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A $4.5 \mathrm{~cm}$ needle is placed $12 \mathrm{~cm}$ away from a convex mirror of focal length $15 \mathrm{~cm}$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
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Height of object, $\mathrm{h}=4.5 \mathrm{~cm}, \mathrm{u}=-12 \mathrm{~cm}$.
$\mathrm{f}=15 \mathrm{~cm}, \mathrm{v}=$ ? $\mathrm{m}=$ ?
$$
\begin{aligned}
&\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}=\frac{1}{15}-\frac{1}{-12}=\frac{1}{15}+\frac{1}{12} \Rightarrow \frac{1}{\mathrm{v}}=\frac{4+5}{60}=\frac{9}{60} \\
&\therefore \mathrm{v}=\frac{60}{9}=6.7 \mathrm{~cm} . \quad \mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{\mathrm{v}}{\mathrm{u}} \\
&\therefore \mathrm{h}^{\prime}=-\frac{\mathrm{v}}{\mathrm{u}} \times \mathrm{h}=-\frac{6.7}{-12} \times 4.5=2.5 \mathrm{~cm} .
\end{aligned}
$$
The image is virtual and erect. As the needle is moved farther from the mirror the image moves away from the mirror.
$\mathrm{f}=15 \mathrm{~cm}, \mathrm{v}=$ ? $\mathrm{m}=$ ?
$$
\begin{aligned}
&\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}=\frac{1}{15}-\frac{1}{-12}=\frac{1}{15}+\frac{1}{12} \Rightarrow \frac{1}{\mathrm{v}}=\frac{4+5}{60}=\frac{9}{60} \\
&\therefore \mathrm{v}=\frac{60}{9}=6.7 \mathrm{~cm} . \quad \mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{\mathrm{v}}{\mathrm{u}} \\
&\therefore \mathrm{h}^{\prime}=-\frac{\mathrm{v}}{\mathrm{u}} \times \mathrm{h}=-\frac{6.7}{-12} \times 4.5=2.5 \mathrm{~cm} .
\end{aligned}
$$
The image is virtual and erect. As the needle is moved farther from the mirror the image moves away from the mirror.
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