16 
Mr Barlow On the Practical Construction 
The result here is positive ; but had the convex side been 16 
and the concave 10, the focal length would have been the same, 
but the rays would have diverged, or the result would have 
been negative. 
5. Having the focal length of a double convex or plano-convex 
lens given , as also the negative focal length of a double con- 
cave lens, or of a concavo-convex lens, to fnd the focal 
length of the combined object-glass. 
Rule. — Multiply the two focal lengths together ; divide the 
product by their difference, and the quotient will be the focal 
length of the compound object-glass. 
Note . — If the negative focal length be the lesser, the result- 
ing focus will still be negative, but if greater, it will be 
positive, and the rays will converge. 
Example . — The focal length of a double convex lens is 6 
inches, and of a concavo-convex lens 9 inches, negative. Requir- 
ed the focal length of the compound object-glass, formed by 
combining the two. 
From negative focus =9 9 
Subtract positive focus =6 6 
3)54 
54 
18 focal length required. 
From these rules are drawn several others which may be of fre- 
quent use ; for example, 
6. The index of refraction and one of the radii of a double con- 
vex lens being given, to fnd the other radius, so as to produce 
a given focal length. 
Rule. — Multiply the proposed focal length, the decimal part 
of the index (a), and the given radius together, for a dividend ; 
and subtract the former part of this product from the given ra- 
dius, for a divisor. Then divide the dividend by the divisor, 
and the quotient will be the other radius. 
Example . — The index of a piece of flint-glass is 1600, and 
one of its curvatures is to a radius of 10 inches. What must 
the other be, to give a focus of 12 inches ? 
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