When an object is placed at distance of 50 cm from concave mirror the magnification produced is 1 2

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-

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

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

$\frac{1}{f}=\frac{-2-1}{50}$

$\frac{1}{f}=\frac{-3}{50}$

$f=-\frac{50}{3}$

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).

Given,
Magnification `(m)=-1/2`

Distance of the object (u) = −50 cm

We have to find the distance of the object when the magnification is `=-1/5`
Therefore, using the magnification formula, we get

`m=-v/u`

v=-mu

`v=-(-1/2xx-50) `

`v=-25` cm

Now, using the mirror formula, we get

`1/f=1/u+1/v`

`1/f=1/-"mu"+1/u`

`1/f=1/-25+1/-50`

or` 1/f=-2/50-1/50=3/-50`