THE AID OF THE ACHROMATIC FRINGES. 31 



Furthermore, if the angle a is small and S is displaced over an angle a or 

 a distance r = da to the right, the original triangle may be regarded as 

 restored. Hence the same number of fringes roughly should pass back 

 again. In the second case, supposing 2bo? can be removed by compensation, 

 r = da and a = \d/2b 2 , nearly, or 



or the object should be located to 30 cm. at a kilometer for each fringe pass- 

 ing. In this case d need not be known, since 



n\ 



and n fringes are observed to pass for the angle of rotation a in the compen- 

 sated apparatus. The direct rays without compensation would of course 

 give indefinitely better results if d is known; for the angle per fringe has been 

 found as a = 2.9Xio~ 7 when r = da = 0.03 cm. per fringe if d is i kilometer. 

 Unfortunately, however, the method of figure 14 can not be rigorously 

 carried out experimentally. For in any practical apparatus the mirrors 

 M and N would have to rotate at a fixed distance from each other, apart 

 from the micrometers; i. e., the two mirrors rotate on a rigid radius or rail 

 and are therefore both rotated and displaced. It is this displacement which 

 is relatively of much importance and by it all terms involving the first order 

 of distance d are wiped out, so that terms of the second order in b/d only 

 remain. 



17. Equations. To derive these equations certain intercepts of the rays, 

 figure 14, in addition to x and y, b and b' may be defined. PI PZ is the trace 

 of the vertical plane of symmetry of the right-angled prism, if rotated at 

 an angle a to the right. In this case the reflected ray n'P^q on the right cor- 

 responds to the reflected ray m'P\ on the left, both terminating in the common 

 wave-front P\qs before entering the telescope. 



Let n'P 2 = c' = b sin /3/cos a sin (f} a)=z l /sin a cos a 

 m'Pi = c =b sin /3/cos a sin (/3+a) =z/sin a cos a 

 P 2 t = z' = b sin a sin /3/sin (/3 a) 

 tq = z =6 sin a sin /3/sin (/3+a) 

 and nn' = x = b sin a/sin (/3 a) = z'/sin /3 

 mm' = y =b sin a/sin (/3+a) =2/sin /3 



since the original angles at the ends of the base are /3 and the rotation a. 

 The angles between incident and reflected rays are respectively /3 2a at n', 

 /3+2a at m', 90 20. atP 2 , and go + 2a at P\. Most of the angles are indicated 

 in the figure. The new radii m'P = b' = b sin /3/sin (0 a); n'P = b" = b sin 

 /3/sin 03+ a). 



The rays, however, do not reach the planes of symmetry, but are reflected 

 by the faces of the right-angled prism, and this may be sketched in, in the 

 rotated position (angle a) at Pipp'. The path of the reflected rays from n' 



