ELECTRIC WAVES ROUND A PERFECTLY REFLECTING OBSTACLE. 137 



therefore 



It = 2''>3-'"7T-V<'[{II (-)}'+6/tn (-1) II (-) cos ^r + Q^ll (-)}"] cos 2 TT. 



Now, 



n(-fH2-V-*n(-i)n(-f), 



whence 



{n (-$)} = 2 1 /3- 1 V / 'ii(-;|). 



similarly, 



{n (-)}' = 2*3- V'n(-i); 



hence 



R = 3-'V-V/'[n (-) + 2'/>II (-1) . 3 F + 2'/>n (-i) (S/i)*] . . . (iv.). 

 Again, 



therefore < = ^Trj3, where /8 is given by 



whence 



= 3/xc(l -I/me) sin fa-, 



where 



and 



<^ = ^TT- 3/nc(l-f/Lic) sin I-TT ....... (v.). 



The value of v, where 



is also required; it is, however, more convenient to calculate < + x- From the 

 relations 



dll du dv du dv 

 -- = J w- + v-- r- u-- 



it follows that 



tan x ~ 

 Now 



. 

 uvuv 



therefore 



, u 



tan (f> = - , 

 v 



v' v ,, , . 



- - - -4 - i- , that is tan -- 



w tan > tan 1 



to obtain <^> + x to the second power of /x, the third power of jj. must be retained in 

 u and v, hence 



tan 



n (-A) sin ^TT cos iir + |n () (3p.) 2 sin frr cos ITT' 



h These expressions for R and < are to the order given, the same as those given by LORENZ. 

 VOL. CCX. A. T 



