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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).