Question

A diverging lens with magnitude of focal length $25cm$ is placed at a distance of $15cm$ from a converging lens of magnitude of focal length $20cm$. A beam of parallel light falls on the diverging lens. The final image formed is

• A. Real and at a distance of $6cm$ from the convergent lens.
• B. Real and at a distance of $40cm$ from the convergent lens.
• C. Virtual and at a distance of $40cm$ from the convergent lens.
• D. Real and at a distance of $40cm$ from the divergent lens.

Answer 1

The correct option is B Real and at a distance of $40cm$ from the convergent lens.

For concave lens,

$\frac{1}{f_1}=\frac{1}{V_1}-\frac{1}{u_1}$

$\Rightarrow \frac{-1}{25}=\frac{1}{V_1}-\frac{1}{\infty }$

$\Rightarrow V_1=-25cm$

This image acts as an object for convex lens placed at $40cm$ from its principle axis.

For convex lens

$\frac{1}{f_2}=\frac{1}{V_2}-\frac{1}{u_2}$

$\Rightarrow \frac{1}{20}=\frac{1}{V_2}-\frac{1}{(-40)}$

$\Rightarrow \frac{1}{20}=\frac{1}{V_2}+\frac{1}{40}$

$\Rightarrow \frac{1}{V_2}=\frac{1}{20}-\frac{1}{40}$

$\Rightarrow V_2=40cm$