A 3 cm tall object is placed 18 cm in front of a concave mirror of focal length 12 cm. At what distance from the mirror should a screen be placed to see a sharp image of the object on the screen. Also calculate the height of the image formed.
CBSE DELHI - SET 2 2017
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$h_1=+3 cm, f =-12 cm, u =-18 cm, v =?, h_2=? \frac{1}{ f }=\frac{1}{ v }+\frac{1}{ u }$
$\Rightarrow$ $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{\text{-12cm}}-\frac{1}{\text{-18cm}}$
$\therefore\text{v}=-36\text{cm}$
$\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$ \because\text{h}_2=\text{-h}_1\times\frac{\text{v}}{\text{u}}$
$=\text{-3cm}-\frac{-36\text{cm}}{-18\text{cm}}$
$=-6\text{cm}$
$\Rightarrow$ $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{\text{-12cm}}-\frac{1}{\text{-18cm}}$
$\therefore\text{v}=-36\text{cm}$
$\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$ \because\text{h}_2=\text{-h}_1\times\frac{\text{v}}{\text{u}}$
$=\text{-3cm}-\frac{-36\text{cm}}{-18\text{cm}}$
$=-6\text{cm}$
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