An image $\frac{2}{3}\text{rd}$ the size of object is formed by a convex lens at a distance of 12cm from Find the focal length of the lens.
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$\text{v}=12\text{cm}$For real and inverted image the magnification will be negative.
Therefore, $\text{m}=\frac{-2}{3}$
$\text{m}=\frac{-2}{3},\text{ v}=+12\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{-2}{3}=\frac{12}{\text{u}}$
$-2\text{u}=36$
$\text{u}=-18\text{cm}$
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{1}{12}-\frac{1}{-18}=\frac{3+2}{36}$
$\text{f}=\frac{36}{5}=7.2\text{cm}$
Therefore, $\text{m}=\frac{-2}{3}$
$\text{m}=\frac{-2}{3},\text{ v}=+12\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{-2}{3}=\frac{12}{\text{u}}$
$-2\text{u}=36$
$\text{u}=-18\text{cm}$
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{1}{12}-\frac{1}{-18}=\frac{3+2}{36}$
$\text{f}=\frac{36}{5}=7.2\text{cm}$
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