A dentist’s mirror has a radius of curvature of 3cm. How far must it be placed from a small dental cavity to give a virtual image of the cavity that is magnified five times?
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R = -3cm (concave mirror) m = (virtual image)$\text{f}=\frac{\text{R}}{2}=\frac{3}{2}=1.5\text{cm}$
and
$\text{m}=5=-\frac{\text{v}}{\text{u}}$
$\Rightarrow\text{v}=-5\text{u}$
We have$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{(5\text{u})}+\frac{1}{\text{u}}=\frac{1}{(-1.5)}$
$\Rightarrow \frac{4}{5\text{u}}=-\frac{1}{1.5}$
$\Rightarrow \text{u}=-\frac{4\times1.5}{5}=-1.2\text{cm}$
The mirror should be placed 1.2cm away from the dental cavity.
and
$\text{m}=5=-\frac{\text{v}}{\text{u}}$
$\Rightarrow\text{v}=-5\text{u}$
We have$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{(5\text{u})}+\frac{1}{\text{u}}=\frac{1}{(-1.5)}$
$\Rightarrow \frac{4}{5\text{u}}=-\frac{1}{1.5}$
$\Rightarrow \text{u}=-\frac{4\times1.5}{5}=-1.2\text{cm}$
The mirror should be placed 1.2cm away from the dental cavity.
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