Question

A convex lens produces an image of an object on a screen with magnification of $\frac{1}{2}$. When the lens is moved $30cm$ away from the object, the magnification of the image is $2$. The focal length of the lens is

• A. $20cm$
• B. $25cm$
• C. $30cm$
• D. $35cm$

Answer 1

The correct option is A $20cm$

By magnifation formula $\Rightarrow \frac{y}{\sqrt{4}}=\frac{v}{2}$

care $2v=u-\left(3\right)$

$\Rightarrow \frac{v}{(u+30)}=2=\frac{v^{\prime }}{(30+u)}\frac{cose(2)}{2(30+u)=v^{\prime }}$

$\begin{array}{c}\text{By lens formula}\to \\ \Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{\frac{1}{M}-\frac{1}{u}}{=}\frac{1}{f}\text{for case}\left(1\right)\\ \Rightarrow \frac{1}{v^{\prime }}-\frac{1}{(u+30)}=\frac{1}{f}\Rightarrow \frac{1}{2(30+u)}-\frac{1}{(30+u)}=\frac{1}{f}]-6\end{array}$

(5) $\to \frac{1}{4}=\frac{1}{f};u=f(-7$

putting $7$ in 6 $\frac{-1}{2(30+4)}=\frac{1}{f}\Rightarrow$ $\frac{-4}{2(30+f)}=\frac{1}{f}$

$\begin{array}{cc}-f=2(30+f)\Rightarrow & \therefore -f=\sin m+2f\\ \Rightarrow & -3f=60\therefore f=20cm\end{array}$

Hence obtion $A^{\prime }$ is correct