 2, -1.5) -- cycle;
    \draw[thick, <->] (12, -1.5,-0.5) -- (12, -1.5, 0.5); %Polarization direction

\draw[thick,blue] plot[domain=0:1080,smooth,samples=540] ({6+\x/180},{-1*cos(\x)},{1*sin(\x)});
\foreach \x in {0,45,...,1080}
{   \draw[blue,->] ({6+\x/180},{0},{0}) -- ({6+\x/180},{-1*cos(\x)},{1*sin(\x)});
}

\end{tikzpicture}
\end{document}

Output

enter image description here

If you want the circular part to start where the flat part ended, simply change the factors in front of the sin and cos terms to 0.5. This will however look odd, so you'll then probably have to change the angles of the coordinate axes (the parameters of the tikzpicture).

Also nice to see someone else is using kerkis ;)

Improve this answer
5
  • 1
    Thank you very much for your answer!!! It realy works! It was exactly what I was trying to do! Btw, kerkis is my favorite font! That was the reason I learned about LaTeX! And when I first used it, It was the only available greek font, apart from the standard babel!!!
    – Thanos
    Commented May 13, 2013 at 9:04
  • 1
    @Thanos: You're welcome!. Do you know the fonts of the Greek Font Society? Artemisia I use for personal correspondence, and Neohellenic I sometimes use for headings. They're all available for LaTeX!
    – Tom Bombadil
    Commented May 13, 2013 at 10:28
  • You must be joking! I do know the Greek Font Society! I once saw a document written in GFS neohellenic and I was really thrilled! Searching in GFS I facvoured two fonts: Neohellenic and Artemisia! I love them! But I have to use XeLaTeX in order to use them, which I don't! It is really weird to share the same taste! It's funny!!!
    – Thanos
    Commented May 13, 2013 at 13:04
  • @Thanos: Hehe, this is really nice ;) But you don't have to use XeLaTeX, it works with pfdLaTeX as well: Here's what you have to do for Artemisia and for Neohellenic. The LaTeX Font Catalogue is really nice for findling a font for any occasion!
    – Tom Bombadil
    Commented May 13, 2013 at 13:37
  • You are awesome! It works that way! Perhaps I'll ask it as a question(I was really intending to) so that future viewers will be able to see it! Thank's for this extra!
    – Thanos
    Commented May 14, 2013 at 6:12
64

Wasn't sure if "something like that" meant "as near to that as possible". Either way I went with:

\documentclass{standalone}
\usepackage{tikz}
\renewcommand{\familydefault}{\sfdefault}

\begin{document}
\colorlet{crystal}{blue!75}

\def\zangle{-20}
\def\xangle{20}

\begin{tikzpicture}[x=(\xangle:0.75cm), y=(90:1cm), z=(\zangle:1.5cm),
    >=stealth, line cap=round, line join=round,
    lines/.style={gray!50, thick}, 
    axis/.style={black, thick},
    plate/.style={fill, opacity=0.875},
    markers/.style={orange, thick}]

\node [yslant=tan(\zangle), above=0.25cm, align=center,font=\small] at 
    (1,1,1.5){Left Handed \\ Circularly Polarized Light};

\draw [lines] (-1,-1,0) -- (-1,1,0) -- (1,1,0) -- (1,-1, 0) -- cycle;
\draw [lines] (1,0,0) \foreach \t in {0,5,...,355}{
        -- (cos \t, sin \t, 0) } -- cycle;

\draw [lines] (1,1,0) -- (1,1,3.125);
\draw [lines] (-1,-1,0) -- (-1,-1,3.125);
\draw [axis, ->] (0,0,3.125) -- (0,0,0);

\foreach \k [evaluate={%
    \i=\k*5.625; 
    \j=\i>0 ? \i-5.625 : 0; 
    \a=90-\i; 
    \b=90-\j; 
    \c=int(mod(\k,4));}] 
    in {0,...,192}{
        \ifnum\c=0
            \draw [->] (0,0,\i/360) -- ++(cos \a, sin \a, 0);
        \fi
        \draw [red] (cos \a, sin \a, \i/360) -- (cos \b, sin \b, \j/360);
    }

\begin{scope}[shift={(0,0,3.125)}]

\node [yslant=tan(\zangle), above=0.25cm, align=center,font=\small] at 
    (1,1,1.5){Linearly Polarized Light};

\begin{scope}[xscale=1.5, yscale=1.5]
    \path [crystal!25, plate] 
        (-1,-1,0) -- (-1,1,0) -- (1,1,0) -- (1,-1,0) -- cycle;
    \path [crystal!50, plate] 
        (-1,-1,0) -- (-1,-1,-0.125) -- (-1,1,-0.125) -- (-1,1, 0) -- cycle;
    \path [crystal!75, plate] 
        (-1,1,0) -- (-1,1,-0.125) -- (1,1,-0.125) -- (1,1, 0) -- cycle;
    \node [yslant=tan(\xangle), text=crystal!50, below, font=\small] at 
        (-1.125,-1,0){Quarter Wave Plate};
\end{scope}

\draw [markers] (0,1) -- (0,-1) (-0.5,0) -- (0.5,0);
\draw [lines] (1,1,0) -- (1,1,3);
\draw [lines] (-1,-1,0) -- (-1,-1,3);

\draw [axis] (0,0,0) -- (0,0,3);

\foreach \k [evaluate={%
    \i=\k*5.625; \j=\i>0 ? \i-5.625 : 0; 
    \a=90-\i; 
    \b=90-\j; 
    \c=int(mod(\k,4)==0 && sin \a != 0); 
    \d=int(\k+1/4);}] in {0,...,192}{
    \ifodd\d
        \ifnum\c=1
            \draw [->] (0,0,\i/360) -- ++(sin \a, sin \a, 0);
        \fi
        \draw [red] (sin \a, sin \a, \i/360) -- (sin \b, sin \b, \j/360);
    \else
        \draw [red] (sin \a, sin \a, \i/360) -- (sin \b, sin \b, \j/360);
        \ifnum\c=1
            \draw [->] (0,0,\i/360) -- ++(sin \a, sin \a, 0);
        \fi
    \fi
}
\end{scope}

\begin{scope}[shift={(0,0,6.125)}]

\node [yslant=tan(\zangle), above=0.25cm, align=center,font=\small] at 
(1,1,1.5){Unpolarized Light};

\begin{scope}[xscale=1.5, yscale=1.5]
    \path [crystal!25, plate] 
        (-1,-1,0) -- (-1,1,0) -- (1,1,0) -- (1,-1, 0) -- cycle;
    \path [crystal!50, plate] 
        (-1,-1,0) -- (-1,-1,-0.0625) -- (-1,1,-0.0625) -- (-1,1, 0) -- 
        cycle;
    \path [crystal!75, plate] 
        (-1,1,0) -- (-1,1,-0.0625) -- (1,1,-0.0625) -- (1,1, 0) -- cycle;
    \node [yslant=tan(\xangle), text=crystal!50, below, font=\small] at 
        (-1,-1,0){Linear Polarizer};
\end{scope}

\draw [markers] (-1.25,-1.25) -- (1.25,1.25);

\draw [lines] (0,1.414,0) -- (0,1.414,2);
\draw [lines] (1.414,0,0) -- (1.414,0,3);
\draw [lines] (1,1,0) -- (1,1,1);
\draw [lines] (-1,-1,0) -- (-1,-1, 0.5);
\draw [axis] (0,0,0) -- (0,0,3);

\foreach \k [evaluate={%
    \i=\k*5.625; \j=\i>0 ? \i-5.625 : 0;
    \a=90-\i; 
    \b=90-\j; 
    \c=int((mod(\k,4)==0 && sin \a != 0) || (\k==65) || (\k==129)); 
    \d=int(\k+1/4);
    \r=(\k>64) ? 1.414 : 1;
    \xa=(\k > 64) && (\k < 129) ? 0 : sin(\a)*\r;
    \xb=(\k > 64) && (\k < 129) ? 0 : sin(\b)*\r;
    \ya=(\k < 129) ? sin(\a)*\r : 0;
    \yb=(\k < 129) ? sin(\b)*\r : 0;
    }] in {0,...,192}{
        \ifodd\d
            \ifnum\c=1
                \draw [->] (0,0,\i/360) -- ++(\xa, \ya, 0);
            \fi
            \draw [red] (\xa, \ya, \i/360) -- (\xb, \yb, 
            \j/360);
        \else
            \draw [red] (\xa, \ya, \i/360) -- (\xb, \yb, 
            \j/360);
            \ifnum\c=1
                \draw [->] (0,0,\i/360) -- ++(\xa, \ya, 0);
            \fi
        \fi
    }

\draw [ultra thick, ->] (0,0,3.5) -- (0,0,3);

\end{scope}

\end{tikzpicture}

\end{document}

enter image description here

Improve this answer
2
  • 17
    Wow!!! I didn't have exactly that on my mind but WOW!!! In case I haven't mentioned:WOW!!! You are awsome!
    – Thanos
    Commented May 13, 2013 at 13:07
  • As an addendum, if one wants to have various angles for the linearly polarized light, one can use (\cos(\angle)*sin \a, \sin(\angle)*sin \a, \i/360) in place of (sin \a, sin \a, \i/360) (and nearby positions). Took me far too long to figure this out. You may need to remove the 1.414 factor, or add it into that if you want to keep the sizes right.
    – IllustriousMagenta
    Commented Sep 19, 2021 at 4:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .