C. Barns — Interference of Reversed Spectra. 429 



cos(o-/2)' 



are useful. 



With these postulates, the path difference AP, after a some- 

 what cumbersome reduction, finally becomes 



„ iVcos o- + n 2 sin 0. 7 , 



&P = - —, — T7T- T7T C ™, 



cos (cr I 2 cos 2 2 



nearly. Hence the angular breadth of an interference fringe 

 would be (AP = A) 



A cos 2 2 cos(o-/2) 



2 sin 6 2 JVgoso- +n' ' 



and if D is the grating space and sin 6 = X / D', 

 de= (V' i -)C)co8(<r/2) 



2D 1 {Ngo$<t + n) 



In case of a single grating a/2 — 2 = 6 : N — n 1 cos a = 

 2 cos 2 — 1, or 



„ 2Ncos i 6 2sin<9 _. __ 



AP= — cW = 4i\^tan 8d6, 



cos COS"0 



a result which may be reduced more easily from figure 11. 

 Hence if the angular distance apart of the fringe is (AP = A) 



X DcosO \/D* - A 2 



a 9 = 



4iY r tan 9 4iV 4JV 



it" D is the grating space. To find the part of the spectrum 

 (dX) occupied by a fringe since sin 6 = X/D 



dO dX D 



and 



cos ~ Z> cos 2 _ 4iV' 

 from the preceding equations, finally, 



D 2 - A 2 



dX = 



4iV 



where dX is the wave length breadth of the fringe, remember- 

 ing that the fringes themselves are homogeneous light.* 



In the grating used, D = 351 X 10 -6 cm., A = 60XlO~ 6 cm., 

 n = 100 cm., or dX = 3 X 10" 10 cm. This is but 1/200 of the 

 distance, dX — • 6 X 10" 8 cm., between the D lines. Hence these 

 fringes would be invisible. Moreover, dd cr.l/JV ; the fringes 

 therefore, should grow markedly in size as JV is made smaller. 



* I shall return to this question elsewhere, the results in the test being in 

 some respects inadequate. 



