A concave lens has focal length 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.

Given:

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

​Image distance, $v$ = $-$10 cm                          (image distance is taken negative because image formed by a concave lens is on the left side of the lens)

To find: Object distance $(u)$, and magnification $(m)$.

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-

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

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

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

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

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

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

$u=-30cm$

Thus, the object is at a distance of 30 cm from the concave lens, and the negative sign implies that it is on the left side of it. 

Now,

Foo magnification we know that-

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

Substituting the given values in the formula, we get-

$m=\frac {-10}{-30}$

$m=\frac {1}{3}$

$m=0.33$

Thus, the magnification $m$, produced by the lens is 0.33 which is less than 1, therefore the image will be smaller than the object.