An object is placed at a distance of 6cm from a convex mirror of focal length 12cm. Find the position and nature of the image.
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u = -6cm, f = 12cm, v = ? We know that $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{-6}=\frac{1}{12}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{12}+\frac{1}{6}=\frac{3}{12}=\frac{1}{4}$
$\therefore\text{v}=\frac{1}{4}\text{cm}$
Image will be formed 4cm the mirror. Since the image is formed behind the convex mirror, it is virtual and erect.
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{-6}=\frac{1}{12}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{12}+\frac{1}{6}=\frac{3}{12}=\frac{1}{4}$
$\therefore\text{v}=\frac{1}{4}\text{cm}$
Image will be formed 4cm the mirror. Since the image is formed behind the convex mirror, it is virtual and erect.
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