Production of Circularly Polarized Light. 525 



the critical angle from glass to air, and since we must have 

 from the geometry o£ the figure, the relation 



satisfied, we must have <p> 65° 51', taking <£/ = 41° 42' as the 

 critical angle. 



Consider the component of the incident vibration which is 

 polarized perpendicular to the plane of incidence. Let <j> be 

 the angle of incidence, 6 the absolute phase-difference pro- 

 duced on the component for this angle of incidence, 0' the 

 corresponding phase-difference for angle of incidence <p'. 



Then, with the usual notation, we have 



e — _ , 



cos <p + rjv 



and for the three reflexions indicated in fig. 3 : 



>(2e+e')t 



("cos <f> 



— 7}L~\ 2 COS ( ft'— 1)1 



+ 7n~' cos^'-hr/'* 



LCOS (f>-{-rji- 



For the component polarized parallel to the plane of 

 incidence, let 8 and 8' correspond to 6 and 0' respectively. 



correspond 

 Then as before 



e W+S')i — f ootj-^ "! 2 1 cotJ/-~7?'^ 

 *-cot<#> + 7?t-' cot^' + V 6 

 If A' be the relative retardation, then will 





cos 



A' = R ' 



* pincft- 

 Lsin (j> 



~l 2 



sin <f>'—7)'i 





+ 7?^ 



sin (ft'+V 6 



where 





COsh 7? = 



sin </> 



> 



cosh >/' 



sin(/)' 



= ■ , 





and 





*4 



7T 



+ 4- 





From 



equation (8) 









(8) 



A , -p ["sin . cosh 77 — t cos eft. sinh 77-| 



3 A =xt — — ; -, -. -. — r 



Lsin </> . cosh rj + i cos 9 . smh 77 J 



[" sin ^ . cosh vf - ^co s eft . sinh V H 



Lsin <j>' . cosh 77' +Tcos <// . sinh 77' J 



__ -p (sin </) . cosh 7? — t cos (ft . sinh ??) 4 



(sin 2 ^ . cosh 2 77 + cos'"' <j£> . sinh 2 77) 



(sin <j>' . c osh 7/ — t cos (/>' . sinh V) 2 



(sin 2 ft . cosh 2 77' -f cos 2 (/>' . sinh 2 rf) ' 



R is used to denote the real part of the function which follows it. 



