348 MR. R. T. GLAZEBROOK ON PLANE WAVES IN A BIAXAL CRYSTAL. 
Now we know that if O x O' are the points where the optic axes cut the unit sphere 
K, the point where any wave normal meets the sphere if 
KO =0 
K0'=6' 
2v*=— s +~ — (— i - — — 2 )cos {6-0') 
Ha He \Ha He ' 
~i(l+ cos { e ~ e ')) 
r*c 
— \ —^(1— COS {6—6')) 
H Ha 
i je-e'\ i i 
— COS" —- =—-2 
He \ 4 / /X“ Ha 
(4) 
If then we can determine 6—6' since /x is given by experiment, this formula deter¬ 
mines the value of /x c . 
Now along the plane X Y, 6—6' is very small, and a considerable change in the 
positions of 0 O' will produce but small variations in the value of 6 — 6' and .'.of He- 
Let us, therefore, take the value of the angle between the optic axes given by the 
values already determined for Ha and Hb, viz. : 
Ha— 1 ‘68560 
//./,= 1‘68115 
and the approximate value of /x,. 
He= 1-53013 
If 2 <f>' is the angle between the optic axes 
whence 
<£' = 9° 4' 5" 
AO —AO'=80° 55' 55" 
AL= 2° 59' 10" 
OL = 83° 55' 5" 
0'L=77° 56' 45" 
X=ALK = 88° 13' 
XL=1° 43' 10" 
Hence 
