360 
MR. R. T. GLAZEBROOK ON PLANE WAVES IN A BIAXAL CRYSTAL. 
Taking the same figure as previously. 
Let O' K be an arc perpendicular to A C. 
Let 0 / 0/ be the new positions of O O'. 
Then 
0 0 = 0'0'=Se 
Let O, m, O' n be perpendicular on N O, N 0/ respectively. 
0'L = 7° approximately 
O'LP = 60° approximately 
. ’. LK =14° approximately 
Let N fall between P and K 
86= — Om= — Se cos NOL 
89'=0 'n=8e cos NO 'A 
/ ✓ 
S(0+6') = -8e (cos NOL- cos NO/A) 
Thus 8(9-\-O') is negative and decreases numerically as we approach P ; for NOL, 
N 0/ A become more nearly equal. 
Thus between K and P 9-\-9' decreases with e. 
Between L and K both 0 and 0' decrease, therefore 9-{-O' decreases d fortiori; but 
8(9-{-9') is always numerically less than 2Se. 
Let us see how these changes affect Hv 
We have [Section IV. (1)] 
2 
—+ —— 
2 1 2 
fJ'a He 
If a c is decreased, is increased, 
He" Ha He" 
Ha. 
is increased, 0-\-9 is decreased. 
Therefore cos (0-\-0') is increased so long as 9-\-9' is less than 90°. 
