522 Prof. Arthur Schuster on 



We have to find the value of 



\pcos qs cos (pt-—/cs)ds, 



and must for that purpose divide the integral into three parts. 

 During the first period the train of waves will have encroached 

 on the grating, and the limits of the integral during that 

 period will be — Z and that value of s, viz. aty — t^/y, at which 

 f(at—ys) ceases to vanish. During the second period the 

 complete train of waves will intersect the grating, and the 

 limits of the integral will be a(t—t 2 )/y and a (t — t^/y. During 

 the third period the disturbance will pass off the grating. 



It is not necessary to give the details of the calculation. 

 During the first period the disturbance is expressed by 



-^e sin qa^-tj/y--^, sin {pt + tcl) ; 



that is, the disturbance may be expressed as a superposition 

 of two trains of waves, one having a period equal to that of 

 the incident light and the other determined by the direction 

 of the diffracted beam. If the telescope is pointed towards 

 the principal maximum of light, the expression reduces to 



~ [ait-tj /y + 1] cos pt, 



and represents a vibration with variable amplitude. The 

 factor in square brackets vanishes when the disturbance first 

 reaches the grating, and after m complete vibrations becomes 



{mirp/ic) cos pt. 



From that time onwards we have to change the limits of 

 integration, and the amplitude of the disturbance is then 

 expressed by 



p cos qs cos (pt—f€s)ds 



, rcos (qat/y—d)) cos (qat/y—ylr)! 



= p sm irmq/K ■ — - — -\ — — . 



1 q—K q+K J 



In this equation (p stands for — ~-\ -, and i/r for 



^ — 1 -\ *. The difference in phase between the two 



vibrations is therefore 2/cat 1 /y=2pti. But as cosptx — l, this 

 only means that the two vibrations have the same phase. 

 Hence the amplitude will be 



2pq . . 



> o sin vrmq/K. 



(f- — K 



