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# A concave mirror produces a real image 1 cm tall of an object 2.5 mm tall placed 5 cm from the mirror. Find the position of the image and the focal length of the mirror.

Given:

Distance of the object from the mirror $u$ = $-$5 cm

Height of the image, $h_{2}$ = $-$1 cm

Height of the object, $h_{1}$ = 2.5 mm = 0.25 cm

To find: Distance of the image $(v)$ from the mirror, and the focal length of the mirror $(f)$.

Solution:

From the magnification formula, we know that-

$m=\frac{{h}_{2}}{{h}_{1}}=-\frac{v}{u}$

Substituting the given values in the magnification formula we get-

$\frac{-1}{0.25}=-\frac{v}{(-5)}$

$\frac{-1}{0.25}=\frac{v}{5}$

$v=\frac{-5}{0.25}$

$v=\frac{-5\times {100}}{25}$

$v=-20cm$

Thus, the distance of the image, $v$ is 20 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-

$\frac{1}{f}=\frac{1}{(-20)}+\frac{1}{(-5)}$

$\frac{1}{f}=-\frac{1}{20}-\frac{1}{5}$

$\frac{1}{f}=\frac{-1-4}{20}$

$\frac{1}{f}=\frac{-5}{20}$

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

$f=-4cm$

Thus, the focal length of the mirror, $f$ is 4 cm, and the negative sign implies that it is in front of the mirror (on the left).