104 THE ROYAL SOCIETY OF CANADA 
the opposite side of the plate to that which would exist if no plate 
were present is given by 
—kh 
IT, =e + kh Ei (— kh) =f( kh). 
Lo 
If the radiation be confined to a cone of semi-vertical angle #, we have 
i f( xh) — cos f( Kh sec) 
Be 1 — cos 
These results follow from the solutions of the problems already con- 
sidered by making use of the reciprocaltheorem stated at the end of Art. 1 
of the present paper. This theorem explains how the two different 
types of absorption problems lead to the same mathematical expressions. 
It is perhaps not without interest to notice that the intensity at a 
point on the axis of a circular slab of matter containing radioactive 
can be evaluated. The result is easily obtained from (5) by taking the 
; T ; 
integrals from oto ginstead of from oto —, ¢ being the angle subtended 
; 2 
~ 
by the radius a of the plate at P. Then, by (1), we obtain for the 
ionization at P 
n = 778 : L709 — f(z + hk) — cose! (> ue ) aT (ae ia 
cose (eo 
(13) 
This result neglects the contribution of the ring of matter generated 
by the revolution of the shaded area in the figure around the axis of 
