At what distance from a converging lens of focal length 12 cm must an object be placed in order that an image of magnification 1 will be produced?

Focal length, $f$ = $+$12 cm

Magnificatiom, $m$ = 1

To find: Distance of object $u$ from the lens.

Solution:

According to the magnification formula, we know that:

$m=\frac {v}{u}$

Substituting the given values in the formula we get-

$1=\frac {v}{u}$

$u=v$

Now, according to the lens formula, we know that:

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

$\frac {1}{(-u)}-\frac {1}{u}=\frac {1}{12}$    [image distance is taken negative $(-)$ as the magnification is 1, which means the image formed will be real and inverted and real-inverted image is formed on the right side of the convex lens]

$-\frac {1}{u}-\frac {1}{u}=\frac {1}{12}$ 

$\frac {-1-1}{u}=\frac {1}{12}$ 

$\frac {-2}{u}=\frac {1}{12}$ 

$u=12\times {(-2)}$ 

$u=-24cm$

Thus, the object $u$ should be at a distance of 24 cm from the convex lens, and the negative $(-)$ sign for object distance implies that the object is on the left side of the convex lens.

Hence, the object is placed at a distance of 24 cm, and on the left side of the lens.