AN INTERRUPTED MEDIUM. 53] 



/ -?\ , l+r 2<r (a + b)p t (h r sin a + e i r cos a) 



or '* r 2 ) sm (p - + - * s 



r-<r <-p 



- -- ' r (Ir sin a ki r cos a) = 



r * 



(I) 



And by eliminating the second term, we get 



" fi'~ r"'\ tin (h 4- s !_>_ 



-^ (i 3 -r 2 ) sin + '^J^--* (At + r cos a + 



> 



er sin a) 



-f tan </> - (1 ) (*'+ cos a k r sin a) = . . . . . (II) 



r g 



By treating (7) and (8) in a similar manner, we obtain 



^ p cos 2 d) + * / p sin 2 d)} (a + b}er,, 



3 a * / _t _i / f If jt SlTl ft / 



(t p) COS (/> r ff 



+ tan </> (1 ) -5- (e r sin a + /* i+r cos a) = (Ill) 



--- (z 2 r 2 ) cos rf) + - " y " ' (kir cos a lr sin a 



v - v 



r tf 



+ tan(/> (1 ) -* (ei r cos a + hr sin a) = ........ (IV) 



TA.,, r. T /?> ON j. (1+*) r 2 f . (a + fr) p t (hr sin a + e ir cos a) 

 rrom 1. u 2 r 2 ) sm ffl ^ - - -- h - ^- - - - = 



re tQ 



II. (a p + b t~) (i 2 r 2 ) sin <p + (a + b) p t (h i + r cos a + e r sin a) = 



Now, our first approximation gave /? = ^ 



f TT .~r 



Let /? then be equal to s + Q where B is a small quantity : /3' = ^ + 4- 



* 



Hence cos (8= sin 6 , cos ft' sin 4/ 



sin /3 = cos 6 , sin |S' = cos 4- 



A= (sin 0-sin 4.) = - (64) + -g (6 s -4-') approximately 



e = cos a 



cos ^ = 



s-- 



*- -t^-% 

 2 



/ = (7 _ Y ) _ 1 ( 7 3_Y) 



We return now to our original equations, and determine the values of t, p, 

 . from them. 

 By (1) and (i!) we obtain 



t sin (/3 /3') = (i r cos a) sin </> sin /3 + r sin a sin < cos (3 

 VOL. XV. PART IV. 7 D 



