Q. 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) A convex mirror of suitable focal length
B) A concave mirror of suitable focal length
C) A convex lens of focal length less than 0.25 m
D) A concave lens of suitable focal length
Explanation

(c) Image 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
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\Rightarrow \frac{1}{2f + x} + \frac{1}{v} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{2f + x} = \frac{f + x}{f(2f + x)}$
$\Rightarrow v = \frac{f(2f + x)}{f + x}$
It is given that $u + v = 1.0\ m$
$2f + x + \frac{f(2f + x)}{f + x} = (2f + x)\left\lbrack 1 + \frac{f}{f + x} \right\rbrack < 1.0$m
Or $\ \frac{(2f + x)^{2}}{f + x} < 1.0$m
Or ${\ (2f + x)}^{2} < (f + x)$
This will be true only when $f < 0.25\ m$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.