352 MR. R. T. GLAZEBROOK ON PLANE WAVES IN A BIAXAL CRYSTAL. 
LN=LP—PN = LP —f=35° O' 19"-f.(9) 
LN+X=40° 22' 6"—<f> .(10) 
LN-X'=:31 0 3' 27"-f.(11) 
Substituting these values we can calculate 6 and O', and hence find the values of 
v x and v. 2 or their reciprocals p, and /x 2 . 
Tables Y. and VI. give the results of this substitution : on the right hand for the 
inner sheet, on the left hand for the outer. 
The centre columns give the values of L N + X, L N— X'. 
Then on either side, 6-\-9' for the inner sheet, and 0 — 6' for the outer. Then the 
theoretical values of p 2 p 2 , then the experimental values taken from Tables I. and II., 
and finally, in the outside column, the excess of theory above experiment. 
The next step is to discuss the theory for the plane P Q. 
With the same figure and notation as before, if this plane cut the plane A C in L' 
we have [Section IV. (3 and 4)] X / X[ being values of X X' in this case. 
Fig. 11. 
y-i COS OL /~\~r | \ \ „\ 
cos 6= - cos (L N+X) . (12) 
cos \ " v ' 
tan X = tan OL' cos AL'N 
/ 
cos O'L' /T . /x . . 
cos 0 = - 7 cos (L N—X ).(13) 
cos + v v ’ 
where 
tan X/= tan O'L' cos AL'N 
CO = CO'=9° 4' 5" [Section V. (5)] 
CL'=1° 15' 40" [Section III. (7)] 
OL'=10° 19'45" .(14) 
0'L'=7° 48' 25".(15) 
AL'N = x'=59° 13' 2" [Section III. (8)] 
whence 
