Given : $u=-30\,\text{cm}, f=-15\,\text{cm}$. (for concave lens)We know that,$\,\frac{1}{v}-\,\frac{1}{u}=\,\frac{1}{f}$ $\,\frac{1}{v}-\,\frac{1}{-30}=\,\frac{1}{-15}$$\,\frac{1}{v}=\,\frac{-1}{15}-\,\frac{1}{30}$$\,\frac{1}{v}=\,\frac{-2-1}{30}=\,\frac{-3}{30}=\,\frac{-1}{10}$$v\,=-10\,\text{cm}$ Characteristics of image :(i) The image is formed at a distance of $10\,\text{cm}$ 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.