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A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror:(i) Calculate the image distance.(ii) What is the focal length of the mirror?
Given:
Distance of the object from the mirror $u$ = $-$20 cm
Height of the object, $h_{1}$ = 1 cm
Height of the object, $h_{2}$ = $-$4 cm
Focal length of the mirror, $f$ = $-$18 cm
To find: Distance of the image $v$, and the focal length of the mirror $f$.
Solution:
A concave mirror is a converging mirror.
So, 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-
$\frac{-4}{1}=-\frac{v}{(-20)}$
$\frac{-4}{1}=\frac{v}{20}$
$v=-80cm$
Thus, the distance of the image $v$ is 80 cm, and the minus sign implies that the image is on the same side as the object (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-
$\frac{1}{f}=\frac{1}{(-80)}+\frac{1}{(-20)}$
$\frac{1}{f}=-\frac{1}{(80)}-\frac{1}{(20)}$
$\frac{1}{f}=\frac{-1-4}{80}$
$\frac{1}{f}=\frac{-5}{80}$
$\frac{1}{f}=\frac{-1}{16}$
$f=-16cm$
Thus, the focal length $f$ of the concave mirror is 16 cm, and the minus sign implies that the focus is on the left side.