Page:Outlines of Physical Chemistry - 1899.djvu/102

 82 OUTLINES OF PHYSICAL CHEMISTRY

dark, whilst the half to the left still allows light to pass through. If, on the contrary, the analyser be so rotated that its plane of vibration is perpendicular to ob' (as indicated by cc f in fig. 16), then the left half of the field is dark and the half to the right illuminated.

To obtain equality of illumination over the whole field, it is necessary to have the plane of vibration of the analyser perpendicular to oa (fig. 17). In this position the index-pointer on the analyser must be at the zero mark on the graduated circle c (fig. 18). If now a tube containing an optically active liquid be interposed, the vibrations ob and ob' will be both rotated through the same angle and in the same direction, and the equality of the illumination of the field will be disturbed. This equality is restored by rotating the analyser through an angle equal to that through which the vibrations ob and ob' have been rotated by the active liquid. This angle as indicated by the pointer is read off on the graduated circle.

The angle of rotation is proportional to the thickness of the layer of active liquid through which the light passes.

The value ~, where a is the measured angle of rotation,

V

and I the length in decimetres of the column of liquid, is constant for a substance, provided that we always work at the same temperature and with the same light.

In order to compare the effects produced by different active substances, the deviation per unit length (one

decimetre) ^ is divided by the density. The expression

V

thus obtained -^ is termed the specific rotation, and is

usually represented by [a] D or [a\ according as the deter- mination is carried out with yellow sodium light or with white light (in some saccharimeters daylight is used and equality of tint, instead of equality of illumination, is established).

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