50 ROYAL SOCIETY OF CANADA 
copy it through a screen. The illumination 7 of the plate is, we have 
seen, proportional to the intensity of the light transmitted through the 
negative and to the area of the fraction 4 of the diaphragm not hidden by 
the screen. 
N being a constant. We may take for 7 the value of the illumination on 
the outlines of the dots and calculate ¢ : 
_ Mpli 
PE TUNER 

(10) 
Assuming again that the tone Z of the subject is to be represented by 
pure white paper in the print, there must be no black dot when p is equal 
to one, and consequently the illumination under the centre of a transparent 
square, where the light is given by the full effective aperture Q of the 
diaphragm, must be equal to 7, or: 

Fe M Li 
— N *. 
Introducing this value in equation (10), it becomes 
iP (11) 
This equation means that any tone pL of the original is transformed into 
dots, from the outline of which the fraction p of the diaphragm is visible. 
For correct representation, the area of the white dots must be pro- 
portional to p, and the area of the black dots to 1 — p; therefore the area 
of the curves of equal illumination must be proportional to g under the 
opaque squares of the screen, and to 1 — 4 under the transparent squares. 
Such curves are given by the chess-board screen and the single or double 
exposure diaphragms described in the first part of this paper. The dots 
are shown in Figs. 12 and 14, and are practically correct. 
The negative must be what is termed a “soft” one, and must be 
produced by full exposure and short development. It should be much 
less dense than for printing by contact. 
The formule indicate that there are white dots in the deeyfest shadows 
of the print ; there is no solid black. In theory this is absolutely correct, 
because the clear glass of the negative does not indicate absence of light 
in the subject, but only an illumination less than the inertia of the plate 
on which the negative was made. We must remember, however, that we 
bave assumed the ink to be perfectly black, which is not the case; we 
should therefore close the white dots in the deepest shadows. With a 
negative of average density, these dots are below printing size; they do 
not appear in the print; with a thinner negative it is sufficient to use a 
diaphragm very slightly larger than the correct size. 
