388 Sir David Brewster on the Compensations of Polarized Light. 



tan = ^^iiyi^ = 42» 31'. 

 cos (l — I ) 



But in the refracted light thus compensated, we must have 0' = 90° — 42° 31 

 = 47° 29', and, therefore, we must determine the angle of incidence i, at which 

 the original light suffered refraction. The expressions from which we obtain i 



are cot 0' = cos (J, — i'), and sin i' = , which give 



. . m / 



sm t = - — -J J 

 tan f/> V 



•^aJ 



tan '0' 

 sm 



tan V ^2 _^ 1 _ 



2 m 



from which we obtain, when 0' ^ 47" 29', i = 56° 45', the maximum polarizing 

 angle. 



Hitherto we have supposed the compensation to be produced by one refrac- 

 tion, or by one reflexion ; but it may be effected by several. In the case of re- 

 flected light this is not necessary, because we have all degrees of polarization by 

 reflexion, from 0° of incidence to the polarizing angle, and from this again to 90° 

 of incidence. 



When the compensation, however, is made by successive reflexions at the 

 same angle of incidence, or when light which is compensated has been so reflected, 

 we may find the angle of incidence ^, when n is the number of reflexions, by 

 means of the formulfe 



^ , cos"(^■ + ^■') ■. ., sin z , «.- — - — co?, {i-\-i') 



tan = „;. .,; , sm i = , and v tan = ~ :7^,* 



^ cos"(z — z') m ^ cos{t — t') 



which give 



when i -}- i' is less than the polarizing angle, and tan positive. But when i -f- i 

 is greater than 90°, and tan negative, we have 



sini- / K+l)(l+^tan0)^ ^ / ( 2m n^ 4^ta«0^ 



* See Phil. Trans. 1830, p. 80, 81. 



