FOCUS AND OFPMiIGsS 47 
CoRRECTION FOR NON-ACHROMATIC (SPECTACLE) LENSES. 
—After focussing, the distance between the lens and plate 
must be decreased by approximately 1/40 the focal length of 
the lens, in the normal use of the lens when photographing 
distant objects. In portraiture, copying, etc., the correction is 
greater, as follows: 
Ratio of reduction Infin. 1/10 1/5 3/10 4/10 % 6/10 7/10 4/5 same 
Correction (% of ee 
focal length) 2 2.2 .2.4 2.6 Pee) eC) Pee 3.4 3.6 4 
TELEPHOTO ForMULAS.—Let WM be the magnification, 1. e., 
the number of times the image produced by the complete 
lens is larger than that produced by the positive lens alone; 
F the focal length of the complete lens; f’ the focal length of 
the positive lens; f’’ the focal length of the negative lens; 
F the camera extension from negative lens to plate. To find 
the magnification M: Divide the camera extension by the 
focal length of the negative lens, and add 1, or M = (E = 
f’’) +1. To find the camera extension: Multiply the focal 
length of the negative lens by the magnification minus 1, or 
FE =f’ (M—1). The focal length of the whole lens for 
distant objects equals the focal length of the positive lens 
multiplied by the magnification, or F = Mf’. For near ob- 
jects when reducing N times, F = (mE + f’) + (mN +1), 
iawnicn a — ithe ratio, of |f’ to 7”, \1.\.e% F771) (Dall 
meyer ). 
Assume the same notation as above. The separation of the 
positive from the negative lens = (f’ —f”) + (f”+WM). 
Example: Suppose the positive lens has a focus of 7 in., the 
negative a focus of 3 in., what separation will be required for 
4 magnifications? Ans—(7—3) + (3+4) =434. An 
alternative formula giving the separation necessary to obtain 
a telephoto combination of a desired focal length is (f’ — 
f’) +f G¢’ +F). Example: suppose the positive lens is 
9 in. focus and the negative 4 in., what separation is required 
