CONTEMPORARY ADVANCES IN PHYSICS 309 



which set the limits for the powers of telescope and microscope, and 

 perhaps for perception altogether; and which are responsible for the 

 haloes and the parhelia of the sky. 



Imagine then that the waves which come up to the screen from be- 

 hind are plane-parallel and monochromatic, and travel in the positive 

 sense of the x-direction, the screen itself occupying the entire plane 

 re = 0. In making the "second approximation" aforesaid, we are to 

 regard this plane as a surface where 5 and its gradient are zero every- 

 where save over certain patches — to wit, the apertures — and over 

 these are given by the expressions : 



5 = cos {nt — nix)x=o = cos nt, 

 ds/dx = m sin (nt — mx)x=o = m sin nt. 



(71) 



We are then to determine the value Sq of 5 at any field-point P any- 

 where before the screen — anywhere in the region x > — by forming 

 Kirchhofif's integral over these apertures: 



4x50 = I ao cos {n,r) -— 



J L dr r r dn 



(72) 



Over the rest of the plane :x; = the integrand vanishes. Since, how- 

 ever, Kirchhoff's theorem involves an integration over an entire closed 

 surface surrounding P, we ought in strictness to extend the integral 

 over some far-flung surface completing the enclosure; as for instance 

 a hemisphere seated upon the plane x = 0, sufficiently great in radius 

 to contain P and all the apertures. This is always neglected, possibly 

 because in practice the wave-motion over such a surface would as a 

 rule be too chaotic to produce any regular effect at P.^ 



In the integrand of (72), r stands for the distance from P to any area- 

 element dS of an aperture; the positive r-direction is measured from 

 P through dS in the direction from front to back; the positive w-direc- 

 tion is the forward- pointing normal to dS, and therefore is identical 

 with the positive x-direction. Remembering the definitions of U and 

 its derivatives, one easily sees that: 



U = cos (nt — nir) ; 



(73) 

 dU/dr = dU/dx = dU/dn = m sin (nt — mr). 



It will be convenient to give the symbol d to the angle between the posi- 



* Certainly it cannot be argued that the effect from a distant surface is necessarily 

 too small to be noticed at P; we have just seen that in a field of plane-parallel waves 

 it is the same for any spherical surface, no matter how great the radius. 



