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120

The apparent lengths with the naked eye, and with the lens, are therefore, as $1 / 120$ and $1 / 8$ , or as 8 and 120, of which numbers the latter being fifteen times the former, the object is said to be magnified in length fifteen times.

In order to generalize this, let $c$ be the nearest distance for correct vision,

and let	$k = OA$ ,	the distance of the eye from the lens,
	$∆ = AQ$ ,
	$F = AF$ ;
	$δ = Aq$ .

Then since $δ = ∆ F / F -∆$ , the linear magnitudes of the object and image are as $∆: δ$ , that is, as $F -∆: F$ .

The angular magnitudes, that is, the angles subtended by the object and image at $O$ , are as $F -∆ / ∆+ k : F / δ + k$ , but the fairer way of stating the matter is to compare the angular magnitudes of the object at the distance $c$ and the image at the distance $δ + k$ : these are as

$F -∆ / c : F / δ + k$ ,

and the magnifying power of the lens is

$F / δ + k \cdot c / F -∆$ .

This of course is increased by diminishing $k$ . If we make this $= 0$ by placing the eye close to the lens, the magnifying power becomes

$F / δ \cdot c / F -∆$ , which is equal to $c / ∆$ ,

and is inversely as the distance $AQ$ , which may consequently be diminished with advantage, as long as $Aq$ is left greater than $c$ .

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