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# An arrow 2.5 cm high is placed at a distance of 25 cm from a perging mirror of focal length 20 cm. Find the nature, position and size of the image formed.

The mirror is a diverging mirror, which means a **convex mirror.**

Distance of the object from the mirror, $u$ = $-$25 cm

Focal length of the mirror, $f$ = 20 cm

Height of the object, $h_1$ = 2.5 cm

To find: Distance or position of the image, $v$, height of the image $h_2$ and its magnification $m$.

Solution:

From the mirror formula, we know that-

$\frac {1}{f}=\frac {1}{v}+\frac {1}{u}$

Substituting the given values we get-

$\frac {1}{20}=\frac {1}{v}+\frac {1}{(-25)}$

$\frac {1}{20}=\frac {1}{v}-\frac {1}{25}$

$\frac {1}{25}+\frac {1}{20}=\frac {1}{v}$

$\frac {1}{v}=\frac {4+5}{100}$

$\frac {1}{v}=\frac {9}{100}$

$v=\frac {100}{9}$

$v=+11.1cm$

Thus, the distance of the image $v$ is 11.1 cm from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

Now, from the magnification formula, we know that-

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

Substituting the given values we get-

$m=-\frac {11.1}{(-25)}$

$m=\frac {11.1}{25}$

$m=+0.44$

Thus, the magnification is $0.44$ which is less than 1, which means the image is small in size, and the positive sign implies that the image is virtual and erect.

Hence, the image is virtual, erect, and small in size.

Again, using magnification formula we get-

$m=\frac {h_2}{h_1}$

$0.44=\frac {h_2}{2.5}$

$h_2=2.5\times {0.44}$

$h_2=+1.1cm$

Thus, the height of the image $h_2$ is** 1.1cm**, and the positive sign implies that it is above the principal axis (upwards).