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A rod of length $10 \mathrm{~cm}$ lies along the principal axis of a concave mirror of focal length $10 \mathrm{~cm}$ in such a way that its end closer to the pole is $20 \mathrm{~cm}$ away from the mirror. The length of the image is
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Verified Answer
The correct answer is:
$5 \mathrm{~cm}$
By mirror formula, image distance of $A$
$$
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
$$
$$
\begin{aligned}
\frac{1}{v_A}+\frac{1}{u} & =\frac{1}{f} \\
\frac{1}{v_A}+\frac{1}{(-30)} & =\frac{1}{-10} \\
v_A & =-15 \mathrm{~cm}
\end{aligned}
$$
Also image distance of $C$
$$
v_C=-20 \mathrm{~cm}
$$
The length of image $=\left|v_A-v_C\right|$
$$
=|-15-(-20)|
$$
$$
=5 \mathrm{~cm}
$$
$$
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
$$
$$
\begin{aligned}
\frac{1}{v_A}+\frac{1}{u} & =\frac{1}{f} \\
\frac{1}{v_A}+\frac{1}{(-30)} & =\frac{1}{-10} \\
v_A & =-15 \mathrm{~cm}
\end{aligned}
$$
Also image distance of $C$
$$
v_C=-20 \mathrm{~cm}
$$
The length of image $=\left|v_A-v_C\right|$
$$
=|-15-(-20)|
$$
$$
=5 \mathrm{~cm}
$$
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