1893.] Phenomena someiuhat analogous to Newton s Rings. 89 



And for the transmitted wave in the second part of the conductor 



2 = ac sin (fit fax + $2) .............. (5), 



where p/pi = v l and {3/j3 2 v z . 



Further, remembering that i = Q X (v) = C0 X (*0, we nave 

 from (4) and (5) 



i, = C l v l aam(ptp l x) C l v l db&iiL(pt + p l x-}-S l ) . .. (6), 

 and iz = C z v 2 ac sin (j3t p z x + &0 .................... (?) 



11. Now take B, the junction of the two parts of the conductor, as 

 the origin of abscissas. Then, for x = 0, we have 0i = 2 and ii = i z ; 

 unless 90/3^ and di/dx become infinite at that point. Applying this 

 relation to equations (4), (5), (6), and (7) yields 



sinj8J + &sin (pt + Sj = c sin (fit + S 2 ) ........ (8). 



dvi[Binpt b sin (pt + $i)~\ = C 2 v 2 c sin (pt + S z ) .... (9). 



12. Now equations (8) and (9) hold good for all values of i ; hence 

 in each of them we may equate the coefficients of sin pt and of cos fit 

 respectively after expanding the sines. This leads to four equations 

 from which to determine the four unknowns. The solution may be 

 written as follows : 



! = 0, . , = 0, 1 



, __ 



v/(C 2 /L 2 ) )>.... (10). 



~ 



13. The following results of (10) may be noted : 



(1) Energy of the original wave is proportional to a constant 

 X CiVi<z 2 , that of the reflected wave to a constant x CiVtcr^ 2 , and that 

 of the transmitted wave to a constant X C 2 v 2 a 2 c 2 . We ought, therefore, 

 to have 



d?! = dvJP + GjVd* .............. . (11). 



And this equation is satisfied by the values of b and c in (10). 



(2) If L 2 = L! we have 



= vi+z *ir+l 



and 



= 



v/C 2 



[Compare Preston's * Theory of Light,' 1890, pp. 285286.] 

 (3) If C 2 L 2 = CiLu then v z = i, A^ = X 1} and we have 



