transmission through grooved surfaces. Ill 



more remote from H than AB is, and for which BG will 

 be equal to twice, thrice, four times the quantity X. We 

 shall then have the angle of deviation d by the expression 



sin. d = — , m being the whole number which marks the or- 

 der of the spectrum. 



By examining the simple relation which exists between the 

 deviation of a ray and the length of an undulation X, on which 

 the tint depends, that the least refrangible rays for which X is 

 greatest will also be the most deviated : Thus in each spec- 

 trum the red will be exterior, and the violet nearest to the 

 disc of image. We see also that the spectra nearest the 

 direct image, for which d is not too great, will be equidistant on 

 account of the proportionality of the arc to its sine. All the 

 other circumstances of the phenomenon are equally deducible 

 from the formula which express its law. * * * 



If we receive upon the system of grooves MN, rays such as 

 S'A', ST, S'B', so that the eye placed in C may receive by 

 reflexion the rays which it gives, it is easy to see that the 

 differences of the paths of the rays being the same as in the 

 preceding case, the same tints should be observed at the same 

 part of MN, which is conformable to experiment. 



If we suppose the rays not to be parallel, but to proceed 

 fromF, Fig. 2, then the colour in AB will depend on EB + BG, 

 for if this quantity is equal to one or to several undulations 

 of a certain tint, this tint will be seen in this direction by the 

 eye placed at C. Let d, as formerly, denote the angle HCB or 

 HCA, and a the angle HFA or HCA, we have 



^^Sin.5; H = Sin.a 

 AB AB 



Whence BG = c Sin. d; EB = c Sin. a 



Whence BG -f- EB = c Sin. 5 + c Sin. a 



but this quantity ought to be a multiple of X. Hence 



mXz=.c Sin. h -\- c Sin. a 



m X 

 or — Sin. b + Sin. a 



c ' 



Figure 3 shows the case where the plane of the system of 

 grooves is oblic[ue to the rays which form the direct image. 

 When the rays proceed from G and reach the eye at C from 

 the part AB, we shall have GAC — GBC = w X. 



1 



