Miller Index Problem - Is it (102) or (112)?

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The Miller index for the tilted plane in question is definitively (1-12), which corresponds to the plane's surface normal vector [1-12]. This plane is part of the family of {112} planes, indicating that it shares characteristics with other planes in this group. The discussion clarifies that the y-intercept is at -1, which is derived from extending the vector to find the intersection with the y-axis. Understanding these concepts is essential for accurately describing planes in cubic systems.

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Can somebody tell me what is the miller index for the tilted plane?

Is it (102) or (112)?

Nothing seems to fit...

I wonder if we can even describe it with miller index?

Please download the picture here in pdf format:

http://www.megaupload.com/?d=F8J344BH

Thanks!
 
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or see the attached file if you can...
 

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Every plane has a Miller index. A common way to find the Miller index for a plane in a cubic system is to take the reciprocal of the axis intercepts and normalize the result so it contains only integers. Negative intercepts are treated by putting a bar over the number. For example, the y-intercept in your figure is at -1.
 
thanks for the reply. But could you explain why is the y intercept for the triangular plane -1? it does not seem to intersect with the y axis?
 
The plane continues on to infinity; if you follow the line in the y-z plane, you'll see that it (and therefore the plane) intersects the y-axis at -1. Use the same approach for the other axes.
 
um..so are you saying that I can extend the vector so it eventually intersects with the y-axis?

so the miller index should be (1-12)?
 
Yes; a (1\bar 1 2) plane (a member of the family of \{112\} planes), with surface normal vector [1\bar 1 2] (a member of the family of \langle 112\rangle directions).
 
But, for example, this picture is also (1,-1,2). so are they a family?
 

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More than another member of a family; that's the same plane.
 
  • #10
ok. I think I am getting it, thanks a lot.
 

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