110 M. Babiiiot On the Law bftne Cdhitrs seeti hy 



is destroyed by its interference with the ray which goes from I 

 to C, and which differs from it half an undulation, besides, all 

 the successive elementary rays which would have for their origin 

 the different points from A to I, will be destroyed by the rays 

 emanating from points similarly situated from T to B, and re- 

 spectively retarded half an undulation in relation to the rays 

 emanating from points situated between A and I. The point 

 AB, therefore, will appear completely deprived of light. But 

 if we now conceive the opaque part KB of the interval AB, 

 and if we take IL equal to AK, the rays whose origin is be- 

 tween I and K will no longer be destroyed by those which are 

 between I and L, and which would have differed from them 

 by half an undulation, since these last rays are suppressed by 

 the opacity of LB. The first rays will then subsist and will 

 convey to C a light, the more or less, as AK approaches in 

 equality to AI or to the half of AB. But we must not 

 increase beyond I the interval AK, or if, for example, AL 

 were the transparent and LB the opaque part of the system : 

 In this case, indeed, a certain portion of the rays near A will 

 be destroyed by the rays whose origin would have been be- 

 tween I and L, and which the opaque part of the system 

 would not have suppressed. This particularity, which relates 

 to the intensity of the light emanating from AB, has escaped 

 Fraunhofer, and deserves to be confirmed by precise experi- 

 ments. 



As the tint for which the length of an undulation is x»f 

 ought to subsist in the part AB of the system in which we 

 have B G = X, it is easy to determine the angle HCA or rather 

 HCB, which the direct ray SC makes with the ray AC or BC, 

 which propagates this tint to the eye at C. The two right 

 angled triangles HCB and BAG have the angle at B com- 

 mon, and consequently are similar. The ratio of HB to BC, 

 or the sine of the angles HCB, which we shall call a, will 

 therefore be equal to the ratio of BG to BA, that is, to the 

 ratio of X to the quantity AB, which we shall call c. We shall 



then have sin. 3 = - 

 c 



In like manner it may be shown that this same tint whose 



length of undulation is \ will still subsist for the intervals 



