Question

An object is placed $0.40\, m$ from one of the two lenses $L_{1}$ and $L_{2}$ of focal lengths $0.20\, m$ and $0.10\, m$ respectively, as depicted in the figure. The separation between the lenses is $0.30\, m$.

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,
$u=-0.40\, m,\, f=+0.20\, m$

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

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 $L_{1}$, $u=-(2 f)$ so image is formed at $2 f$ distance.
Now, image distance for $L_{2}$ is $0.1 m$ and focal length of $L_{2}$ is also $0.1 m$. Hence, object for $L_{2}$ is at focus, so its image is formed at infinity.