At what distance from a converging lens of focal length 12cm must an object be placed in order that an image of magnification 1 will be produced?
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$\text{f}=12\text{cm}$$\text{m}=1$
$\text{m}=\frac{\text{v}}{\text{u}}=1$
$\text{v}=\text{u}$
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
Putting the value of v, u and f
$\frac{1}{\text{u}}-\frac{1}{-\text{u}}=\frac{1}{12}$ (image distance is negative)
$\frac{2}{\text{u}}=\frac{1}{12}$
$\text{u}=24\text{cm}$
The object should be placed at a distance of 24cm to from the lens (on the left side).
$\text{m}=\frac{\text{v}}{\text{u}}=1$
$\text{v}=\text{u}$
Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
Putting the value of v, u and f
$\frac{1}{\text{u}}-\frac{1}{-\text{u}}=\frac{1}{12}$ (image distance is negative)
$\frac{2}{\text{u}}=\frac{1}{12}$
$\text{u}=24\text{cm}$
The object should be placed at a distance of 24cm to from the lens (on the left side).
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