where 

 and 



X-RAYS AND ORIENTATION OF QUARTZ CRYSTALS 



Ni 



323 



tan 84 = 



sni d" = 



N2 

 sin 6 



cos sin"^ N3 

 6[ = 8', 



(3.36) 



(3.290 

 (3.37) 



F'or example, suppose an NT finished crystal is to be checked by the 

 (21-3) atomic plane as suggested in section 3.7B. The components of the 

 unit normal to the atomic plane (2l-3) in terms of the plate edges Pi , P2 , 

 Pz of an XT plate were found (at the end of section 3.8) to be 



Nx = -.11223 

 7V2 = -.99180 

 N, = .060293 



BEVEL 

 Fig. 3.22 — Position 4 for a plate of general orientation 



Substituting in equation (3.26): 



tan 5i = 



.060293 



-.99180 



8[ = -3°28.7' 



and in equation (3.26') 



sin 32°3' 



cos sm" 



.11223' 



r = 32°16' 



and in equation (3.29) 



5^ = tan-' 21,11223 ^ 

 ' - .99180 



and in equation (3.29') 



sm c/ = 



sin 32°3' 



cos sin-i .060293 ' 



r = 32°7' 



