An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image.
CBSE DELHI - SET 1 2017
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$h_1=+4 cm, f=-10 cm, u=-15 cm, v=?, h_2=? \frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\Rightarrow$ $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{\text{-10cm}}-\frac{1}{\text{-15cm}}$
$\therefore\text{v}=-30\text{cm}$
$\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\because\text{h}_2=-\frac{\text{v}}{\text{u}}\times\text{h}_1$
$=-\frac{-30\text{cm}}{-15\text{cm}}\times4\text{cm}$
$=-8\text{cm}$
$\Rightarrow$ $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{\text{-10cm}}-\frac{1}{\text{-15cm}}$
$\therefore\text{v}=-30\text{cm}$
$\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\because\text{h}_2=-\frac{\text{v}}{\text{u}}\times\text{h}_1$
$=-\frac{-30\text{cm}}{-15\text{cm}}\times4\text{cm}$
$=-8\text{cm}$
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