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The size of the image of an object, which is at infinity, as formed by a convex lens of focal length $30 \mathrm{~cm}$ is $2 \mathrm{~cm}$. If a concave lens of focal length $20 \mathrm{~cm}$ is placed between the convex lens and the image at a distance of $26 \mathrm{~cm}$ from the lens, the new size of the image is
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Verified Answer
The correct answer is:
$2.5 \mathrm{~cm}$
Given, $f_{1}=30 \mathrm{~cm}$,
$\begin{aligned}
&f_{2}=-20 \mathrm{~cm}, h_{0}=2 \mathrm{~cm} \\
&u_{2}=(30-26)=4 \mathrm{~cm}
\end{aligned}$
Using lens formula for concave lens,
$\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}=-\frac{1}{20}+\frac{1}{4}=\frac{4}{20}=\frac{1}{5}$
or $v_{2}=5 \mathrm{~cm}$
Magnification, $m=\frac{v_{2}}{u_{2}}=\frac{5}{4}=1.25$
Also, $m=\frac{h_{i}}{h_{o}}=1.25$
$\therefore$ Size of new image,
$h_{i}=h_{0} \times 1.25=2 \times 1.25=2.5 \mathrm{~cm}$
$\begin{aligned}
&f_{2}=-20 \mathrm{~cm}, h_{0}=2 \mathrm{~cm} \\
&u_{2}=(30-26)=4 \mathrm{~cm}
\end{aligned}$
Using lens formula for concave lens,
$\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}=-\frac{1}{20}+\frac{1}{4}=\frac{4}{20}=\frac{1}{5}$
or $v_{2}=5 \mathrm{~cm}$
Magnification, $m=\frac{v_{2}}{u_{2}}=\frac{5}{4}=1.25$
Also, $m=\frac{h_{i}}{h_{o}}=1.25$
$\therefore$ Size of new image,
$h_{i}=h_{0} \times 1.25=2 \times 1.25=2.5 \mathrm{~cm}$
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