8 DISPLACEMENT INTERFEROMETRY BY 



Thus there are three objects seen in the telescope: the direct landscape 

 from K, the reflected landscape from L, and the achromatic interference fringes 

 due to the partial beams abc and adc when the apparatus is in adjustment. 

 To find the latter the method pursued in the preceding report suffices. 

 It is first necessary to find the spectrum interferences when L is a beam 

 of intense white light from a collimator and fine slit and the telescope is 

 provided with a direct-vision prism. These are to be centered by moving the 

 micrometer at M' and adjusting the pair of mirrors M'N. The spectro- 

 scope is now removed (swiveled out) and the slit broadened, whereupon the 

 intense achromatic fringes will appear covering the position of the image 

 of the originally fine slit. If not vertical, the fringes may be made so by 

 further slightly rotating M' and N on a horizontal axis, in the absence of 

 compensators. 



The collimator at L is then removed and the fringes will be found super- 

 posed on the foreground. If they are not bright enough from the light of 

 the landscape, they may be given any intensity by reflecting (by way of 

 adc and abc} a narrow horizontal strip of skylight from white paper into the 

 telescope from L. Intense fringes will be seen transverse to the strip. 



3. Rigorous equations. In addition to the sides of the ray parallelogram 

 b (base) and 2R (R radius of rotation) we shall have to consider the following 

 angles or angular increments: A a the angular rotation of the paired mirrors, 

 A0 the corresponding angular displacement of the fringes, AJV the linear dis- 

 placement of the micrometer mirror in a direction normal to its face, and 

 A<p the angle subtended by two consecutive fringes. If n is the order of the 

 fringe, we may write 



( \ A(? 



(1) A ^= 



An 



Moreover, if i is the angle of incidence (45) of L at the mirrors and X the 

 wave-length in question, 



/ N 2 cos i AN 



(2) =; 



An 



or on substitution, 



/ \ A0 A<f> 2A(f> cos i 



~AN ~ AN /An ~ X 



Moreover, if 5 is the angle at the apex of the distance triangle on the base 6, 

 (4) A s = 



(5) 



And since the distance d = b/2S = b/2 A a, from (5), 



bR F 



(6) 



2 AN cos * 



