A magnified image of real object is to be obtained on a large screen 1 m from it. This can be achieved by 

Formation of real image for real object is possible for concave mirror and convex lens

$$\begin{gathered} \mathrm{Considering} \mathrm{a} \mathrm{concave} \mathrm{mirror} \\ \mathrm{Object} \mathrm{distance}=\mathrm{u} \\ \mathrm{Image} \mathrm{distance}= \mathrm{u}+1 \\ \mathrm{using} \mathrm{mirror} \mathrm{equation} \\ \frac{1}{-\mathrm{u}}+\frac{1}{-(\mathrm{u}+1)}=\frac{1}{\mathrm{f}} \\ {\mathrm{u}}^2+(2\mathrm{f}+1)\mathrm{u}+\mathrm{f}=0 \\ \mathrm{to} \mathrm{have} \mathrm{real} \mathrm{solutions} \\ \mathrm{we} \mathrm{get} 1+{\mathrm{f}}^2>0 \\ \mathrm{So} \mathrm{for} \mathrm{a} \mathrm{concave} \mathrm{mirror} \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{limitation} \mathrm{on} \mathrm{the} \mathrm{focal} \mathrm{length}. \end{gathered}$$Similarly for convex lens,

The lens is placed in between the object and the screen.

Object distance = u

Image distance= 1-u

On using the lens formula and applying the condition to have real solutions

We get $4\mathrm{f}\le 1$, So for convex lens the focal length must be less than 0.25m