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A point source of light is moving at a rate of $2 \mathrm{~cm}-\mathrm{s}^{-1}$ towards a thin convex lens of focal length $10 \mathrm{~cm}$ along its optical axis. When the source is $15 \mathrm{~cm}$ away from the lens the image is moving at
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The correct answer is:
$8 \mathrm{~cm}-\mathrm{s}^{-1}$ away from the lens
$$
\begin{array}{l}
\frac{d v}{d t}=\frac{-v^{2}}{u^{2}} \frac{d u}{d t} \\
\& \frac{1}{v}=\frac{1}{F}+\frac{1}{u} \\
\frac{1}{\mathrm{~V}}=\frac{1}{10}+\frac{1}{-15} \\
\frac{1}{v}=\frac{3-2}{30} \\
\mathrm{v}=30 \mathrm{~cm}
\end{array}
$$
$$
\frac{\mathrm{dv}}{\mathrm{dt}}=\left(\frac{30}{15}\right)^{2} \cdot 2
$$
$=8 \mathrm{~cm} / \mathrm{s}$ away from lens
\begin{array}{l}
\frac{d v}{d t}=\frac{-v^{2}}{u^{2}} \frac{d u}{d t} \\
\& \frac{1}{v}=\frac{1}{F}+\frac{1}{u} \\
\frac{1}{\mathrm{~V}}=\frac{1}{10}+\frac{1}{-15} \\
\frac{1}{v}=\frac{3-2}{30} \\
\mathrm{v}=30 \mathrm{~cm}
\end{array}
$$
$$
\frac{\mathrm{dv}}{\mathrm{dt}}=\left(\frac{30}{15}\right)^{2} \cdot 2
$$
$=8 \mathrm{~cm} / \mathrm{s}$ away from lens
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