Finding 111 plane of Carbon Diamond

Question

All,

I looked at the example on the Mathematica Website with the Carbon Diamond Lattice. Example I am referring to

I am wondering is there a way to highlight an individual plane such as the 111 or 110 plane? I like the example I am just not sure of my orientation.

Cheers, BEn

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Code:

Manipulate[
 tet = PolyhedronData["Tetrahedron", "Faces"];
 tetv = PolyhedronData["Tetrahedron", "VertexCoordinates"];
 sp = {RGBColor[1, 1, 0.3], Sphere[{0, 0, 0}, sizes]};
 bar1 = {RGBColor[0.2, 1, 1], 
   Cylinder[{{0, 0, 0}, {0, 0, Sqrt[2/3] - 1/(2 Sqrt[6])}}, sizec]};
 bar2 = {RGBColor[0.3, 1, 1], 
   Cylinder[{{0, 0, 0}, {-(1/(2 Sqrt[3])), -(1/2), -(1/(2 Sqrt[6]))}},
     sizec]};
 bar3 = {RGBColor[0.4, 1, 1], 
   Cylinder[{{0, 0, 0}, {-(1/(2 Sqrt[3])), 1/2, -(1/(2 Sqrt[6]))}}, 
    sizec]};
 bar4 = {RGBColor[0.5, 1, 1], 
   Cylinder[{{0, 0, 0}, {1/Sqrt[3], 0, -(1/(2 Sqrt[6]))}}, sizec]};
 sp4 = Map[Translate[sp, #] &, tetv];
 base = {bar4, bar3, bar2, bar1, sp4, sp, If[tets, tet, {}]};
 dia = Rotate[
   Rotate[Map[Translate[base, #] &, tetv], -Cos[Sqrt[3]/3/2], {0, 1, 
     0}], Pi/4, {0, 0, 1}];

 cub = If[cubs, 
   Scale[PolyhedronData["Cube", "Edges"], 
    Sqrt[2] {1, 1, 1}, {0, 0, 0}], {}];
 diax = Table[Translate[{dia, cub}, {i Sqrt[2], 0, 0}], {i, n}];
 diay = Table[Translate[{dia, cub, diax}, {0, i Sqrt[2], 0}], {i, n}];
 diaz = Table[
   Translate[{dia, cub, diax, diay}, {0, 0, i Sqrt[2]}], {i, n}];

 Graphics3D[{dia, cub, diax, diay, diaz}, Boxed -> False, 
  SphericalRegion -> True, ImageSize -> 380],
 {{n, 0, "frequency"}, 0, 2, 1, RadioButton},
 {{sizec, 0.025, "cylinder size"}, 0, 0.5},
 {{sizes, 0.1, "sphere size"}, 0, 0.5},
 {{tets, False, "show tetrahedra"}, {True, False}},
 {{cubs, False, "show cubes"}, {True, False}}, 
 TrackedSymbols -> Manipulate]

Answer 1

A convenient resource for the Miller Indices can be found here. This ref provides sufficient information for us to draw the (111) and (110) planes. First, reproduce the graphic from the demonstration. I just made the necessary changes to make it run outside of a Manipulate and did not try to optimize it.

tet = PolyhedronData["Tetrahedron", "Faces"];
tetv = PolyhedronData["Tetrahedron", "VertexCoordinates"];
sp = {RGBColor[1, 1, 0.3], Sphere[{0, 0, 0}, 0.1]};
bar1 = {RGBColor[0.2, 1, 1], 
   Cylinder[{{0, 0, 0}, {0, 0, Sqrt[2/3] - 1/(2 Sqrt[6])}}, 0.025]};
bar2 = {RGBColor[0.3, 1, 1], 
   Cylinder[{{0, 0, 0}, {-(1/(2 Sqrt[3])), -(1/2), -(1/(2 Sqrt[6]))}},
     0.025]};
bar3 = {RGBColor[0.4, 1, 1], 
   Cylinder[{{0, 0, 0}, {-(1/(2 Sqrt[3])), 1/2, -(1/(2 Sqrt[6]))}}, 
    0.025]};
bar4 = {RGBColor[0.5, 1, 1], 
   Cylinder[{{0, 0, 0}, {1/Sqrt[3], 0, -(1/(2 Sqrt[6]))}}, 0.025]};
sp4 = Map[Translate[sp, #] &, tetv];
base = {bar4, bar3, bar2, bar1, sp4, sp};
dia = Rotate[
   Rotate[Map[Translate[base, #] &, tetv], -Cos[Sqrt[3]/3/2], {0, 1, 
     0}], Pi/4, {0, 0, 1}];


diax = Table[Translate[{dia}, {i Sqrt[2], 0, 0}], {i, 1}];
diay = Table[Translate[{dia, diax}, {0, i Sqrt[2], 0}], {i, 1}];
diaz = Table[Translate[{dia, diax, diay}, {0, 0, i Sqrt[2]}], {i, 1}];

Graphics3D[{dia, diax, diay, diaz}, Boxed -> False, 
 SphericalRegion -> True, ImageSize -> 380]

The (110) plane has the coordinates defined as shown below:

Which is easily replicated in Mathematica:

axis = Arrow /@ {{{0, 0, 0}, {0, 0, 1}}, {{0, 0, 0}, {0, 1, 0}}, {{0, 
 0, 0}, {1, 0, 0}}}
p110 = {Opacity[0.5], Black, 
  Polygon[{{2, 0, 0}, {0, 2, 0}, {0, 2, 2}, {2, 0, 2}}]}
p111 = {Opacity[0.5], Black, Polygon[{{2, 0, 0}, {0, 2, 0}, {0, 0, 2}}]}

I add the axis symbol just for reference. Combining these graphics with the original gives

Graphics3D[{dia, diax, diay, diaz, axis, p110}, Boxed -> False, 
 SphericalRegion -> True, ImageSize -> 380]

And for the (111)

You can tweak the values in p111 and p110 to suit your needs for the size of the plane.