814 Mr. C. G. Darwin on the Reflexion of 



end condition now is S TO = 0. The solution is found to be 

 S = — iq T / (A + ika cos 6 . u + e coth me) 



[' <**: 



T„j = e cosech me . T / (h + i^a cos # . u + e coth me 



where e 2 = q 2 + (h + ika cos 6 . u) 2 (6* 3) 



Now, as we saw, 7i is very much smaller than q. If we 

 neglect it altogether we have 



(6-4) 



e = \/q 2 — (ka cos . u) 2 for | w [ < qjka cos -v 



and e = iv [ka cos .u) 2 — q 2 for \u\ > qjkacosO ) 



It is then easy to verify that for all values of u 



|T„,p+|So| 2 =|T p. 



Further, if h is not quite zero, it may be verified that to a 

 first approximation 



| % n |* +\ S Q |»= I T Q | 2 (1- 2mh). . . . (6-5) 



Thus we can always calculate the intensity of the trans- 

 mitted beam by reducing the incident by an amount 

 corresponding to ordinary absorption and subtracting from 

 it the intensity reflected. This will play an important part 

 in the next section, as it gives rise to the secondary 

 extinction. 



Now, consider the reflected beam coming from a point 

 source. Following the line of argument of D. ii., we may 

 resolve the spherical wave into plane, and the whole re- 

 flexion is given by integrating the plane wave formula over 

 all values of u. The exact form of the answer involves sucli 

 matters as the length of the slit in the observing instrument, 

 but here it will suffice to find a quantity that is proportional 

 to it. This is 



J 00 ^*oo 



I S 1 2 du = q 2 1 T 1 2 1 du/\h + ika cos b . u + e coth me | 2 . 



... ((3-6; 



The evaluation of the integral seems to be impossible in 

 general, but our object can be achieved by taking advantage 

 oi" the smallness of h and using (6*4). But there is a com- 

 plication ; for unless h actually vanishes, e will have a small 

 real part in the outer region, and therefore if m is large 



