17 
of Achromatic Object-Glasses, 
a, =■ .600 
Focal length, = 12 
l 9 
7« The index of a piece of glass being given, to fnd what the equal 
convex surfaces must be to produce a given focal length. 
Rule. — Multiply the focal length by double the decimal part 
surface. 
Example. With a similar piece of flint to the above, What 
of 6 inches ? 
Here twice a — 1.2 
Focus — 6 
7*2 inches radius. 
For a plano-convex lens we must multiply the decimal part of 
the index by the focal length for the radius. 
8. The index of r fraction and the convex surface of a concavo-con- 
vex lens being given, to fnd the radius f the concave surface, so 
that the lens may have a given negative focal length. 
Rule. — Find the dividend exactly as in Rule 6. Then add 
the first product to the given radius for a divisor. Divide the 
dividend by the divisor for the radius sought. 
Example. The radius of the convex surface of a concave 
convex lens is 12 inches ; the index of refraction 1.600 ; and the 
negative focal length is to be 5 inches. Required the radius of 
the concave surface. 
of the index, and the product will be the radius of the equal 
must be the equal radii of its two surfaces to give a focal length 
Focal length — 5 
Decimal part (a) — *6 
First product = 8*0 
Given radius — 12 
— = 2A = radius sought. 
VOL. XIV. NO. 27. JANUARY 1826. 
B 
