When an object is placed at a distance of 50cm from a concave spherical mirror, the magnification produced is,$-\frac{1}{2}$ Where should the object be placed to get a magnification of, $-\frac{1}{5}?$
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Case 1:$\text{u}= -50\text{cm},$
$\text{m}=-\frac{1}{2}$
$\text{m}=-\frac{\text{v}}{\text{u}}$
$-\frac{1}{2}=\frac{\text{v}}{-50}$
$\text{v}=-25\text{cm}$
We know$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{-25}+\frac{1}{-50}=\frac{1}{\text{f}}$
$\Rightarrow\frac{-3}{-50}=-\frac{1}{\text{f}}$
$\Rightarrow\text{f}=\frac{-50}{3}\text{cm}$
Case 2:
$\text{m}=\frac{1}{2}$
$\text{f}=\frac{-50}{3}$
$\text{m}=-\frac{1}{5}=-\frac{\text{v}}{\text{u}}$
$\text{v}=-\frac{\text{u}}{5}$
Now,$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{5}{\text{u}}+\frac{1}{\text{u}}=\frac{-3}{50}$
$\frac{6}{\text{u}}=-\frac{-3}{50}$
$\text{u}=\frac{600}{-3}=-100\text{cm}$
$\text{m}=-\frac{1}{2}$
$\text{m}=-\frac{\text{v}}{\text{u}}$
$-\frac{1}{2}=\frac{\text{v}}{-50}$
$\text{v}=-25\text{cm}$
We know$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{-25}+\frac{1}{-50}=\frac{1}{\text{f}}$
$\Rightarrow\frac{-3}{-50}=-\frac{1}{\text{f}}$
$\Rightarrow\text{f}=\frac{-50}{3}\text{cm}$
Case 2:
$\text{m}=\frac{1}{2}$
$\text{f}=\frac{-50}{3}$
$\text{m}=-\frac{1}{5}=-\frac{\text{v}}{\text{u}}$
$\text{v}=-\frac{\text{u}}{5}$
Now,$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{5}{\text{u}}+\frac{1}{\text{u}}=\frac{-3}{50}$
$\frac{6}{\text{u}}=-\frac{-3}{50}$
$\text{u}=\frac{600}{-3}=-100\text{cm}$
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