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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.