Focal length $= \frac{1}{P} = \frac{1}{4} = 0.25 \mathrm{m}$ or $25 \mathrm{cm}$. It is a convex lens as power and focal length are positive.Focal length, $f = +25 \mathrm{cm}$Object distance, $u = -50 \mathrm{cm}$Image distance, $v =$ ?Image height, $I =$ ?Using lens formula, $\; \frac{1}{f} = \; \frac{1}{v} - \; \frac{1}{u}$$\; \frac{1}{25} = \; \frac{1}{v} - \; \frac{1}{-50}$$\; \frac{1}{v} = \; \frac{1}{25} - \; \frac{1}{50}$$v \; = +50 \mathrm{cm}$Also, magnification, $m = - \frac{v}{u}$Hence, $\; \; m = \frac{50}{-50} = -1$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.