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An object $O$ is placed at $7 \mathrm{~cm}$ to the left of a concave mirror of radius of curvature $12 \mathrm{~cm}$ as shown in the figure. The position of the image will be at a distance of
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The correct answer is:
$42 \mathrm{~cm}$ from the mirror to the left.
Given, distance between object and mirror,
$u=-7 \mathrm{~cm}$
Radius of curvature, $R=-12 \mathrm{~cm}$
$\therefore$ Focal length, $f=-\frac{12}{2}=-6 \mathrm{~cm}$
From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Putting the given values,
$\begin{array}{cc} & \frac{1}{v}+\frac{1}{-7}=\frac{1}{-6} \\ & \frac{1}{v}=\frac{1}{-6}+\frac{1}{7}=-\left(\frac{1}{6}-\frac{1}{7}\right) \\ \Rightarrow \quad v & v=42 \mathrm{~cm}\end{array}$
Hence, the position of the image will be $42 \mathrm{~cm}$ from the mirror to the left.
$u=-7 \mathrm{~cm}$
Radius of curvature, $R=-12 \mathrm{~cm}$
$\therefore$ Focal length, $f=-\frac{12}{2}=-6 \mathrm{~cm}$
From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Putting the given values,
$\begin{array}{cc} & \frac{1}{v}+\frac{1}{-7}=\frac{1}{-6} \\ & \frac{1}{v}=\frac{1}{-6}+\frac{1}{7}=-\left(\frac{1}{6}-\frac{1}{7}\right) \\ \Rightarrow \quad v & v=42 \mathrm{~cm}\end{array}$
Hence, the position of the image will be $42 \mathrm{~cm}$ from the mirror to the left.
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