and Minute Magnitudes by Miniatures. 
267 
If now this object-glass be employed at 1 0 inches, with an eye- 
piece C magnifying 10 times, the magnifying power will be 10 times 
increased, or 52 • 6. 
Ex. (2). Suppose it be required to find the size of an image 
formed at 15 inches instead of 10, by a lens nearly ^th of an inch 
focal length, which magnifies 200 times at 10 inches. 
Then first the real focal length (see footnote) / = m + % 
10 5 
~ 200 + 2 ~ iol* 
Then the magnifying power at 15 inches will be 
.. ., , , 5 15 x 101 
15 divided by — less 2 = 2 
= 301 - 2 
= 299, 
* A question often arises such as the following : — If a given lens magnifies 
20 times at 5 inches, what would it magnify at 10? Now this will not be just 
double. The law of images is somewhat complicated, and I find that if (d) the 
distance between the image and object (say 10 inches) be divided by the focal 
length (/), it will for a single small lens give the magnifying power m too great 
by 2. But for a lens of considerable focal length, such as £ inch, m = new 
distance divided by focal length less 2, to be further diminished by reciprocal of 
this result, which will afford a closer approximate value. Those who desire to 
see the solution of the problem will be able to follow the following equation, 
given by the writer, ‘Phil. Trans.,’ II., 1870 (p. 594): — 
„ Id/ , d \ 
+ 2 + - = - , f or nearly m = - — 2 1 
— m 
Completing square, 
/ 
m 3 + 2 m + 1 = 
(H ' 
+ 1 = 0. 
Taking the root (positive sign), 
( d 2 V 
( d 2 V 
i / v- 
(/ V 
y~ ( — 2 
)-*• 
4 1/ 
(I-) 
(II.) 
The formula will give the magnifying power, in the same way for a distance 
d = 20 inches between object and image; for a focal length 1-8 as found in 
Ex. (3) : which is a satisfactory application of the principle of verification : — 
Thus, since d = 20 and / now = 1 • 8, 
-i(S-)+^GRr- 
16-4 . v /_256_ _ 1+4 
3-6 
+ 
(3 • 6)- 3-6 
16 _ 32 + 
3 6 ~ 3-6 
~ 9. 
VOL. VIII. 
X 
