At what distance from a concave mirror of focal length 10cm should an object 2cm long be placed in order to get an erect image 6cm tall?
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$f=-10 cm, h_1=2 cm, h_2=6 cm$ (erect image) u = ? We know that:$\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{6}{2}=3$
and
$\text{m}=-\frac{\text{v}}{\text{u}}=3$
$\Rightarrow3\text{u}=-\text{v}$
$\Rightarrow\text{v}=-3\text{u}\ ....\text{(A)}$
We have,$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{(-3\text{u})}+\frac{1}{\text{u}}=\frac{1}{10}$
$\Rightarrow\frac{1}{\text{u}}-\frac{1}{\text{3u}}=-\frac{1}{10}$
$\Rightarrow\frac{2}{\text{3u}}=-\frac{1}{10}$
$\Rightarrow\text{u}=-\frac{20}{3}= 6.66\text{cm}$
The object should be palced at a distance of 6.66cm on the left side of the mirror.
and
$\text{m}=-\frac{\text{v}}{\text{u}}=3$
$\Rightarrow3\text{u}=-\text{v}$
$\Rightarrow\text{v}=-3\text{u}\ ....\text{(A)}$
We have,$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{(-3\text{u})}+\frac{1}{\text{u}}=\frac{1}{10}$
$\Rightarrow\frac{1}{\text{u}}-\frac{1}{\text{3u}}=-\frac{1}{10}$
$\Rightarrow\frac{2}{\text{3u}}=-\frac{1}{10}$
$\Rightarrow\text{u}=-\frac{20}{3}= 6.66\text{cm}$
The object should be palced at a distance of 6.66cm on the left side of the mirror.
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