Determine how far an object must be placed in front of a converging lens of focal length 10cm in order to produce an erect (upright) image of linear magnification 4.
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f = 10cm m = +4 (upright image)$\text{m}=\frac{\text{v}}{\text{u}}=4$
$\text{v}=4\text{u}$
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{4\text{u}}-\frac{1}{\text{u}}=\frac{1}{10}$
$\frac{-3}{4\text{u}}=\frac{1}{10}$
$\text{u}=-7.5\text{cm}$
The object must be placed 7.5cm in front of the converging lens.
$\text{v}=4\text{u}$
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{4\text{u}}-\frac{1}{\text{u}}=\frac{1}{10}$
$\frac{-3}{4\text{u}}=\frac{1}{10}$
$\text{u}=-7.5\text{cm}$
The object must be placed 7.5cm in front of the converging lens.
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