APPLICATION TO AUGITE, ETC. 377 



A careful drawing is generally a sufficient check on the calculation and 

 accurate as the measurements, and no numerical work is necessary but the 

 finding of S. 



Determination of optical Axes from a random Section. 



§ 8. It is interesting to note that from such a random section we can 

 theoretically determine the position of the optical axes, i. e., a and V. Let 

 e = < m SB, / = < m SA, g = < m SB^, /i = < m SA^ (CA = a -\- F, 

 CB = a — F), figure 2. Then, in a series of spherical right ti'iangles, we 

 have — 



tan e sin Sm = tan ( Cm — a — F), 

 tan f sin Sm = tan ( Cm — a -f- F), 

 tan g sin Sm = tan {Cm -f- « — F), 

 tan h siii Sm = tan ( Cm -\- a + V ) ; 



(1) 



and — 



sin e sin SB = sin {Cm — a — F), 

 sin f sin SA = sin ( Cm — a + F), 

 sin g sin SB^ = sin {Cm 4- a — F), 

 si)i h sin SA-^ = si)i {Cm + '>'- -f V). 



(2) 



Moreover — 



— ^ — = m A^A., = X.,, 



where x^ and .i-^, are the extinction angles and may be determined by observa- 

 tion. Lastly, if / — a is the retardation of one half and /' — «" that of 

 the other half of the augite twin, their ratio, which may be determined 

 by comparing polarization colors, is — 



/ — of sin SA sin SB 

 /-^/'^sinSA^inlSB^ . ^^-^ 



From equations (1) and (3) we can find tan 2x^ by clearing of fractions and 

 expressing tangents in terms of sines and cosines in the form — 



tan 2x^ = sin Sm X 



sin Cm — a -f- F cos Cm — a — F -f- sin Cm — a — F cos Cm — « -f- F 



sin^ Sm cos Cm — a -\- Vcos Cm — a — F — sin Cm — a -{- Vsi)i Cm — a — F 



(5) 



