Wave-trains through a Conducting Dielectric. 331 



The remaining portion of I r is 



Y 2 dt = AW^'sin 8 [p 2 t-ty)dt 



J o 



A 2 & 2 

 = -j-{secx-cos(%-2^r)}. 



Dividing ont by I , 



g' = y[seo X -oo8( X -2^)] ( ' 



I sec%-cosx ^ 



This last expression is independent of the thickness of the slab 

 of electrolyte. 



The total expression for ~ is (65) + (69) + (70), or rewriting 



these equations, 



Ir- c 2 / 2 .t» 2 g 2 f (l-g 4 )sec % 



I ~" 6 2 (sec % -cos % ) 1(1-2? cos2S+f 4 )(l-a> 2 P) 



C(l-f)[cos(<£-2>Jr)-ro 2 f cos((H2f)] 1 ^ 



1 - ag 8 sin (28 - 4jr) [sin (ft - 2#) - 6) 2 g 2 sin (ft jjjj] f I 

 (l-2a> 2 g 2 cos4^ + w 4 f 4 )(l-2f 2 cos (28-4^) +f*) j 



c/. f 2 f sec % [cos (yjr + a — 28) — f 2 cos (yfr + a)] 



{ 



cos ( A + 28 - ty) - P cos A 



sec% — cos%|. 1 — 2£ 2 cos2S + f 4 



l-2f cos (28-4^)+ £ 



:4 



} 



?> 2 [secx-cos(%-2^)] ^ 



sec X — cos x * 



As a matter of arithmetic this expression is fairly quickly 



calculable if =A has been already worked out ; a large number 



of the constants, &c, are the same in the two cases. The 

 curve (2) (fig. 3) has been calculated from the same data as 



were given on p. 326 ; the two curves j- and —■ may be thus 



compared. In the former the maxima were slightly shifted 

 back, in the latter the minima are slightly shifted forward. 

 The proportion of energy absorbed by the plate is 



1* + !^ 



/ JU + V V 

 I Io )' 



The curve thus deduced from -^and / is shown in fig 4, 



In i-n 



L x 



