When an object is placed at a distance of 50 cm 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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Given,
Magnification $\left(m\right)=\frac{-1}{2}$

Distance of the object (u) = −50 cm

We have to find the distance of the object when the magnification is $m=-\frac{1}{5}$.
Therefore, using the magnification formula, we get

$m=\frac{-v}{u}$

v = −mu
$v=-(-\frac{1}{2}\times -50)$
$v=-25\mathrm{cm}$

Now, using the mirror formula, we get

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{-mu}+\frac{1}{u}$

$\frac{1}{f}=\frac{1}{-25}+\frac{1}{-50}$

$$\begin{gathered} \mathrm{o}\mathrm{r}\frac{1}{f}=-\frac{2}{50}-\frac{1}{50}=-\frac{3}{50} \\ \mathrm{or}f=-\frac{50}{3}\mathrm{cm} \end{gathered}$$
Now,
Magnification = $m=-\frac{1}{5}$

Therefore, using the mirror formula, we get
$$\begin{gathered} \frac{1}{f}=\frac{1}{-mu}+\frac{1}{u} \\ \mathrm{or}\frac{-3}{50}=\frac{-5}{-5}+\frac{1}{u} \\ \mathrm{or}\frac{-3}{50}=\frac{5}{u}+\frac{1}{u} \\ \mathrm{or}\frac{-3}{50}=\frac{6}{u} \\ u=-\frac{50\times 6}{3}=-100\mathrm{cm} \\ \mathrm{The}\mathrm{object}\mathrm{should}\mathrm{be}\mathrm{placed}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}100\mathrm{cm}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{magnefiction}\mathrm{of}-\frac{1}{5}. \end{gathered}$$