32 PHYSIOLOGICAL CHEMISTRY 



the apparatus. If the tube / contains only water, the polarized 

 ray passes through it without altering the direction of its 

 vibration. On arriving at the second Nicol prism g the ray 

 will pass through unaltered provided the prism is in a posi- 

 tion corresponding to that of the first prism. If, on the other 

 hand, the prism g has been rotated to the right or the left so 

 that its position no longer corresponds to that of d, only a por- 

 tion of the ray will pass through, and the intensity of the ray 

 reaching the eye at i will be diminished. If the prism g is 

 rotated 90 from the position corresponding to that of d, no 

 light will pass through. Beyond this point the illumination in- 

 creases until at 180 it again is at its maximum. At 270 the 

 field again is dark. 



Imagine the apparatus set with the two Nicol prisms in cor- 

 responding positions. Light will pass through to the eye. Now 

 if a tube of sugar solution is placed at /, the ray of light 

 passing from d will be rotated about its axis by the sugar solu- 

 tion, which is optically active. The result will be that it strikes 

 g in a position in which it will not pass through without losing 

 some of its intensity. If we rotate prism g through the same 

 angle through which the ray of light has been rotated by the 

 sugar solution, the ray again will pass through at maximum 

 intensity. By observing the angle through which the prism 

 must be rotated to bring this about, the angle of rotation of 

 the ray caused by the sugar solution is determined. 



In practice it would be difficult to observe a changing illu- 

 minated field and select the point at which the illumination was 

 at its maximum. A mechanism has been devised to obviate this 

 difficulty. This is represented in the diagram by e which is 

 a thin quartz plate covering one-half the visible field. This 

 plate so alters the light passing through it that when the half 

 of the field not covered by the plate is at its maximum illu- 

 mination, the half of the field covered by the plate is somewhat 

 darker, and vice versa. By rotating the prism g a point can 

 be found intermediate between the two, at which the two 

 halves of the field are equally light. This is taken then as the 



