qftresnaVs Optical Theory of Crystals. 



81 



and of course if we call the angle between the two prime normals 

 2E 



sm 



^ 



i„(E + 'L+^.)sin(E-'l±A') 



smi/ 



Cor. (1.) When i, or </^ = tan . P O I assumes the form — 



which may be interpreted analogously to the method used in the 

 reverse problem, but may be more elegantly illustrated by 



Cor. (2.) Which is that the meridian plane P O T (i. e. the 

 plane in which both normal and radius bisects lie) the angle formed 

 by R O P, Q O P, and therefore that formed by the planes drawn 

 through the normal and the two prime normals to which these two 

 are perpendicular. 



Now we have found (Cor. 4, page 23) that it also bisects the an- 

 gle formed by the two planes passing through the radius and the two 

 prime radii. Hence when the ray is given, we may find by the 

 easiest geometry the normal and the tangent plane, and vice versa. 



Thus suppose (N, N') (R, R') 

 to be the projections of the 

 prime perpendiculars and prime 

 radii on a sphere concentric with 

 the wave surface. 

 R' Let n be the projection of 

 any given perpendicular on the 

 same sphere, join w N, w N', 

 bisect N w N' by w M, which will be the meridian plane. 



Draw from R', R' T V perpendicular to n M and make R' T 

 = TV produce R V to meet Mw in r, then RrM = R'r M, and 

 therefore r is the projection of the radius. Just in the same way 

 when r is given we may find n. 



Now suppose n to come to N, then the position of the meridian 

 plane n M becomes indeterminate, and r from a point becomes a 

 locus, subject to the condition that R' r N = R rw 



from r draw r D perpendicular to r N. 



Then it is clear that because r N bisects R w R' 



sin. R D sin. R r sin. R N 



sin. R' D sin. R' r sin. R' N 

 and .*. D is a fixed point and N D a fixed length, and 



cos. r N D = tan. r N • cot. N D 

 PhiL Mag, S. 3. Vol. 12. No. 71. SuppL Jan. 1838. 



M 



