X-Raysfrom Imperfect Crystals. 811 



are to be made infinite later, the result of the I and m 

 integrations is 



fJA,rf* Jf(w, ^, ±g*) dw§*dvir 



jn 2 / 2 cos e 



Fsin 2 ' 



sin £(# —a/) sin (9 . 2/^ sin /fc (?/ —3/') sin 6 . 2m 

 x-x ~J^V~ 



x exp — ik(z— z')(u + v) cos 0. 



In this a?' has been equated to x and y' to y in all the terms 

 that do not involve Z^, ??i oo , Next integrate for v. With 

 exactly the same argument, we may put —u for v in F, and 

 this makes the integration possible. The result is 



jn 2 /* 2 f°° f cw 



/Asiir 2 ^ * 2V ) *+ J F( W ' M ' *+ C0t 6) dW | w ^ V <* V ' 



—00 



sin &(#—#') sin # . 2^ sin k(y—y') sin # . 2???^ 

 ^— 0?' y-y 



sin k(z—z') cos # . ^ 

 x . 



z — z 



The argument o£ § 4 now shows that the double volume 

 integration is equal to 7r 3 W. Thus the whole effect is 



JN 2 / 2 \ 3 cosec 20 . V f dW f "wF(W, u, m')dm'. 

 Let J -- 



Q (W f dm' WF (W, u, to') = G-(m) . . . (5-4) 



Then, using (4*7), the power reflected is 



JVG(w), (5'5) 



and by (5*2) 



".G(t«)rfu=Q (5-6) 



j 



The relation (5*5) is practically equivalent to saying that the 

 incident beam is reflected by those blocks which are at the 

 proper angle and no others*. 



In order to test the validity of the assumptions made in 

 this work, we shall simplify the problem by supposing that 

 all the blocks are rectangular of sides f, 77, f and orientated 



* With a little modification the same argument proves the result 

 deduced in general terms in D. ii, p. 686. 



