Explanation

(a) Lens formula is given by
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\ \ \ \ \ \ \ldots(i)$
Where $f$ is focal length of lens,$\ v$ is image distance and $u$ is object distance.
Given $f = 10\ cm$ (as lens is converging)
$u = \ - 8\ cm$ (as object is placed on left side of the lens)
Substituting these values in Eq. (i) , we get
$\frac{1}{10} = \frac{1}{v} - \frac{1}{- 8}$
$\Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{8}$
$\Rightarrow \frac{1}{v} = \frac{8 - 10}{80}$
$\therefore v = \frac{80}{- 2} = - 40\ cm$
Hence, magnification produced by the lens
$m = \frac{v}{u} = \frac{- 40}{- 8} = 5$