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Miller index problem

  1. Sep 21, 2008 #1
    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!
     
  2. jcsd
  3. Sep 21, 2008 #2
    or see the attached file if you can...
     

    Attached Files:

  4. Sep 21, 2008 #3

    Mapes

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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.
     
  5. Sep 21, 2008 #4
    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?
     
  6. Sep 21, 2008 #5

    Mapes

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    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.
     
  7. Sep 21, 2008 #6
    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)?
     
  8. Sep 21, 2008 #7

    Mapes

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    Yes; a [itex](1\bar 1 2)[/itex] plane (a member of the family of [itex]\{112\}[/itex] planes), with surface normal vector [itex][1\bar 1 2][/itex] (a member of the family of [itex]\langle 112\rangle[/itex] directions).
     
  9. Sep 21, 2008 #8
    But, for example, this picture is also (1,-1,2). so are they a family?
     

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  10. Sep 21, 2008 #9

    Mapes

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    More than another member of a family; that's the same plane.
     
  11. Sep 21, 2008 #10
    ok. I think I am getting it, thanks a lot.
     
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