Double Refraction of Quartz. 119 



the others will also terminate in points of the dark curve. This condition 

 can be reconciled with the spiral form of the curve, only by supposing that 

 the curve consists of four similar spirals, each of which is turned 90° from 

 the position of that adjacent to it. The general form of the curves will 

 be therefore four spirals, in positions differing by 90", and all turned the 

 same way, intersecting a series of circles. This very remarkable form is 

 precisely the form given by observation. 



4. The intersection of the spirals and circles is found by making both 

 the equations 



sin- — = and sin'- (— - x j = o. 

 to hold at the same time. This gives 



X = 0, or =7r, or = 2tt, &c. : 

 and consequently 



<f> = 0, or = - , or = 



7r, or = 



3 



7T 



That is, the intersections of the spirals and circles will all lie in the lines 

 through the center parallel and perpendicular to the plane of reflexion. 

 This is exactly true in the experiment. 



5. And since the successive circles require values of ~ successively in- 

 creasing by *■, and the successive points of the spirals on the same radius 

 vector also require values of — successively increasing by *, every circle 

 will be intersected by the spirals in the lines above mentioned, and there- 

 fore every circle will be intersected at every quadrant. This is verified by 

 experiment. 



6. If we consider the parts near the center and not in the circumference 

 of one of the circles, the angle <j>' corresponding to a dark point will be nearly 



