ABSORPTION OF LIGHT IN GASEOUS MEDIA. 
37.9 
Writing k = 47r/A the above equation gives 
- = (a + 'c)I(G 0).(11) 
If we write 
K = a + zf,.(12) 
the solution of (11) is 
I(r,0) = I(O,e)(!-t'^'‘',.(13) 
which puts into evidence the variation of a and k with the density of the gas. 
Consider now the radiation incident on an element of volume at {x, y, z ):—- 
(i) The external illumination E which contributes to the scattered radiation 
from Sv an amount 
M ( 0 ) E {x, y, z) Sv So ); 
(ii) An element of volume St/ at (x\ ?/', 2') gives rise to an intensity 
I (x\ y\ 2', r', O') Sv' at the point {x, y, z): this contributes to the scattered 
radiation from Sv the amount 
/uL {rr') I {x',}/, z', r', O') Sv' dco Sv r'^^, 
rr' denoting the angle between r and r'. The total contribution to the 
scattered radiation from Sv due to self-illumination by the entire volume 2 is 
S(jd Sv J /w (vv') I (x', y', 2', r'. O') dv', 
the integral being taken throughout the entire volume enclosed liy the 
surface 2. 
By definition the sum of contributions (i) and (ii) is I {x, y, 2 , 0, 0), so that we 
obtain the following integral equation for the scattered radiation at and from any 
point, 
l{x,y,z,0,0) = n{(l)^{x,y,z)+ fx{rr')l{x', y',z ,0, O') r' “c . (14) 
I {x', y', 2 ', r', O') being expressed in terms of I {x', y', z', 0, 0') by means of (13). 
A differential equation involving E {x, y, 2 ) is obtained by considering the rate of 
accumulation of energy in an element of volume. 
As soon as I {x, y, 2 , 0, 0) is known as a function of the position of the point O, 
{x, y, z), in the volume bounded by the surface 2, the radiation scattered to any point 
P within the bonndary 2 contained in a small solid angle w is given by the formula 
To, =oJ’'“l(a:,^,2,O,0)e--f>'"'dr,.(15) 
J 0 
where r = PO, 7’(, = PQ, the radius vector to the boundary, and T is the intensity at 
3 c 2 
