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An object is 2 m from a lens which forms an erect image one-fourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this?
Magnification, $m$ = $+\frac {1}{4}$ $(\because image\ is\ erect,\ 'm'\ will\ be\ positive)$
Object distance, $u$ = $-$2 m
To find: Focal length, $f$.
Solution:
From the magnification formula, we know that-
$m=\frac {v}{u}$
Substituting the given values, we get-
$\frac {1}{4}=\frac {v}{-2}$
$4v=-2$ (by cross multiplication)
$v=-\frac {2}{4}$
$v=-\frac {1}{2}$
$v=-0.5m$
Thus, the image distance $v$ is 0.5 cm from the lens, and the negative $(-)$ sign implies that the image is formed in front of the lens (on the left).
Now,
From the lens formula, we know that-
$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$
Substituting the given values, we get-
$\frac {1}{(-0.5)}-\frac {1}{(-2)}=\frac {1}{f}$
$-\frac {1}{0.5}+\frac {1}{2}=\frac {1}{f}$
$\frac {1}{f}=\frac {1}{2}-\frac {10}{5}$
$\frac {1}{f}=\frac {1}{2}-2$
$\frac {1}{f}=\frac {1-4}{2}$
$\frac {1}{f}=-\frac {3}{2}$
$f=-\frac {2}{3}$
$f=-0.666m=-66.6cm$
Thus, the focal length of the lens is 66.6 cm, and the negative $(-)$ sign implies that the lens is diverging in nature. Therefore, it is a concave lens.