A concave mirror of focal length 1.5m forms an image of an object placed at a distance of 40cm. Find the position and nature of the image.
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$\text{f}=-15\text{m}$$\text{u}=-40\text{cm}$
$=\frac{-40\text{cm}}{100}$
$=\frac{-2}{5}\text{m}$
$\frac{1}{\text{v}}=\frac{1}{\text{F}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{10}{-15\text{m}}-\frac{5}{-2}$
$\frac{1}{\text{v}}=\frac{-2\text{m}}{3}+\frac{5}{2}\text{m}$
$\frac{1}{\text{v}}=\frac{-4\text{m}+15\text{m}}{6}$
$\frac{1}{\text{v}}=\frac{-11\text{m}}{6}$
$\frac{-6\text{m}}{11}=\text{v}$
$\frac{-600\text{cm}}{11}=\text{v}$
$=54.54\text{cm}=\text{v}$
$=\frac{-40\text{cm}}{100}$
$=\frac{-2}{5}\text{m}$
$\frac{1}{\text{v}}=\frac{1}{\text{F}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{10}{-15\text{m}}-\frac{5}{-2}$
$\frac{1}{\text{v}}=\frac{-2\text{m}}{3}+\frac{5}{2}\text{m}$
$\frac{1}{\text{v}}=\frac{-4\text{m}+15\text{m}}{6}$
$\frac{1}{\text{v}}=\frac{-11\text{m}}{6}$
$\frac{-6\text{m}}{11}=\text{v}$
$\frac{-600\text{cm}}{11}=\text{v}$
$=54.54\text{cm}=\text{v}$
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