PHYSICS: C. BARUS 
123 
^^^^ 
We may now take the above case {b = 10 cm.) from the graph i, /x, for a 
small cylinder, making 100 turns per second. 
R = 5.3 cm; M = 1.63; co = 628;/ = 70.6°; r = 35.3°; 6 = 10 cm. 
In accordance with equation (10), therefore, the uncorrected path difference 
is 
A P' = 2.95 X 10-6 cm. 
and the corrected path difference finally, 
A P = 1.84 X 10-6 cm. 
The fringes which appear in the above interferometer are primarily those 
of reversed spectra. If the yellow parts of the spectra (X = 60 X 10"^) are 
superposed, 0.031 of a fringe would pass for the given radius of cylinder 
(R =* 5.3 cm.) at 100 turns. A cylinder 30 cm. in diameter (about a foot) 
would therefore show .28 fringe, and since this may be doubled by reversing 
the rotation of the cyhnder, (by which strains due to centrifugal force are 
also eliminated) something short of | of a fringe should be observed. 
With an ocular micrometer divided in 1/10 milHmeter, it should be possible 
to secure fringes as much as 3 mm. apart, so that a displacement of 20 scale 
parts may be expected, ten for each of the directions of rotation. 
4. Equations. Two reflections. — The equations for this case are somewhat 
more involved than the preceding; but it suffices to accept for the angle of 
incidence i at the cylinder G, figure 3, the value given by the old-fashioned 
theory of the rainbow; viz., 
8 cos2 i = - 1 (12) 
The chord C from c to d, etc., and its distance h from the axis a will be, as 
before, C = 2R cos r, h = R sin r, where r is the angle of refraction and R the 
radius of the cylinder. Finally equation (8), for the average speed v along a 
chord, also applies. Hence with the inclusion of equation (4), the path 
difference on rotation may be written, c being the velocity of light, 
3 X 2 C (v/c) (1 - ' (13) 
since there are three chords, C, in sequence. This expression may be reduced 
by the equations for C, v, and equation (12), eventually to a form con- 
venient for computation. 
{9R'c/c) (3/m' + 1) \/(1-1/m')/(9V-1) 
