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# If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm, find the position, nature and height of the image.

**Given:**

**To find:**Distance of the image from the mirror, $v$ and height of the image, $h_{2}$

**Solution:**

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

Substituting the given values in the mirror formula we get-

$\frac{1}{-12}=\frac{1}{v}+\frac{1}{(-36)}$

$\frac{1}{-12}=\frac{1}{v}-\frac{1}{(36)}$

$\frac{1}{v}=\frac{1}{36}-\frac{1}{12}$

$\frac{1}{v}=\frac{1-3}{36}$

$\frac{1}{v}=\frac{-2}{36}$

$\frac{1}{v}=-\frac{1}{18}$

$v=-18cm$

Hence, the distance of the image from the mirror, $v$ is** -18 cm, **which means that the position of the image is 18 cm **in front of the mirror.**

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-

$\frac{{h}_{2}}{10}=-\frac{(-18)}{(-36)}$

$\frac{{h}_{2}}{10}=-\frac{1}{2}$

${h}_{2}=-\frac{10}{2}$

${h}_{2}=-5cm$

Hence, height of the image, $h_{2}$, is **-5 cm, **which means that the** image is real and inverted.**

Again using the magnification formula, we get-

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

$m=\frac{-(-18)}{36}$

$m=\frac{-1}{2}$

Hence, the magnification, $m$ is $\frac{-1}{2}$, which means the image is **small in size.**

Thus, the image is real, inverted, and small in size.