INTERFERENCE OF DIRECT WITH REFLECTED LICJHT. 



241 



certain points and neutralize at others, according to known laws. 

 Any point P of the plane of adjustment, which here represents 

 the intercepting screen, will appear bright if the rays coming from 

 both the sources of light have at this point the same phase, but 

 dark if the undulations are opposed 

 to each other. Since the reflected 

 ray is retarded by the reflexion 

 itself by half a wave-length, then 

 the first condition is equivalent to 

 an arithmetical progression of 1, 3, 

 5, 7 . . . . , and the latter, to one 

 of 2, 4, 6 .... half wave-lengths. 

 The distance of the first dark line 

 from the reflecting surface corre- 

 sponds, therefore, to such a position 

 of the point P that its distance from 

 the slit is shorter by a wave-length 

 than from the reflected image (7. 



If we denote the distance A B of 

 the plane of adjustment from the 

 slit by a, the distance B of the 

 latter from the plane of the mirror 

 by d, and the angle B A by ; if 

 we further draw from the points and 0' concentric circles 

 (Fig. 132), which correspond to the waves of light so that the 

 dotted arcs represent wave-depressions, and the lined arcs wave- 

 elevations (in which the retardation caused by the reflexion is to be 

 considered as half a wave-length), we obtain the following relations : 

 The triangle A Q P, whose vertex P corresponds to the first dark 

 line (which may be regarded, on account of the smallness of the arcs 

 A P and P Q, as a straight line), is similar to the triangle A C/ 0. 

 Therefore, if we draw a perpendicular P JR, which gives the distance 

 of the point P from the reflecting surface, the triangle R Q P 



FIG. 131. 



is also similar to the triangle ABO, and we get P R : R 

 = aid; consequently PR = R Q . - , or, since P R = ^ . - 



Cv ^ COS 



if X denotes one wave-length, 



PE = " X . 



d 2 cos (b 



Q 



This expression admits in all 



cases, where it is a question of 



R 



