1 4 Dr. Brewster on the absorption of polarised light 
colours, and in some cases they are equal to nearly one half 
of the transmitted light. 
When the rhomboid is exposed to polarised light, a series 
of still more interesting phenomena is exhibited. In the 
position where O vanishes, E is an orange yellow , exactly 
the same as it appeared by common light ; and in the position 
where E vanishes, O is a yellowish white , as before. Now it 
is obvious, that in the first of these positions the image E 
was not strengthened by the white light of the vanished image 
O, otherwise the image E would have had the same colour 
as O -{- E, or the natural tint of the spar ; and that in the 
second position, the image O had not received the whole of 
the vanished image E, otherwise it would have had the tint 
expressed by O + E. It therefore necessarily follows, that a 
portion of the pencil Ohas been absorbed in the first position, 
and a portion of the pencil E in the second. The quantity of 
light absorbed is a maximum in the two positions where 
a is o° and 90°, and is equal to the quantities m and n, which 
the two images interchange. At different angles with the 
axis, therefore, it is measured by sin.* <p m, sin.* p n. When 
this angle is given, the absorbed light varies with the azi- 
muthal angle a, and may be found from the following for- 
mula, viz. T = O cos.* a -f E sin.*tf, which supposes that m 
and n are equal to E and O. Hence when a — o° T = O, or 
the whole of the pencil E is absorbed. When a = 45°,T = \ 
0 -|--jE, or one half of O and E is absorbed, and when 
a = go°, T = E, or the whole of the pencil Ois absorbed. When 
the absorbing crystal is viewed by a doubly refracting prism, 
the tints of the two pencils Pe andPo will be given by the formu- 
lae Pe = O cos. 3 a + E sin*, a , and Po = E cos. 3 a -j- O sin. 2 a . 
