550 On the Double Refraction of Electric Waves. 



This value substituted in (4) gives 



v-v.-™ « 



Equation (6) is the relation between the velocity of the 

 waves in a given weakly- conducting dielectric and their 

 absorption by the dielectric. We are supposing here that the 

 medium is not crystalline in the sense of having different 

 dielectric constants along different axes, but that it is merely 

 heterogeneous in that it has different conductivities along- 

 different axes. In a medium of this kind, Y is constant 



for a given specimen, while the absorption-coefficient ^ 



z o 

 depends upon the direction of the electric force in the 

 specimen. In that orientation of electric force in which 

 absorption is greatest, V is smallest, and therefore index of 

 refraction is greatest. 



This agrees with the results of my experiments on woods. 

 With all the specimens tested, when the wood is so oriented 

 that its grain is parallel to the electric displacement, the 

 index of refraction is greater than with the grain perpendicular. 

 In the former orientation also the absorption is greater than 

 in the latter. 



Returning now to the experimental data, let Vj and q l be 

 the values found for V and q with the first orientation; V 2 and 

 q 2 the corresponding values with the second orientation. 

 When these values are substituted in (6) we have two new 

 equations which combined give 



Vi— Vq _ log 2 9i 

 V,-V log 2 ?/ 



Dividing numerator and denominator of the left-hand 

 member by v the velocity of the waves in air, we obtain 



1/ri! — 1/ wq _ log 2 <h / 7 ) 



l/n 2 -l/n log 2 q 2 ' 



The right-hand member can be calculated from the ab- 

 sorption data. Call it R 2 , then 



l/ % = R> 2 -(R 2 -l)//i .... (8) 



where ?i 1 = index of refraction in the first orientation, 



n 2 = index of refraction in the second orientation, 

 and 7i =n///,. 



