MR. R. T. GLAZEBROOK OR PLANE WAVES IN A BIAXAL CRYSTAL. 
365 
the experimental value for /% corresponding to the incident wave which gave rise to 
the value of we have just been discussing, we have 
/x, 2 = 1*68447 
Whence by means of the formula 
we find 
0-0'= 9° 
~V) cos {6—6') 
H'a 1 
15' 
The value of 6—6' given by theory is Table VI., line 21, 
6—6'= 9° 30' 
We have to find the change in the value of y or A L N which will produce this. 
In this variation 6-{-O' is constant 
... S6=-S6' 
but 
8(0-0')=-15' 
86=—7' 30" 
80'= 7' 30" 
From these results we can find the increment in y from the equations 
n cos OL 
cos 0= -cos 
cos X 
(LN-X) 
tan X— tan OL cos y 
cos 6= cos OL cos LN + sin OL sin LN cos y 
sin 686 
sin OL sin LN sin y 
Substituting values we are led to a value of S y, nearly equal to 1°. So that the new 
value of y=60° 20' [Section II. (17)]. 
This change in the value of y produces a corresponding change in the values of X, X'. 
The new values are for the prism P It 
X = 5° 12' 20".(1) 
X'=3° 50' 0".(2) 
