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# An object 3 cm high is placed at a distance of 10 cm in front of a converging mirror of focal length 20 cm. Find the position, nature and size of the image formed.

**Given:**

Distance of the object from the mirror $u$ = $-$10 cm

Height of the object, $h_{1}$ = 3 cm

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

**To find: **Distance of the image $(v)$ from the mirror, and the height of the image $(h_2)$.

**Solution:**

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-

$\frac{1}{(-20)}=\frac{1}{v}+\frac{1}{(-10)}$

$-\frac{1}{20}=\frac{1}{v}-\frac{1}{10}$

$\frac{1}{10}-\frac{1}{20}=\frac{1}{v}$

$\frac{1}{v}=\frac{2-1}{20}$

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

$v=+20cm$

Thus, the distance of the image $v$ is 20 cm, and the positive sign implies that the image forms behind the mirror (on the right).

Now, from the magnification formula, we know that-

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

Substituting the given values in the magnification formula we get-

$\frac{{h}_{2}}{3}=-\frac{20}{(-10)}$

$\frac{{h}_{2}}{3}=\frac{20}{10}$

$\frac{{h}_{2}}{3}=\frac{2}{1}$

${{h}_{2}}=3\times {(2)}$

${{h}_{2}}=+6cm$

Thus, the height of the image $h_{2}$ is **6 cm**, and the positive sign implies that the image forms above the principal axis (upwards).

Again, using the magnification formula, we get-

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

$m=-\frac{20}{(-10)}$

$m=\frac{20}{10}$

$m=+2$

Thus, the magnification of the image $m$ is 2 cm, and the positive sign implies that the image is **virtual and erect.**

Hence, the position of the image is **20cm behind the mirror**, its nature is virtual and erect, and its size is large.