244 Lord Rayleigh on the Reflexion of Light 



relation between force and displacement, 



P, Q, E=4^V 2 (/, g, h), .... (9) 

 and in the lower, 



P, Q, E=4ttV 1 2 (/, ff , A), .... (10) 



V, Vi being the two wave-velocities, whose ratio gives the 

 refractive index. Since h = 0, R = Q ; and since R = 0, 

 dF'dz = 0, it follows by (6) that 6 = 0. The only conditions 

 (A) requiring further consideration are thus the continuity 

 of/, Q or V 2 g, and c. 



As the expression for the incident wave we take 



y = g e i(px+qy+st)^ g r= _p e i(px+qy+st) _ ^ ^ M]\ 



the ratio of the coefficients being determined by the con- 

 sideration that the directions/, g, h and p, q, r are perpen- 

 dicular *. In like manner for the reflected wave we have 



and for -the refracted wave 



f=q0 ie i{ P^^ +st \ g=—p l 6 1 e i( Pi x+ w +st \ . (13) 



The coefficient of y is the same for all the waves, since their 

 traces on the plane x^O must move together. The multi- 

 pliers #', 1 determine the amplitudes of the reflected and 

 refracted waves, and may be regarded as the quantities whose 

 expression is sought. The velocity of propagation in the 

 first medium is s/^/ (p 2 + q 2 ), so that 



V 2 (/+<? 2 )=Vi 2 0i 2 + ? 2 ) (14) 



We have now to consider the boundary conditions. The 

 continuity of/ when # = 0, requires that 



l + e' = 6,; (15) 



and the continuity of Y*g requires that 



V 2 p(l-6')=Y 1 %6 1 (16) 



These two equations suffice for the determination of 6', X ; 

 and we may infer that the third boundary condition is super- 

 fluous. It is easily proved to be so ; for in the upper 

 medium, 



dc = dP_dQ = Y2 (df_dg\ 



dt dy dx X. dy dx J 



= V 2 (l + ff) (p 2 + q 2 ) ««*+■? 

 * In the present case r=0. 



