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)

$4$

Given distance between object and image is $0.72m$

Hence, $|u|+|v|=0.72...........\left(i\right)$

Also, given magnitude of magnification is $2$

$\therefore |m|=\frac{|v|}{|u|}=2...........\left(ii\right)$

Solving (i) & (ii), we get

$|u|=0.24,|v|=0.48$

As image is real, $u=-0.24,v=0.48$

Using lens formula,

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

$\frac{1}{0.48}-\frac{1}{-0.24}=\frac{1}{f}$

Solving, $f=0.16m$

New object distance is: $u=-0.24+0.04=-0.2m$

Using lens formula,

$\frac{1}{v^{\prime }}-\frac{1}{u^{\prime }}=\frac{1}{f}$

$\frac{1}{v^{\prime }}-\frac{1}{-0.2}=\frac{1}{0.16}$

$⟹v^{\prime }=0.8m$

Correspondingly, magnitude of magnification is:

$m=\frac{|v|}{|u|}=\frac{0.8}{0.2}=4$