154 Prof. R. Birkeland : An Attempt to explain 



Introducing this value of ? ls we obtain the following equa- 

 tion, determining £ x : — 



t 1 \c ( 2 -v 2 )-2vlcos<i>t 1 -l 2 = <d. 

 Hence 



_ vl cos $ 4- yv 2 l 2 cos 2 $ + 1 2 {C4 2 — v 2 ) 



The other solution is negative and must be rejected. On- 

 the way back to S a time t 2 is consumed and the light has 

 to cover the distance S/ &' = l 2 , the mirror S being at S" 

 after the elapse of the time t x + t 2 . We find, the assumption 

 •as to the light's velocity being the same, 



c 2 t 2 = 1% = I 2 + v 2 t 2 — 2vt 2 l cos cf> ; 

 hence 



— vl cos (/> 4- >sjv 2 l 2 cos 2 <£ + l 2 (c 3 — v 2 ) 



f 9 = _ . 



The total time, consumed in going from S to Si and 

 back, is 



T _* , t _ _ 2/ A , r 2 cos 2 (ft. 



Introducing the value of c , we obtain 



^-, 2 = c 2 [l-J(l+sin^)]. 



With an accuracy of the second order with respect to - we 

 -obtain G 



1 -[l + i~(l+sm 2 6)~\ 



A V 2 COS 2 A V 2 „ , _, , v 2 9 , 



V 1+ c 2 -^ 2 = v 1 + ?" cos ^ =1 + *? cos ^ ; 



hence 



T = t 1 + t 2 =^(l + J) (1) 



