﻿in a Rectangular Bar by means of Polarized Light. 13 



a horizontal straight line drawn at a distance V below it. 

 Let the radius vector CP ( = r r ) cut DEQ at Q, and let 

 PQ=r 1 ', where the segment is to be taken with proper sign, 

 the positive sense being from C to P. Then 



and the equation to the isoclinic line takes the form 



sin 2((// — i) — — r'r/qsm 2i. . . . (13) 



Now (Phil. Trans. A. vol. cci. p. 96) H =--2417, 

 H!=— -0598, so that as M increases from -co to +oo, 

 q and qb' begin by being negative (i. e. q<0, b' >0) : q 

 changes sign first, and we have q>0, b'<0; then qb ' changes 

 sign and we have q>0, b f >0. In the first and third cases 

 the figure is as in fig. 3, in the second case as in fig. 4. 



Fig. 4.— 6'<0. 



^. 



£ 







2> 





^ 





-jr— If. — 



H 



^ 



£\ 



\? 



F 



& 







e 1 





to trace the form of the loci 



given by 



We have now 

 equation (13). 



First we notice that since r' + r x ' = b' sec </> / = CQ, a second 

 point P' of the locus is found by interchanging r' and ?•/, so 

 that the two points P, P' on CQ divide the segment CQ 

 symmetrically. Hence if r f r x ' > P,P' are between C and Q ; 

 if r'r{ < P, P' are outside CQ; and this holds in all cases. 



From equation (13) we have, since i is an acute angle, 

 sin 2i>0 ; therefore if q>0, P and P' are between and Q 

 if sin 2((fi' — i) is negative, that is if CQ fall inside the quail- 

 rant DCE (fig. 3) bounded by <// — /, cj)' — i— x : outside this 



