Given: $u=12\ \text{cm}$ (negative), $f=8\ \text{cm}$ (positive) We know, by lens formulae, $\;\;\frac{1}{f} = \;\frac{1}{v} - \;\frac{1}{u}$ $\;\;\frac{1}{v} = \;\frac{1}{u} + \;\frac{1}{f}$ or $\;\frac{1}{v} = \;\frac{1}{u} + \;\frac{1}{f}$ or $\;\;\frac{1}{v} = \;\frac{1}{-12} + \;\frac{1}{8}$ $\;\;\frac{1}{v} = \;\frac{-2+3}{24} = \;\frac{1}{24}$ $\therefore \;\; v=24\ \text{cm} \;\text{ (positive) }\;$ Thus, the image is formed at a distance of 24 $\text{cm}$ behind the lens. We know, $\;\; m = \;\frac{v}{u} = \;\frac{24}{-12} = -2$ Thus, the image is Real, inverted and magnified in size.