An object of size 7.0cm is placed at 27cm in front of a concave mirror of focal length 18cm. 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.
[Hint. Find the value of image distance (v) first. The screen should be placed from the mirror at a distance equal to image distance.]
[Hint. Find the value of image distance (v) first. The screen should be placed from the mirror at a distance equal to image distance.]
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Given: $h _1=7 cm, u =-27 cm, f =18 cm$ We know that$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{1}{(-18)}-\frac{1}{(-27)}$
$\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}=\frac{1}{54}$
$\therefore\text{v}=-54\text{cm}$
The screen shoud be palce at a distance of 54cm in front of the concave mirror.$\text{m}=-\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$-\frac{(-54)}{(-27)}=\frac{\text{h}_2}{7}$
$\text{h}_2=-14\text{cm}$
Image is 14cm in size, real and inverted.
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{1}{(-18)}-\frac{1}{(-27)}$
$\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}=\frac{1}{54}$
$\therefore\text{v}=-54\text{cm}$
The screen shoud be palce at a distance of 54cm in front of the concave mirror.$\text{m}=-\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$-\frac{(-54)}{(-27)}=\frac{\text{h}_2}{7}$
$\text{h}_2=-14\text{cm}$
Image is 14cm in size, real and inverted.
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