386 Sir David Brewster on the Compensations of Polarized Light. 



table, the values of -|- and — in the case of reflexion, and of ± and 90" — 

 0' in the case of a reflexion, and a refraction, are nearly equal ; yet it requires to 

 be proved, that when the planes of polarization are inclined at an angle, ± x, to 

 the plane of incidence, greater or less than 45°, another reflexion at another 

 angle, which would give ± 0, or 90° — 0', of the same value, will restore the planes 

 to their original inclination. 



When X = 45°, and when one reflexion has turned the planes of a ray polar- 

 ized 45° into 37° 21', or given the planes a rotation of 45° — 37° 21' = 7° 39', 

 the action of a refracting surface which produces the same rotation, or 

 52° 39' — 45° = 7° 39' will bring the planes back to 45°, or restore the partially 

 polarized light to common light. Call x ■=. 37° 21', then in order that the 

 refraction may restore the ray to 45° we must have 0' = 45° or cot 0' = cot x 

 cos (^ — i') zz 1. Now, cot 0' = cot x cos {i — ^'), and when x = 45° and 



0' = 52° 39', cot 0' = cos (i — i'). But x = 37° 21' = 90° - 0, hence — "— = 



^ '^ cot .r 



cot0', and = cos {i — i'), consequently cot x cot {i — i') = 1. In like 



manner 0' will be restored to 45° by a reflexion which gives such, that + 0' 

 = 90°, or tan = cot 0'. That is when x = 45°, and = 37° 21', 



cos (i "4— i I 



tan = tan x 7-7- ^r ^ 1 . The general formula 



^ cos (^ — ^ ) ° 



cos (i -\-i') , 1 



tan = tan x ^. n^ becomes, when x zz 45 , 



cos (i — ^ ) 



cos (i + »■') 



tan = f-. ^, 



cos (?. — I ) 



But when x zz 52° 39' = 90° — 0, we have 

 1 



= tan 0, and 

 tan X 



1 cos (^■ + i') 



tan a; cos (i — i')' 



cos (i -\- i) , 

 tan X ■ — , . .; - = 1 . 

 cos (e — i) 



Consequently, 



