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# An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of image.

Given:

Distance of the object from the mirror $u$ = $-$27 cm

Height of the object, $h_{1}$ = 7 cm

Focal length of the mirror, $f$ = $-$18 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-

$\frac{1}{(-18)}=\frac{1}{(v)}+\frac{1}{-27}$

$-\frac{1}{18}=\frac{1}{(v)}-\frac{1}{27}$

$\frac{1}{27}-\frac{1}{18}=\frac{1}{(v)}$

$\frac{2-3}{54}=\frac{1}{(v)}$

$\frac{-1}{54}=\frac{1}{(v)}$

$v=-54cm$

Thus, the distance of the image $v$ is** 54 cm, **and the minus sign implies that the image is on the same side as the object (in the left).

Therefore, the screen should be placed at a distance of 54 cm in front of the concave mirror so that a sharply focussed image can be obtained.

**
**

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-

$\frac{{h}_{2}}{7}=-\frac{(-54)}{(-27)}$

$\frac{{h}_{2}}{7}=-\frac{2}{1}$

$\frac{{h}_{2}}{7}=-14cm$

Thus, the height of the image $h_{2}$ is** 14 cm, **and the minus sign implies that the image is inverted or below the principal axis.

Again, using the magnification formula, we get-

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

$m=-\frac{(-54)}{(-27)}$

$m=-2$

Thus, the magnification of the image $m$ is 2, and the minus sign implies that the image is real and inverted.

Hence, the size and nature of the image are real, inverted, and large in size.

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