We can determine the indices of the plane (hkl) by using the following relations:
h ¼
v 1 w 1
v 2 w 2
, k ¼
w 1 u 1
w 2 u 2
, l ¼
u 1 v 1
u 2 v 2
For the first set of directions: [131] and 0 11
½
Š, we have
h ¼
3 1
À1 1
= 4 , k ¼
1 1
1 0
¼ À1; l ¼
1 3
0 À1
= À1
⟹ (hkl) ð4 1 1Þ.
For the second set of directions: [102] and 111
½
Š, we have
h ¼
0 2
À1 1
= À 2; k ¼
2 1
1 À1
= -3, l ¼
1
0
À1 À1
¼ 1
⟹ (hkl) ð 2 31Þ.
Example 5 Determine the Miller indices (hkl) of a plane marked P in Fig. 4.22, which
is sandwiched between two sets of planes (i) 1 1 1
ð
Þ and (111), (ii) 1 11
ð
Þ and 11 1
ð
Þ.
Solution: Given: Two sets of planes: (i) 1 1 1
ð
Þ and (111), (ii) 1 11
ð
Þ and 11 1
ð
Þ.
(hkl) = ?
According to the addition rule, we can obtain:
For the first set of planes: 1 1 1
ð
Þ and (111), we have
ðhklÞ ¼ ð1 1 1Þ þ ð111Þ ¼ ð200Þwith m ¼ 1 and n ¼ 1
Similarly, for the second set of planes: 1 11
ð
Þ and 11 1
ð
Þ, we have
ðhklÞ ¼ ð1 11Þ þ ð11 1Þ ¼ ð200Þwith m ¼ 1 and n ¼ 1
Therefore, the required (hkl) plane is (200) 2(100).
Fig. 4.22 Determination of
required plane in the
polyhedron
168
4 Unit Cell Representations of Miller Indices