336 Mr. It. Hargreaves on Radiation 



The exact relation between v and v" is 



(l-v 2 )/(l-r 2 ) = /. 2 (l-v" 2 )/(l-f> 2 ), 

 and v has the range to 1. The equations are then 

 R\+H.'\'=f J ,K"\", (H\-H'\ / >=H"W, > 

 the solution is 



and 



(45) 



A 2 =(H'X') 2 : ,(H\) 2 =(/zv-V') 2 : i^y + y") 2 



The solution in terms of nn' n" is of course readily found : 

 the object of using (45) is to present A 2 (the fraction 

 determining the partition of energy, cf. § 6 and end of § 11) 

 in simple shape. 



(ii.) Electric induction along a?, i. e. Y = Z = 0, a = 0. 

 The scheme is 



£ = ?iH, 7=-mH, X = H//a, 

 •V=0, /3'=H(»-r) X'-f(l-=) Y'=0 



For continuity we require, 



H" / rn"\ 

 H(l-m) + H'(l + rn') = 5L(l- ^-) 



H(Vnr-w)-H / (Vn' + w)=A*H' , (— -- s ) 



or with X, 



HX + HV = H"X", (H\mH'X>=H'V'.f" 



(46) 



V . (47) 



and .*. 



and 



HX : HV : H."\" = v + fiv" : v-pv" : 2v 



A 2 =(i/-/zv") 2 : {v + fiv'y. 



The quantity A 2 is invariant for the successive actions 

 which take place when a wave enters a dielectric plate with 

 parallel faces and is repeatedly reflected ; for perpendicular 

 incidence it is (jLt — 1) 2 /(/a + 1) 2 . 



(iii.) For the general case when u and v exist, the polari- 

 zations may be taken in conjunction. Case [i.) is altered to 



