A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved by approximately placing:

a convex lens of focal length less than 0.25 m

mage can be formed on the screen if it is real. Real image of reduced size can be formed by a concave mirror or a convex lens.

Let u = 2f + x, then

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$⇒\frac{1}{2f+x}+\frac{1}{v}=\frac{1}{f}$

$⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{2f+x}=\frac{f+x}{f(2f+x)}$

$⇒v=\frac{f(2f+x)}{f+x}$

It is given that u + v = 1m.

$2f+x+\frac{f(2f+x)}{f+x}=(2f+x)+\frac{f}{f+x}<1m$

or, $\frac{f(2f+x)^2}{f+x}<1m$

or, $(2f+x)^2<(f+x)$

This will be true only when f < 0.25 m.