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.

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Updated on: 10-Oct-2022

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