482 Royal Society : — 



a geometric series, 



r^ + Q^tf. (4) 



Similarly, if t be the intensity of the transmitted light, 



1-pV* () 



aud we easily find 



1— 99 



which is in general less than 1, but becomes equal to 1 in the limiting 

 case of perfect transparency, in which case g=l. 



The values of \i, i, and q in any case being supposed known, the 

 formulae (I), (2), (3), (4), (5) determine r and t, which may now 

 therefore be supposed known. The problem therefore is reduced to 

 the following : — There are m parallel plates of which each reflects 

 and transmits given fractions r, t of the light incident upon it : light 

 of intensity unity being incident on the system, it is required to find 

 the intensities of the reflected and refracted light. 



Let these be denoted by (p(jn), \p(m). Consider a system of m + n 

 plates, and imagine these grouped into two systems, of m and n plates 

 respectively. The incident light being represented by unity, the 

 light §{m) will be reflected from the first group, and \p(m) will be 

 transmitted. Of the latter the fraction \[>(n) will be transmitted by 

 the second group, and 0(^) reflected. Of the latter the fraction 

 \p(m) will be transmitted by the first group, and (p(m) reflected, and 

 so on. Hence we get for the light reflected by the whole system, 



(f)(m) + (\pmy<p(n) + (\pmy<p(m)(<pny-\- . . ., 



and for the light transmitted, 



which gives, by summing the two geometric series, 



*»+»)=*»)+ <*"**"> t .... (6) 



1 — <p(in)<l)(n) 



Mm-\-n)= Yy / v 7 (7) 



We get from (6) 



(p(m + n){l-~ <p(m)(j)(n)} = <l)(m) + (})(n){(\pm) 2 ~((pmy} ; 



and the first member of this equation being symmetrical with respect 

 to m and n, we get, by interchanging m and n and equating the results, 



which is therefore constant. Denoting this constant for convenience 



