ABSORPTION PRODUCED BY FLUORESCENCE. 



91 



by the surface a^ is then made up of two parts, i.e., the fluorescent light of A l and 

 the fluorescent light of A 2 transmitted by A x . 



Hence the intensity of the light given out by a^ is E (1 + a). The light from A.\ 

 consists of its fluorescent light only, the internal reflection being the same as in 

 the other two blocks, and is thus = E . 



Denoting the widths of the slits by s and s', and remembering that these quanti- 

 ties are inversely proportional to the intensities, we obtain the equation 1 + a = s'/s, 

 from which a may be determined. 



In order to avoid errors arising from want of symmetry, the process is reversed by 

 screening A 3 instead of A' 3 , arid similarly a is given by the equation 1 + a =. cr/cr', 

 a- and cr' being now the widths of the slits s, s'. If there are n observations with 

 the screen in each of these two positions, 



The (ft) Set of Experiments. 



To determine ft for A 1( we screen A x and A' e from the incident light, as in fig. I. 

 We can then compare the fraction of the light emitted by 

 A 2 which is transmitted by Aj (when the latter is not 

 fluorescing) with that emitted by A.\, which is the same 

 as that from A 3 ; the constant of reflection being of 

 course involved, as in the previous case. 



The light coming from a\ is then 



= E u> 

 whilst that from a x 



ft = ,'/. 



^8 = cr/cr', 



Hence 



If the process be reversed, 

 and after n such, observations 



The removal of A' a is found to make no appreciable difference. The cube Aj and 

 the glass screen may be moved before the determination is made, to make sure that 

 A 2 and A\ radiate equally. 



The ft/(l + ) Set of Experiments. 



The ratio ft/ (I + a) can be obtained independently by the arrangement shown in 



iig. 5. 



N 2 



