Ca]^aciUj of the Microscojpe. By Prof. Helmholtz. 27 
Let s indicate the distance of vision, p the radius of the pupil, 
then the area of its surface will be tt the cross section of the 
pencil of light tt s^. sin.^ ay and the general relation will be 
H : Ho = s2 . sin.2 : p\ 
Or using equation [7], 
S2 ^2 ^2 
H = Ho . — • — • sm.2 a. 
The last medium in front of the eye must necessarily be air, 
therefore 71^=1, and if we indicate by the angle of divergence 
of -the instrument measured in air according to Lister's method, 
then sin. a^^ = n . sin. a. Putting the amplification ^ = N, then 
H z= H . 
With an amplification Nq by which the cone of light just fills 
the pupillary opening, and which we shall call the normal ampli- 
fication of the instrument, H = Hq. Hence 
No = %m. ao. [8] 
And if aQ remains constant, 
h:h„ = No2:n^ [8»] 
If as was assumed 
Whilst H = Ho when 
That is to say, 
The brightness of an image seen through the microscope is equal 
to that of light filling the unoccupied eye * when the amplification 
is less (or not greater) than the "normal'' amplification (i.e. 
when the area of the ocular image just fills the pupil) ; other- 
wise, with the same constant divergence of incident rays, the hright- 
7iess is inversely proportional to the amplification of image. 
The normal amplification increases with the increase of the sine 
of the divergence angle whose greatest value is 1 when this angle 
approaches a right angle (as is the case with the widest-angled 
objectives). 
Assuming 1 0 inches as the distance of clear vision for calculation 
* Daylight is of couriie supposed, and a njouocular microscope in usco 
