Peterson 
D = 6 log 10 I A 
BD) =) leee a Ss 
where 6 = constant 
I = incident intensity 
I = transmitted intensity 
As 
Now, since wave amplitude varies as v.(1)° 
us iT 
then pam 0 A a a 
i i t isp -t 
: : Mein - 6/2 
and for a unit amplitude incident wave, D. = I. 
If 6 is positive a photographic ''negative'' is produced. Conversely, 
if 6 is negative a photographic ''positive'’ is produced. 
The holographic process has been analytically described for 
the production of the hologram, If on the hologram a plane wave is 
incident, then we have the same analytical situation previously de- 
scribed, The new boundary condition is the variation in amplitude 
of the incident radiation, 
a. [vay ey ae 
where YW™* is given by equation (12A) and A is the plane wave 
amplitude outside the hologram diffraction pattern. (yw *)-4/2 has 
the form (1-X)Vandif |X| < 1 and sufficiently small, taking 
only the first 2 terms of a series expansion gives 
Ae ees = x (14A) 
Thus, the use of the integral solution to the Helmholtz equation, (4A) 
with the boundary condition given in (14A), will result in an image 
of the original object. 
152 
