518 Mr. E. Edser on the Reflexion of Electromagnetic 



In (13) and (14) write vt for a, and substitute the resulting 

 values of E x and E 2 in (15). Then 



a(c-\-v cos 0) cosm {(c+v cos 6)t+y sin 0\ 

 + a f (c — v cos 0') cosm' {(v cos ] — c)t + y sin 0'} = O. 

 This equation is satisfied i£ 



a(c + v cos 0) + a'(c— v cos 6 f ) = Q (17) 



m{(c+ v cos 0)t+y sin 0} = m'{(c — v cos 0')t — y sin 0'} = 0.(18) 



(18) must be satisfied for all values of t and y; and as these 

 variables are independent, we must have 



w(c-fv cos 6)-=m'(c — v cos 6') . . . (19) 



msin0 =— m'sinfl'. .... (20) 



From (19) and (20) - 



C + VCOS0 C — VCOS0' 



sin 6 sin 0' 



(21) 



a result that might have been obtained directly from (15) 

 and (16) . Equation (21) determines 0' implicitly in [terms 

 of 0. Then from (17) 



. C + VCOS0 , nex . 



a!=-a W/3 k ...... ; (22) 



c—vcos0 / v ' 



and from (19) and (20), 



X> = X.2=^ = -x»?*. . . . (23) 



c + vcosv sm u v ' 



8. To determine the explicit relation between! 0' and 0, 

 rewrite (21) in the form 



c(sin f + sin 0) + v sin (0 ! — 0) = 0. 



.'. 2c sin .cos — — +2vsin — ^~ .cos — — =0. 



. 0' + . e'-0 



/. csm — ^ f-vsin — <p =0. . . . (24) 



Writing the last equation in the form 



. 0' + , 0' ,' 

 sin — - — tan tt + tan - 



2 2 2 _ v 



. 0'-0-\ 0' x 0~~~c' 

 sin — - — tan ~ tan - 



we find that 0' c — v^ /ac ' N 



tan — = — tan — .... (zo) 



2 c+v 2 v y 



