Question

The distance between an object and its real image produced by a converging lens is $0.72m$. The magnification is $2$. What will be the magnification when the object is moved by $0.04m$ towards the lens ?

• A. 2
• B. 4
• C. 3
• D. 6

Answer 1

Answer: (B)

Given, $m=2$ and the image formed is real
$\therefore$ For a real image $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}(\because u$ is negative $)$
or $1+\frac{1}{m}=\frac{u}{f}$
$\Rightarrow 1+\frac{1}{2}=\frac{u}{f}$
$\Rightarrow \frac{3}{2}=\frac{0.72}{f}$
or$f=\frac{0.72\times 2}{3}=0.16m$
As the object is moved by $0.04m$ towards the lens, the new
$u_1=0.72-0.04=0.68m$
Again real image is formed
So, $1+\frac{1}{m}=\frac{u_1}{f}$
$\Rightarrow \frac{1}{m}=\frac{0.04}{0.16}$
$\Rightarrow m=4$