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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?

**Given:**

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.**