Given,$u=-15\ \mathrm{cm},\ f\;=20\ \mathrm{cm}$$\;\frac{1}{f}\;=\;\frac{1}{v}-\;\frac{1}{u}$$\;\frac{1}{20}\;=\;\frac{1}{v}+\;\frac{1}{15}$$\;\frac{1}{v}\;=\;\frac{1}{20}-\;\frac{1}{15}$$v\;=-60\ \mathrm{cm}.$ Four characteristics of the image formed by the lens are :(i) Virtual, (ii) Erect, (iii) At a distance of 60 $\mathrm{cm}$ on the same side of the lens as the object, (iv) Enlarged or magnified image.