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A point object is placed on the axis of a thin convex lens of focal length $0.05 \mathrm{m}$ at a distance of $0.2 \mathrm{m}$ from the lens and its image is formed on the axis. If the object is now made to oscillate along the axis with a small amplitude of $A \mathrm{cm},$ then what is the amplitude of oscillation of the image? $\left[\right.$ you may assume, $\frac{1}{1+x} \approx 1-x,$ where $\left.x< < 1\right]$
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Verified Answer
The correct answer is:
$\frac{A}{9} \times 10^{-2} \mathrm{m}$
According to the question, we can draw the following diagram
Given, $u=-0.2 \mathrm{m}$ and $f=0.05 \mathrm{m}$
As we know that,
$$
\begin{array}{l}
\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\
\frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\
\frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.2} \\
\frac{1}{v}=\frac{100}{5}-\frac{10}{2} \\
\frac{1}{v}=20-5 \Rightarrow v=\frac{1}{15} \mathrm{m}
\end{array}
$$
Now, differentiating eq. (i), we get
$$
\begin{array}{l}
-\frac{d v}{v^{2}}=-\frac{d u}{u^{2}} \quad \therefore \quad d v=d u \frac{v^{2}}{u^{2}} \\
A_{\max }=A \times\left(\frac{1}{15}\right)^{2} \times\left(\frac{1}{(-0.2)}\right)^{2} \\
A_{\max }=A \times \frac{1}{225} \times 25 \\
A_{\max }=\frac{A}{9}
\end{array}
$$
Here $A$ is in $\mathrm{cm} .$
Hence, $A_{\max }=\frac{A}{9} \times 10^{-2} \mathrm{m}$
Given, $u=-0.2 \mathrm{m}$ and $f=0.05 \mathrm{m}$
As we know that,
$$
\begin{array}{l}
\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\
\frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\
\frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.2} \\
\frac{1}{v}=\frac{100}{5}-\frac{10}{2} \\
\frac{1}{v}=20-5 \Rightarrow v=\frac{1}{15} \mathrm{m}
\end{array}
$$
Now, differentiating eq. (i), we get
$$
\begin{array}{l}
-\frac{d v}{v^{2}}=-\frac{d u}{u^{2}} \quad \therefore \quad d v=d u \frac{v^{2}}{u^{2}} \\
A_{\max }=A \times\left(\frac{1}{15}\right)^{2} \times\left(\frac{1}{(-0.2)}\right)^{2} \\
A_{\max }=A \times \frac{1}{225} \times 25 \\
A_{\max }=\frac{A}{9}
\end{array}
$$
Here $A$ is in $\mathrm{cm} .$
Hence, $A_{\max }=\frac{A}{9} \times 10^{-2} \mathrm{m}$
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