Focal length $=\frac{1}{\mathit{P}}=\frac{1}{4}=0.25\mathrm{m}$ or $25\mathrm{cm}$. It is a convex lens as power and focal length are positive.Focal length, $\mathit{f}=+25\mathrm{cm}$
Object distance, $\mathit{u}=-50\mathrm{cm}$
Image distance, $\mathit{v}=$ ?
Image height, $\mathrm{I}=$ ?
Using lens formula, $\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}$
$\frac{1}{25}=\frac{1}{\mathit{v}}-\frac{1}{\left(-50\right)}$
$\frac{1}{\mathit{v}}=\frac{1}{25}-\frac{1}{50}$
$\mathit{v}=+50\mathrm{cm}$
Also, magnification, $\mathit{m}=-\frac{\mathit{v}}{\mathit{u}}$
Hence, $\mathit{m}=\frac{\left(50\right)}{-\left(50\right)}=-1$
Hence, image is formed at $50\mathrm{cm}$ in front of the lens. The image is of the same size as the object as magnification is 1 . The negative magnification shows that image is inverted.