# 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.

Updated on: 10-Oct-2022

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