If an object of 10cm height is placed at a distance of 36cm from a concave mirror of focal length 12cm, find the position, nature and height of the image.
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$h_1=10 cm, u =-36 cm, f =-12 cm$ We know that
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{(-36)}=\frac{1}{(-12)}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{36}-\frac{1}{12}=\frac{1-3}{36}=-\frac{2}{36}=-\frac{1}{18}$
$\text{v}=-18\text{cm}$
$\therefore$ The position of the image is 18cm in front of the mirror.
Magnification, $\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\Rightarrow\frac{\text{h}_2}{10}=-\frac{(-18)}{(-36)}$
$\Rightarrow\text{h}_2=-5\text{cm}$
The image formed is real and inverted.
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{(-36)}=\frac{1}{(-12)}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{36}-\frac{1}{12}=\frac{1-3}{36}=-\frac{2}{36}=-\frac{1}{18}$
$\text{v}=-18\text{cm}$
$\therefore$ The position of the image is 18cm in front of the mirror.
Magnification, $\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\Rightarrow\frac{\text{h}_2}{10}=-\frac{(-18)}{(-36)}$
$\Rightarrow\text{h}_2=-5\text{cm}$
The image formed is real and inverted.
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