A concave mirror has a focal length of 4cm and an object 2cm tall is placed 9cm away from it. Find the nature, position and size of the image formed.
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$h_1=2 cm, u =-9 cm, f =-4 cm$ We know that$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{1}{(-4)}-\frac{1}{(-9)}$
$\frac{1}{4}+\frac{1}{9}=\frac{-9+4}{36}=\frac{5}{36}$
$\text{v}=-7.2\text{cm}$
The image is formed at a distance of 7.2cm in front of the mirror. Again $\text{m}=-\frac{\text{v}}{\text{u}}=-\frac{(-7.2)}{-9}=-0.8$$\Rightarrow\text{m}=\frac{\text{h}_2}{\text{h}_1}=-0.8$ $\Rightarrow\frac{\text{h}_2}{2}=\text{h}_2=-1.6\text{cm}$$$$$
So, Image is 1.6cm in size, real and inverted.
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