THE AID OF THE ACHROMATIC FRINGES. 91 



In figure gi let u be the angular velocity of the cylinder and dx an element 

 of the chord C at a distance p from the axis A . Let the minimum distance of 

 this chord from A be h, and 6 its angle with p. Then 



dx = poxft/cos 6 = p^udt/h 

 if dx is described in the time dt. Hence 



(7) dx/dt = (h*+x 2 )a/h 



To find the mean speed v along C, we may multiply dx/dt by dx, integrate 

 between o and C/2, and divide the result by C/2. Thus 



2 r cl2 



(8) v = g I u(h*+x*)/h . dx = u(h+C 2 /i2h) 



J o 



Reducing this by equations (i), (2), (5), eventually 



or the mean speed along C may be expressed in terms of R, co, /x; v is naturally 

 proportional to R and o>. 



The ratio of the speed in equation 9 (seeing that it is respectively + and 

 for the two interfering rays) to the velocity of light is thus 2V/c. Since these 

 rays traverse a path iC in the rotating cylinder in opposite directions, the 

 path difference resulting will be 

 f \ A TV f i \ r C v 4M<*>/? 2 i M 2 /6 



(10) AP'=(2fl/c)2C=4- -7= == 



c c Vi-M 2 /4 



so that the path difference for a given y. and co increases with the square of 

 the radius R of the cylinder or disk. 



But equation (4) introduced another factor (i i/V 2 ),so that finally the 

 path- difference is 



We may now take the above data (6 = 10 cm.) from figure 91, for a small 

 cylinder, making n= 100 turns per second. 



R=S. 3 cm. ^=1.63 40 = 628 i = 7o.6 ^ = 35-3 6=iocm. 



In accordance with equation (10) therefore, since \/i / 2 /4=o.s8 and 

 i M 2 /6 = o.44 nearly, the uncorrected path- difference is 



= 



3 Xio 10 0.577 



The corrected path-difference AP = AP'(i i//i 2 ) thus is finally 



= 2.95Xio- 6 Xo. 623 = 1. 84Xio- 6 cm. 



The fringes which appear in the above interferometer are primarily those 

 of reversed spectra. If the yellow parts of the spectra (X = 6oXio~ 6 cm.) are 



