The magnification of an object placed in front of a convex lens of focal length $20cm$ is $+2$. To obtain a magnification of $-2$, the object has to be moved a distance equal to:

The correct option is B $20cm$

Given that,

Convex lens focal length $f=20cm$

Magnification $M=2$

Then, we know that

Magnification

$M=\frac{v}{u}$

$2=\frac{v}{u}$

$v=2u$

Also, we know that,

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

$\frac{1}{20}=\frac{1}{2u}-\frac{1}{u}$

$\frac{1}{20}=-\frac{1}{2u}$

$\frac{1}{u}=-\frac{1}{10}$

$u=-10cm$

Now, According to given question,

$m^{\prime }=\frac{v^{\prime }}{u^{\prime }}$

$v^{\prime }=2u^{\prime }$

Also,

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

$\frac{1}{20}=-\frac{1}{2u^{\prime }}-\frac{1}{u^{\prime }}$

$-\frac{3}{2u^{\prime }}=\frac{1}{20}$

$u^{\prime }=-30cm.$

So, the object has to moved by $|-30|-|-10|=20cm.$

Hence, this is the answer.