PROFESSOR STOKES, ON THE DYNAMICAL THEORY OF DIFFRACTION. 31 



The displacement represented by these equations takes place along the same line as before ; and 

 if we put ^ 2 for the displacement, and write cos 9 for I, we get 



£ 2 = cos 6 sin (pf'(b( - r) (44). 



33. By combining the partial results obtained in the preceding article, we arrive at the 

 following theorem. 



Let £ = 0, ti = 0, £ = f(bt - at) be the displacements corresponding to the incident light ; 

 let O l be any point in the plane P, dS an element of that plane adjacent to Oj ; and consider 

 the disturbance due to that portion only of the incident disturbance which passes continually 

 across dS. Let be any point in the medium situated at a distance from the point O x which 

 is large in comparison with the length of a wave ; let 0^0 = r, and let this line make angles 6 

 with the direction of propagation of the incident light, on the axis of x, and (p with the 

 direction of vibration, on the axis of %. Then the displacement at O will take place in 

 a direction perpendicular to O t O, and lying in the plane zO^O; and if £' be the displacement 

 at 0, reckoned positive in the direction nearest to that in which the incident vibrations are 

 reckoned positive, 



In particular, if 



we shall have 



<C = (1 + cos 0) sin <pf(bt - r)*. . . . (45). 



2ir 

 f{bt — x) = c sin — (bt — x), 



^ = (l + cos 6) sin (p cos (bt - r). . . ■ (46). 



34. On finding by means of this formula the aggregate disturbance at O due to all 

 the elements of the plane P, O being supposed to be situated at a great distance from P, we 

 ought to arrive at the same result as if the waves had not been broken up. 



To verify this, let fall from O the perpendicular 00' on the plane P, and let OO 1 = p, 

 or = - p, according as O is situated in front of the plane P or behind it. Through O' draw 

 OV, 0y, parallel to 0,#, y y, and let 0'0 X = /, O.O'y = u>. Then dS = r'dr'da> - rdrdw, 

 since r 8 = p 2 + r' 2 , and p is constant. Let £' = s sin (p. The displacement <^ takes place in the 

 plane ssOx 0, and perpendicular to X ; and resolving it along and perpendicular to x z, we 

 get for resolved parts s sin 2 (p, s sin (p cos (p, of which the latter is estimated in the direction 

 OM, where M is the projeetion of O t on O'y. Let MOO' = Y, % ^ e ' n g reckoned positive 

 when M falls on that side of O 1 on which y is reckoned positive ; then, resolving the displace- 

 ment along OM parallel to O'x', O'y', we get for resolved parts — s sin <p cos cos y,, ssincp 

 cos (p sin y_. Hence we get for the displacements £, q, £ at O 



% = - s sin <p cos (p cos y_, n = s sin <p cos <p sin %, £ = s sin 2 (p. 



* The corresponding expression which I have obtained for I provided we suppose b to be the velocity of propagation of 

 sound differs from this only in having cos 8 in place of sin <p, sound, and 5' to represent a displacement in the direction O, O. 



