A concave lens forms the image of an object such that the distance between the object and image is 10cm and the magnification produced is ¼. The focal length of the lens will be:

Concave lens forms the virtual image of a real object. So let

$$\begin{gathered} m=\frac{-v}{u} \\ \frac{1}{4}=\frac{-v}{u}\Rightarrow v=-4u \end{gathered}$$

$u=-4xandv=-x$  then

 $$\begin{gathered} u-v=10 \\ -4x-\left(-x\right)=10 \end{gathered}$$ 

 $3x=10cm$ or          

  $x=\frac{10}{3}cm$

$u=-\frac{40}{3}cm$    and   $v=-\frac{10}{3}cm$

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

We get $\frac{1}{f}=\frac{-3}{10}+\frac{3}{40}$

Or $\frac{1}{f}=-\frac{40}{9}$

Or $f=-4.4cm$