Question

An object is placed $0.40m$ from one of the two lenses $L_1$ and $L_2$ of focal lengths $0.20m$ and $0.10m$ respectively, as depicted in the figure. The separation between the lenses is $0.30m$.

The final image formed by these two lenses system is at

• A. 0.13 m to the right of the second lens
• B. 0.05 m to the right of the second lens
• C. 0.13 m to the left of the second lens
• D. infinity

Answer 1

Answer: (D)

Image of first lens acts like object for second lens.
Now, for first lens,
$u=-0.40m,f=+0.20m$

By lens equation,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
or $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{0.2}+\frac{1}{(-0.4)}$
$\Rightarrow \frac{1}{v}=\frac{1}{0.4}$
$\Rightarrow v=0.4m$
Now, this image acts like a virtual object for second lens.
For second lens, $u=+0.1m,f=-0.1m$

By lens equation,
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-0.1}+\frac{1}{0.1}$
$\Rightarrow \frac{1}{v}=0\Rightarrow v=\infty$
Hence, final image is formed at infinity.
Alternate Method For lens $L_1$, $u=-(2f)$ so image is formed at $2f$ distance.
Now, image distance for $L_2$ is $0.1m$ and focal length of $L_2$ is also $0.1m$. Hence, object for $L_2$ is at focus, so its image is formed at infinity.