Given : $\mathit{u}=-30\mathrm{cm},\mathit{f}=-15\mathrm{cm}$. (for concave lens)
We know that,
$\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}=\frac{1}{\mathit{f}}$
$\frac{1}{\mathit{v}}-\frac{1}{\left(-30\right)}=\frac{1}{\left(-15\right)}$
$\frac{1}{\mathit{v}}=\frac{-1}{15}-\frac{1}{30}$
$\frac{1}{\mathit{v}}=\frac{-2-1}{30}=\frac{-3}{30}=\frac{-1}{10}$
$\mathit{v}=-10\mathrm{cm}$
Characteristics of image :
(i) The image is formed at a distance of $10\mathrm{cm}$ from the concave lens on the left side.
(ii) Image formed is virtual.
(iii) Image formed is erect.
(iv) The size of the image formed is diminished.