Given, $\mathit{f}=+20\mathrm{cm},\mathit{u}=-30\mathrm{cm}$ ${\mathit{h}}_0=4\mathrm{cm}$
We know that,
$\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}$
$\frac{1}{20}=\frac{1}{\mathit{v}}-\frac{1}{-30}$
$\frac{1}{\mathit{v}}=\frac{1}{60}$
$\Rightarrow \mathit{v}=60\mathrm{cm}$
$\frac{{\mathit{h}}_{\mathit{i}}}{{\mathit{h}}_0}=\frac{\mathit{v}}{\mathit{u}}$
$\frac{{\mathit{h}}_{\mathit{i}}}{4}=\frac{60}{-30}$
So,$\Rightarrow {\mathit{h}}_{\mathit{i}}=-8\mathrm{cm}$
Thus, the height or size of the image is $8\mathrm{cm}$. The minus sign shows that this height is in the downward direction, that is, the image is formed below the axis.
Ratio of size of image to object $=-\frac{8}{4}=-2$
So image is enlarged beyond $2{\mathrm{F}}_2$.Object between ${\mathit{F}}_1$ and $2{\mathit{F}}_1$.
Image is formed beyond $2{\mathrm{F}}_2$, real, inverted.