REFLECTED AT THE SURFACE OF A CRYSTAL. 55 



ostf . . : r / (5) 



sm <p sin <p sin 



By adding (1) and (4) we have 



I-R_Tcosfl TsiHfr n . 

 sin<~ sincf)' "' sin(p' ' ( ' > 



By combining (1, 4) and (2) we get 



2 I = T cos 6 - + -T sin <p' -- 



\cos 9 sm 9 / \cos 9 sm 9 



(6) 



- 



cos sin 



* : (7) 



-- 

 cos cp sin <p 



In the same way, by combining (3) and (5) we have 



..." - (8) 



cos 



.- ,' (9). 



sm <p cos <p / 



Now let the incident light be supposed to be polarized in a plane, making the 

 angle u with that of incidence, then I= psina, I'=pcosu, .-. 21' = 2Icotanw, 



1= _cos-sncot. + , 

 sm (f)' cos (J) ) \ cos (p sm <p'/ 



or 



sin (j)' cos <j> cos (p sin <j)' 



or 



which gives T' (cos & sin 0' cot a cos < </>') = T (cos cot u cos < - 0' + sin 0) . (10). 



Again, if the reflected light be polarized in a plane, making the angle a with 

 that of incidence R'=> cos a, R=p sin a, 



2R'=-2Rcota, 



and 



or (T cos 6 - T' sin &) cos (/> + 0' cot a = (T sin 6 + T' cos 0'), 



or T(-cos0cos</> + </>'cota + sin0)= T'{sin0'cos<p' +</>' cota + cos #} . . (11). 



