80 LIMIT OF VISIBILITY FOR MINUTE MASSES. 
2. How we may ascertain the least number of the par- 
ticles into which that weight must be divided in order to 
affect the eye. 
From these data the actual weight of one such particle 
is readily found. 
The first of these purposes may be accomplished as 
follows : 
Let v represent any convenient volume (cc) of solution. 
Let # represent the smallest weight (g) of substance 
which, dissolved in v, can be detected by the eye. 
Ww 2 ° e . 
Then » represents the weight in one unit volume. 
Let us fix attention upon a definite portion of ». Sup- 
pose it to. be in the form of a cube, and represent its vol- 
ume (cc) by 2’, and let m represent the weight dissolved 
in this volume. It is evident that 
ne (1) 
v 
By assigning known values to v and 27°, the vaiue of w 
can be found by experiment, as will shortly appear, 
when m—the smallest actual weight of the solid in the 
cubical volume 2’—may be found. 
We next proceed to ascertain the least number of par- 
ticles into which the mass of 2 must be divided. 
It may be fairly assumed that the particles are uni- 
formly distributed in the solution ; therefore, the num- 
ber in the volume @° depends on the value of the uniform 
distance by which they are separated, and this distance 
is fixed in any case by the sensitivity of the eye. 
We may represent by d@ the diameter of the smallest 
magnitude which theeye can perceive(we will suppose it to 
be a circular disk for a reason which will shortly appear). 
Then this value, d, would be the uniform distance, meas- 
ured from center to center, of the particles in solution, 
provided, that the whole number required to absorb light 
enough to affect the eye, could be arranged ina single 
30 
