Double Refraction of Quartz. Ill 



2 nd . This multiplier however is small when k is nearly = 1. It is 

 also small when k is small, if sin 20 = 0, that is, when = 0, =90". 

 = 180°, =270°. Hence there will be no rings very near the center: and 

 at a distance from the center, they will be faint in the lines parallel and 

 perpendicular to the plane of reflexion. 



3 rd . The form of the dark rings will be determined by making 



|-A = 6: and that of the bright ones by making it = 1 . The 



cos 



A. 



first of these suppositions gives 



T0 V 7T 37T - TrH 7T Y 



lT + -2 = 2' 0r =T' &C - : and T = 2--2' &C - 

 Now x increases from to 90°, from 90° to 180", &c, while 20 increases 

 from to 90", from 90" to 180", &c. : consequently x never differs much 

 from 20: and therefore for the dark rings 



7T0 it 3ir , 



— — = — 0, or = — 0, &c. nearly. 



That is 9 increases continually as diminishes: and consequently in- 

 creases continually as diminishes. This shews that the curve is a spiral, 

 and that (reckoning from the central fold) it is turned in a negative di- 

 rection; the eye when fixed on a part above the center must turn to the 

 left to trace the curve as it recedes from the center, supposing the crystal 

 right-handed. If the crystal be left-handed, the sign of k must be changed : 

 this changes the sign of x> an( l the spiral is turned in the opposite di- 

 rection. This agrees perfectly with observation. 



4"'. If we take the radius vector in the direction opposite to that cor- 

 responding to any point of the spiral, that is, if we increase by 180°, 

 we find for the new values of 9 in the dark rings 



IT 



5ir 



- 0+180", — - + 180", — - + 180", &c. 



or - T, - 0> q - 0' ~gT _ °» &c - nearly. 



