326 Prof. W. L. Bragg and Messrs. James and Bosanquet : 



crystal per second is I, and it is all intercepted by the slip 

 of crystal, the total energy reflected will be given by 



E _ n 2 \ 2 F2 j^_ pX 

 I ~sin 2 6> 



or 



Eo> WxH 



I " 2 sin 2 6 cos 



•F 2 zVj (3) 



nrc 



In this calculation it has been assumed that the absorption 



of the radiation is inappreciable. 



As a corollary, we can calculate the reflecting power of a 



homogeneous fragment of crystal of volume V. The volume 



of the slip irradiated by a narrow pencil of rays is equal 



S 

 to ~^—x . t. where S is the area of cross-section of the pencil. 

 sin 6 



Erom the above formula 



Ea) N 2 \ 3 -_ e± t 



I ~~ 2 sin 6 cos 0' 'mV sin <9 



Now, I = SI when I is equal to the intensity of the beam 

 irradiating the crystal, defined as the amount of energy 

 falling on one square centimetre per second, whence 



Eo>_ NV y2 j^ Y 

 J "~ sin 20 mV " ' . ' ' W 



This result shows that the " Reflecting power " o f a homo- 

 geneous fragment of the crystal is proportional to its 

 volume, if the fragment be so small that absorption in it 

 is inappreciable. 



We will now assume that the crystal consists of a number 

 of such homogeneous crystalline particles, set approximately 

 parallel to each other, but not exactly so. When the rays 

 are reflected from the face of a crystal, the reflexion by 

 particles below the surface is diminished by absorption. 

 It will be assumed that the linear coefficient of absorption jul 

 is a constant. (This assumption will be discussed more fully 

 below.) 



Bays reflected by a particle at a depth z beneath the 



crystal surface suffer absorption by passing through a 



2z 

 distance - — ^ of the crystal. They are therefore reduced 



in intensity in the ratio 



