The convex lens $\left(L_1\right.$ and $\left.L_2\right)$ of equal focal length $f$ are placed at a distance $\frac{f}{2}$ apart. An object is placed at a distance $4 f$ in the left of $L_1$ as shown in figure. The final image is at

• Correct answer
  A

  $\frac{5 \mathrm{f}}{11}$ right of $\mathrm{L}_2$
• B

  $\frac{5 f}{11}$ left of $L_2$
• C

  $5 f$ right of $\mathrm{L}_2$
• D

  5 left of $L_2$

Solution

$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{v}-\frac{1}{-4 f}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{4 f} \Rightarrow \frac{1}{v}=\frac{4-1}{4 f} \Rightarrow v=\frac{4 f}{3} \\ & u_2=\frac{4 f}{3}-\frac{f}{2}=\frac{8 f-3 f}{6} \Rightarrow u_2=\frac{5 f}{6} \\ & \frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f} \Rightarrow \frac{1}{v_2}-\frac{6}{5 f}=\frac{1}{f} \Rightarrow \frac{1}{v_2}=\frac{1}{f}+\frac{6}{5 f} \\ & \frac{1}{v_2}=\frac{11}{5 f} \Rightarrow v_2=\frac{5 f}{11}\end{aligned}$

Hence, the correct answer is option: 'A'.