An object of 5.0cm size is placed at a distance of 20.0cm from a converging mirror of focal length 15.0cm. At what distance from the mirror should a screen be placed to get the sharp image? Also calculate the size of the image.
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$h _1=5.0 cm, u =-20 cm, f =15 cm, v=?, h_2=?$ We know that$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}+\frac{1}{(-20)}=\frac{1}{(-15)}$
$\frac{1}{\text{v}}=\frac{1}{20}-\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{1}{20}-\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{-5}{300}$
$\text{v}=-60\text{cm}$
The screen should be palced 60cm in front of the mirror. And$\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\frac{\text{h}_2}{\text{h}_1}=\frac{(-60)}{(-20)}$
$\text{h}_2=-15\text{cm}$
Height of image = 15cm.
$\frac{1}{\text{v}}+\frac{1}{(-20)}=\frac{1}{(-15)}$
$\frac{1}{\text{v}}=\frac{1}{20}-\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{1}{20}-\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{-5}{300}$
$\text{v}=-60\text{cm}$
The screen should be palced 60cm in front of the mirror. And$\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\frac{\text{h}_2}{\text{h}_1}=\frac{(-60)}{(-20)}$
$\text{h}_2=-15\text{cm}$
Height of image = 15cm.
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