A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray-diagram.

Given: 

 Focal length, $f$ = $-$15 cm                          (focal length of a concave lens is always taken negaative)​

​Image distance, $v$ = $-$10 cm                (image distance is taken negative as the image is virtual)

To find: Object distance, $u$.

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-

$\frac {1}{(-10)}-\frac {1}{u}=\frac {1}{(-15)}$

$-\frac {1}{10}-\frac {1}{u}=-\frac {1}{15}$

$\frac {1}{15}-\frac {1}{10}=\frac {1}{u}$

$\frac {1}{u}=\frac {2-3}{30}$

$\frac {1}{u}=-\frac {1}{30}$

$u=-30\ cm$

Thus, the object is placed at a distance of 30 cm or at a distance of $2f$ from the concave lens as the focal length is 15 cm (given). The negative sign implies that it is on the left side of it.