Energy in the Electromagnetic Field. 29 



direction of the incident rays, we get 



^S =l + /c 2 -h*' 2 -2*cos-2r6> 2 

 c x c 



— 2*'cos(-2r0 2 + j\ ^- 2k k' cos~, 

 ~ S y = 26 | -« cos *2r<9 2 -*' cos (^2r0 2 -f j) 



+ 2kk'cos|+k 2 + «; /2 1, (29) 



where 2 is neglected in comparison with unity and a and 

 are considered to be the same in the small term S y . Since 

 k, /c ! are each small in the present problem, terms involving 

 their squares and products may be neglected. The expression 

 for tan 6 may then be written 



20 S - K cos- 2r0 2 -K' cos (-2r0 2 + r)\ 

 tan* = f* = 1 ? V? iZJL. (30) 



S * l_2*cos-2r0 2 -2/<: / cos(-2r0 2 4- £) 



c ' c 4/ 



But by equation (20), 



f( 2 ,-«0) + 2tf=(l-«f)tan*. 



Hence (30) becomes, after some reduction, 



- 1 -k cos- 2r0 2 - / c ! cos (~2r0 2 + 7 f) 

 dd c \c 4/ 



Tr=~r' ^— ZX' (31) 



Now 



r 1 - 2/c cos -2r6> 2 - 2^ cosf-2r6> 2 + £) 



/ccos-2^ 2 + «'cos( U 2r^-r^)=Dcos/-2r(9 2 + 6Y 

 if D 2 =(/c + /c7v / 2") 2 + /c' 2 /2 = /c 2 + ^' 2 + ^2/c^ 



and tan e= 



Then (31) may be written 



<w = _, l- Dcos^ +e) 



* "'•'l-2Dcos(;"2,^ + e j' 



