An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Fine the position and nature of the image.

Given: 

​Object distance, $u$ = $-$4 cm                           (object distance is always taken negative, as it is placed on the left side of the lens)

Focal length of the lens, $f$ = $-$12 cm     (focal length of the concave lens is always taken negative)

To find: ​Position or distance of the image, $v$ and its nature.   

Solution:

From the lens formula, we know that-

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

Substituting the given values in the formula, we get-

$\frac {1}{v}-\frac {1}{(-4)}=\frac {1}{(-12)}$

$\frac {1}{v}+\frac {1}{4}=-\frac {1}{12}$

$\frac {1}{v}=-\frac {1}{12}-\frac {1}{4}$

$\frac {1}{v}=\frac {-1-3}{12}$

$\frac {1}{v}=-\frac {4}{12}$

$\frac {1}{v}=-\frac {1}{3}$

$v=-3cm$

Thus, the position or distance of the image $v$ is 3 cm away from the lens, and the negative sign implies that the image is formed in front of the lens (left side of the lens). Also, we know that when the image is formed on the left side, the image is always virtual and erect in nature.

Hence, the position of the image is 3 cm on the left of the lens, and the nature of the image is virtual and erect.