Question

A convex lens is given, for which the minimum distance between an object and its rel image is $40cm$. An object is placed at a distance of $15cm$ from this lens. The liner magnification of adjustment will be

• A. $\frac{5}{3}$
• B. $-2$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{5}{3}$

Given object distance $u=15$ cm

Distance between object and real image produced $=40$cm

Thus image distance $v=40-15=25$ cm

Also we know linear magnification,

$m=\frac{-v}{u}=\frac{-25}{-15}=\frac{5}{3}$