Miller index problem

1. Sep 21, 2008

Carolyn

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?

Thanks!

2. Sep 21, 2008

Carolyn

or see the attached file if you can...

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• miller index.pdf
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3. Sep 21, 2008

Mapes

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.

4. Sep 21, 2008

Carolyn

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?

5. Sep 21, 2008

Mapes

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.

6. Sep 21, 2008

Carolyn

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)?

7. Sep 21, 2008

Mapes

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).

8. Sep 21, 2008

Carolyn

But, for example, this picture is also (1,-1,2). so are they a family?

Attached Files:

• miller.pdf
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9. Sep 21, 2008

Mapes

More than another member of a family; that's the same plane.

10. Sep 21, 2008

Carolyn

ok. I think I am getting it, thanks a lot.