CHAP. I.] THE PROTEIDS. 11 



the crystal, which is cut of such a thickness that one ray is retarded in 

 its passage just half a wave-length of sodium light behind the other, or 

 what amounts to the same, executes half a vibration more than the other 

 while in the crystal. On emergence then, while one vibration is from 

 toy the other instead of being from to x is from to x' in the opposite 

 direction, and the two now unite to form a resultant vibration OB' equal 

 to OB but at an angle AOB' equal to AOB on the other side of OA. 



Now if the tube T (Fig. 4) only contain water or some non-rotating 

 liquid, the two rays will pass through it with their directions of vibration 

 OB, OB' unaltered to the analysing NicoFs prism N. This will only allow 

 rays to pass through it which vibrate parallel to a particular direction. If 

 the prism be turned so that this direction SP (Fig. 5, (2)) is perpendicular 

 to OB, the right-hand ray having no component parallel to SP is 

 extinguished, while the left-hand ray will have a more or less considerable 

 component in that direction and the left-hand side of the diaphragm D will 

 alone be visible in the telescope OH (Fig. 4). 



So if the prism be turned round till SP is perpendicular to OB' 

 (as in Fig. 5, (3)) only the right-hand side of the diaphragm is visible. 



But if SP be turned so as to be perpendicular to OA, vibrations 

 parallel to OB, OB' have equal components parallel to SP, and the two 



\ *'k-S 



(1) (2) (3) (4) 



Fig. 5. 



halves of the diaphragm appear equally illuminated (as in Fig. 5, (4)). In 

 this position of the analyser the instrument should read 0. 



It is possible to adjust SP perpendicular to OA with very great accuracy, 

 for when OB, OJ? make small angles with OA a very small rotation of the 

 analyser makes a great difference in the relative illumination of the two 

 halves of the field 1 . 



When $P is thus adjusted perpendicular to OA and the instrument reads 

 0, let a liquid possessing the power of rotating the plane of polarization be 



1 This will be seen at once from the mathematical expression for the intensity of 

 the component parallel to SP. 



Let AOB Fig. 5 (2) =a, BOP 90-0, where = a in the position of equality of 

 illumination. Let the intensity of the resolved part of the ray OB parallel to SP = I. 



Then 1= OB 2 cos 2 BOP = OB* sin 2 0, and ~ = 20 & sin cos 0. 



da 



Therefore i ~ = 2 cot 0. 

 I dO 



This, which expresses the proportion between the change of intensity and the original 

 intensity, is greatest when is least, and therefore o should be as small as possible. 



