The distance between an object and a diverging lens is m times the focal length of the lens. The linear magnification produced by the lens is.
A.$m$
B. $\dfrac{1}{m}$
C. $m + 1$
D. $\dfrac{1}{{m + 1}}$

Hint:Linear magnification can be calculated by knowing the values of image formed distance divided by object placed distance i.e., $m = \dfrac{v}{u}$ and for getting the value of $v$ we should apply the formula $\dfrac{1}{v} - \dfrac{1}{u} = - \dfrac{1}{f}$ , $u$ is given in terms of $f$ .

Formula Used:
 $m = \dfrac{v}{u}$ here, ‘$v$’ is the image distance and ‘$u$’ is the object distance and $\dfrac{1}{v} - \dfrac{1}{u} = - \dfrac{1}{f}$ here $f$ is the focal distance.

Complete step by step solution:Our objective is to find the linear magnification produced by the lens which is the ratio of image formed distance.
According to the question
$u = - mf$(in divergence lens ‘$u$’ is negative) but ‘$v$’ is not given, so we have to calculate the ‘$v$’ (image formed distance).
By using lens formula,
$\dfrac{1}{v} - \dfrac{1}{u} = - \dfrac{1}{f}$ (for divergent lens $f$ is negative)
$\dfrac{1}{v} - \left( { - \dfrac{1}{{mf}}} \right) = - \dfrac{1}{f}$
$\dfrac{1}{v} = - \dfrac{1}{f} - \dfrac{1}{{mf}}$
$ = - \dfrac{1}{f}\left( {1 + \dfrac{1}{m}} \right)$
$ = - \dfrac{1}{f}\left( {\dfrac{{m + 1}}{m}} \right)$
$\dfrac{1}{v} = \dfrac{{m + 1}}{{ - mf}}$
$\dfrac{1}{v} = \dfrac{{m + 1}}{u}$
$\therefore \dfrac{v}{u} = \dfrac{1}{{m + 1}}$
and m(linear magnification)
$ = \dfrac{v}{u} = \dfrac{1}{{m + 1}}$

$\therefore $ Linear magnification $ = \dfrac{1}{{m + 1}}$.

Note:We have used the lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ , but for the case of divergent lens we always have to take negative sign for object distance and negative sign for focal length.