(31). 



! 



(33). 



422 Sir William Thomson on the 



dimensional solution of (14), to take 



dy* dx 



Thus we have 



u = d± + d± v= _dj> + p? . . . (32) 



dy dx dx dy 



which, by (6) and (7), give 



We may now, to represent the two refracted waves, assume, 

 in the lower medium, 



<l> = sm((0t + l / x + my) ; , yjr = C / sm((ot + X / x + my) . (34); 



and to represent the incident wave, supposed distortional, and 

 the two reflected waves, in the upper medium 



</> = F sin (at + Ix + my) + Gsin {(ot — lx + my) ; 



-^r = C sin(&>£— \x + my) . (35), 



where I, l n and m, still given by (22) and (24), verify (15), 

 which give 



W = B(2 2 + m 2 ) ; and $,»■ = B y (1/ + m 2 ) . (36) ; 



and similarly, X, X y , according to (18), are given by 



?a> 2 = A(X 2 + m 2 ); ? y o> 2 =A / (V + ™ 2 ) . . (37). 



Also, if by j and j t we denote the angles of reflexion and 

 refraction of the condensational-rarefactional waves, we have, 

 similarly to (25), (26), 



sin j /ot = smj i la l =smi /(3; \ = mcotj ; X l =mcotj l . (38). 



14. The continuity of w, v, P, U, on the two sides of the 

 interface, gives, by (32), (33), (34), (35); 



tw(F + G)-XC= ro+XjC, (39); 



-Z(F-G) + mC=-Z / + mC / (40); 



-BZm(F-G) + [Bm 2 -JA(X 2 + m 2 )]C 



= -B/ y m+[B i m 2 -iA / (\ / 2 + m 2 )]C / . (41); 



B{ (Z 3 -m 2 )(F + G) + 2XmC} = B y (Z, 2 - m 2 - 2X,roC,) (42) . 



