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# When an object is placed 20 cm from a concave mirror, a real image magnified three times is formed. Find:**(a)** the focal length of the mirror.**(b) **Where must the object be placed to give a virtual images three times the height of the object?

(a) Given:

Distance of the object from the mirror, $u$ = $-$20 cm

Magnification, $m$ = $-$3 cm

To find: Focal length of the mirror $(f)$.

Solution:

From the magnification formula, we know that-

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

Substituting the given values in the magnification formula we get-

$-3=-\frac{v}{(-20)}$

$-3=\frac{v}{20}$

$v=-60cm$

Thus, the distance of the image, $v$ is 60 cm, 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-

$\frac{1}{f}=\frac{1}{(-60)}+\frac{1}{(-20)}$

$\frac{1}{f}=-\frac{1}{60}-\frac{1}{20}$

$\frac{1}{f}=\frac{-1-3}{60}$

$\frac{1}{f}=-\frac{4}{60}$

$\frac{1}{f}=-\frac{1}{15}$

$f=-60cm$

Thus, the focal length, $f$ is 60 cm, and the negative sign implies that it is in front of the mirror (on the left).

**(b) **If the image is virtual and 3 times magnified

Given:

Focal length of the mirror, $f$ = $-$15 cm

Magnification, $m$ = 3 cm

Distance of the object from the mirror, $u$ = $-$20 cm

To find: Position or the distance of the object from the mirror, $(u)$.

Solution:

From the magnification formula, we know that-

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

Substituting the given values in the magnification formula we get-

$3=-\frac{v}{u}$

$v=-3u$

Thus, the distance of the image, $v$ is 3u, 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-

$\frac{1}{(-15)}=\frac{1}{(-3u)}+\frac{1}{u}$

$\frac{1}{-15}=-\frac{1}{3u}+\frac{1}{u}$

$\frac{1}{-15}=\frac{-1+3}{3u}$

$\frac{1}{-15}=\frac{2}{3u}$

$3u=2\times {(-15)}$

$u=\frac{-30}{3}$

$u=-10cm$

Thus, the object should be placed at a distance of** 10 cm** to get a virtual image three times the height of the object.

The negative sign with the object implies that the object is placed in front of the mirror (on the left).