X-RAYS AND ORIENTATION OF QUARTZ CRYSTALS 319 



For the atomic plane (21 -3) we find from equation (3.13) that 



N = l ] (3.18) 



\3/l.l/ 



Whence 5=4 + 9/1.21 = 11.438 (3.19) 



and, from equation (3.14) the unit vector normal to (2T-3) 



/.5914\ 



N = 1 I (3.20) 



\.8064/ 



We have now determined the orientation of the atomic plane with respect 

 to the orthogonal axes X, Y and Z. To determine its orientation with 

 respect to the plate edges necessitates the construction of a matrix which 

 expresses the components of the plate edges P1P2P3 (length, thickness, and 

 width respectively) in terms of the A', Y, Z axes. When the components 

 of the unit normal to the atomic plane in terms of the orthogonal axes X, 

 Y, Z are acted upon by this matrix they are converted to the components 

 of the unit normal to the atomic plane in terms of the plate edges Pi , P2 , 

 Pz . (For fuller discussion see Section 5 of "The Mathematics of the Physi- 

 cal Properties of Crystals" by Walter L. Bond, Bell System Technical 

 Journal, Volume XXII, No. 1, pp. 1-72.) Equation (3.8) gives such a 

 matrix. 



To continue with the NT cut as an example, the product of the matrix 

 given in equation (3.9) for the NT cut and the components of the unit 

 vector for the atomic plane (21-3) given in equation (3.20) gives us the 

 components of the unit normal to the atomic plane (2l-3) in terms of the 

 plate edges Pi , P2 , Pz as axes. Thus, 



/ .99027 -.13917\/.5914\ 



Np,p,P, = [-M279 -.10662 -.75852 )( J 



\- .76604 .08946 .63653/ \.8064/ 



/-. 11223 \ 

 = ( -.99180 )..(3.21) 

 \+. 060293/ 



That is, the components of the unit normal to the atomic plane in terms 

 of the plate edges Pi, P2, P3 are: 



Ni = -.11223 

 N2 = -.99180 

 iVs = +.060293 



