A concave mirror has a focal length of 4 cm and an object 2 cm tall is placed 9 cm away from it. Find the nature, position and size of the image formed.

Given:

​Distance of the object from the mirror, $u$ = $-$9cm

Height of the object, $h_1$ = 2 cm

Focal length of the mirror, $f$ = $-$4 cm

To find: Distance of the image $(v)$ and 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-

$\frac{1}{(-4)}=\frac{1}{v}+\frac{1}{(-9)}$

$-\frac{1}{4}=\frac{1}{v}-\frac{1}{9}$

$\frac{1}{9}-\frac{1}{4}=\frac{1}{v}$

$\frac{1}{v}=\frac{4-9}{36}$

$\frac{1}{v}=\frac{-5}{36}$

$v\times {(-5)}=36$

$v=-\frac{36}{5}$

$v=-7.2cm$

Thus, the distance of the image, $v$ is 7.2 cm from the mirror, and the negative sign implies that the image forms in front of the mirror (on the left).

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-

$\frac{{h}_{2}}{2}=-\frac{(-7.2)}{(-9)}$

$h_2=\frac{2\times (-7.2)}{(-9)}$

$h_2=-1.6cm$

Thus, the height of the image, $h_2$ is 1.6 cm, and the negative sign implies that the image forms below the principal axis (downwards).

Again, using the magnification formula, we get-

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

$m=-\frac{(-7.2)}{(-9)}$

$m=-0.8$

Thus, the magnification, $m$ of the image is 0.8, and the negative sign implies that the image is real & inverted.

Hence, the nature of the image is real & inverted, its position is 7.2 cm in front of the mirror (to the left of the mirror) and its size is small.

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Updated on: 10-Oct-2022

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