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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, $-\frac {1}{5}$?
CASE-1
Given:
Distance of the object from the mirror $u$ = $-$50 cm
Magnification, $m$ = $\frac{-1}{2}$
To find: Focal length, $(f)$ of the mirror.
Solution:
From the magnification formula, we know that-
$m=-\frac{v}{u}$
Substituting the given values in the magnification formula we get-
$\frac{-1}{2}=-\frac{v}{(-50)}$
$\frac{-1}{2}=\frac{v}{50}$
$2\times{v}=-50$
$v=\frac{-50}{2}$
$v=-25cm$
Thus, the distance of the image, $v$ is 25 cm from the mirror, and the negative sign implies that the image forms in front of the mirror (on the left).
Now, from the mirror formula, we know that-
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Substituting the given values in the mirror formula we get-
$\frac{1}{f}=\frac{1}{(-25)}+\frac{1}{(-50)}$
$\frac{1}{f}=-\frac{1}{25}-\frac{1}{50}$
$\frac{1}{f}=\frac{-2-1}{50}$
$\frac{1}{f}=\frac{-3}{50}$
$\frac{1}{f}=\frac{-3}{50}$
$f=-\frac{50}{3}$
Thus, the focal length of the mirror, $f$ is $\frac{50}{3}cm$, and the negative sign implies that it is in front of the mirror (on the left).
CASE-2
Given:
Magnificataion, $m$ = $\frac{-1}{5}$
Focal length, $f=-\frac{50}{3}$
To find: Distance of the object $(u)$ from the mirror.
Solution:
From the magnification formula, we know that-
$m=-\frac{v}{u}$
Substituting the given values in the magnification formula we get-
$\frac{-1}{5}=-\frac{v}{u}$
$5\times{(-v)}=-u$
$-5v=-u$
$v=\frac{u}{5}$
Thus, the distance of the image, $v$ is $\frac{u}{5}cm$ from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).
Now, from the mirror formula, we know that-
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Substituting the given values in the mirror formula we get-
$\frac{1}{\frac{-50}{3}}=\frac{1}{\frac{u}{5}}+\frac{1}{u}$
$-\frac{3}{50}=\frac{5}{u}+\frac{1}{u}$
$-\frac{3}{50}=\frac{5+1}{u}$
$-\frac{3}{50}=\frac{6}{u}$
$-3u=6\times 50$
$u=-\frac{6\times 50}{3}$
$u=-100cm$
Thus, the distance of the object, $u$ is 100 cm from the mirror, and the negative sign implies that the object is placed in front of the mirror (on the left).
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