شباب عندي سؤال عن miller indices of hexagonal

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http://www.uwgb.edu/DutchS/symmetry/millerdx.htm

The proof is simple
Imagine a hexagonal axis system as shown and a face cutting each axis with intercepts a, b, and c.
Now construct CE parallel to x1 and CD parallel to x2. Obviously triangles AOB, ADC and CEB are similar.
Therefore
a/b = {a-c}/c = a/c -1.
Dividing by a and rearranging,
we have 1/a + 1/b = 1/c
Define h = 1/a, k = 1/b
and i = -1/c
Then h + i + k =0.
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Note: In the case of an hexagonal lattice, one uses four axes, [FONT=Times New Roman,Times New Roman]a[/FONT][FONT=Times New Roman,Times New Roman]1[/FONT], [FONT=Times New Roman,Times New Roman]a[/FONT][FONT=Times New Roman,Times New Roman]2[/FONT], [FONT=Times New Roman,Times New Roman]a[/FONT][FONT=Times New Roman,Times New Roman]3[/FONT], [FONT=Times New Roman,Times New Roman]c [/FONT]and four indices, ([FONT=Times New Roman,Times New Roman]hkil[/FONT]), called Bravais-Miller indices, where [FONT=Times New Roman,Times New Roman]h[/FONT], [FONT=Times New Roman,Times New Roman]k[/FONT], [FONT=Times New Roman,Times New Roman]i[/FONT], [FONT=Times New Roman,Times New Roman]l [/FONT]are again inversely proportional to the intercepts of a plane of the family with the four axes. The indices [FONT=Times New Roman,Times New Roman]h[/FONT], [FONT=Times New Roman,Times New Roman]k[/FONT], [FONT=Times New Roman,Times New Roman]i [/FONT]are cyclically permutable and are related by
h
+ k + i = 0
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