Question

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

• A. $10\text{cm}$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
• B. $20\text{cm}$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
• C. $30\text{cm}$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
• D. $40\text{cm}$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->

Answer 1

The correct option is B $20\text{cm}$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->

Given,

$f=+20\text{cm}{m}_1=+2$

$\therefore m_1=\frac{v}{u}=\frac{f}{f+u_1}$

$\Rightarrow \frac{20}{20+u_1}=2$

$\Rightarrow 20=40+2u_1$

$\therefore u_1=-10\text{cm}$

Now for, $m_2=-2$

$\Rightarrow m_2=\frac{v}{u_2}=\frac{f}{f+u_2}$

$\Rightarrow \frac{20}{20+u_2}=-2$

$\Rightarrow 20=-40-2u_2$

$\therefore u_2=-30\text{cm}$

$\therefore$ Object must be moved by,

$d=\left|u_2\right|-\left|u_1\right|=|-30|-|-10|=20\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(b\right)$ is correct.

Why this question ? To understand the concept of lens formula $\&$ magnification formula.