Question

A convex lens of focal length $f$ is cut into two halves each of which is displaced equally from the axis as shown an object $O$ is kept at a distance $f/2$ on the axis of the original lens. Distance between the images formed is

• A. $X/4$
• B. $4X$
• C. $2X$
• D. $X/2$

Answer 1

The correct option is C $2X$

$\begin{array}{c}\text{for}\begin{array}{cc}{L}_1& :-\\ u& =-f/2,h_0=-x\end{array}\\ \begin{array}{cc}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}& \\ \Rightarrow \frac{1}{v}=\frac{1}{u}+\frac{1}{f}& \Rightarrow \frac{1}{v}=-\frac{2}{f}+\frac{1}{f}\\ & \Rightarrow v=-f\end{array}\end{array}$

$\begin{array}{c}\frac{hi}{h_0}=\frac{v}{u}=\frac{-f\times 2}{-f}\\ \Rightarrow h_i=2h_0=-2x\end{array}$

for $L_2$ :

$\begin{array}{c}u=-f/2,h_0=x\\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow v=-f\\ \Rightarrow \frac{1}{v}=\frac{1}{u}+\frac{1}{f}\Rightarrow \\ \frac{h_i}{h_0}=\frac{v}{u}=2\\ \Rightarrow h_i=2h_0=2x\end{array}$

Distance betweer both the image as shown

in the figure $=2x$