260 MR. J. C. MAXWELL GAKNETT 



Incident light : 



X = exp {tp (t-i/z/c)}, Y = 0, Z = 0, 



a = 0, /3 = v' exp {ip (tv'z/c)}, y = 0. 



Reflected light : 



X = B exp {ip (t+v'z/c)}, Y = 0, Z = 0, 



a = 0, /3 = z/B exp {ip (t + v'z/c)}, y = 0. 



Light inside absorbing medium : 



X = C exp {ip (t-Wz/c)}, Y = 0, Z = 0, 



= 0, /3 = N'G exp {ip (t-Wz/c)}, y = 0. 



Making X and /3 continuous at Z = 0, we have 



C=1+B, N'C = /(1-B). 

 Hence 



B = 



Taking the square of the modulus, we have, for the value II of the ratio of the 

 intensity of the reflected light to that of the incident light, 



R = (B)* = ="! ........ (12). 



' '' 2IJ 



If, now, the absorbing medium consist of minute spheres of metal embedded in a 

 transparent medium of refractive index v, we have equations (10), namely, 



n' K ' = Sfjiv/3, n'= y(l+f/ta) ....... (10). 



Substituting these values of nV and of n' in (12) we obtain 



...l - - (13), 



in which powers of /t higher than the second have been neglected. 



Suppose first that v' = 1, so that we consider the reflection at the front face of the 

 stained glass. Omitting powers of p. except the lowest which occur, we then have 

 from (13) 



It appears from equation (14) that light is reflected from the stained glass almost 

 as if the stain did not exist, the effect of the stain being slightly to increase the 

 reflection of those colours (in the blue) for which, according to Table II., a is greatest. 



