An arrow 2.5cm high is placed at a distance of 25cm from a diverging mirror of focal length 20cm. Find the nature, position and size of the image formed.
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$h_1= 2.5cm$, u = -25cm, f = 20cm We know that$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{-25}=\frac{1}{20}$
$\therefore\text{v}=\frac{9}{100}\text{cm}=11.1\text{cm}$
The image is formed 11.1cm behind the convex mirror. Now,$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\Rightarrow -\frac{11.1}{(-25)}=\frac{\text{h}_2}{2.5}$
${\text{h}_2}=\frac{11.1\times2.5}{25}=1.11\text{cm}$
The image is virtual, erect and 1.11cm tall.
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{-25}=\frac{1}{20}$
$\therefore\text{v}=\frac{9}{100}\text{cm}=11.1\text{cm}$
The image is formed 11.1cm behind the convex mirror. Now,$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\Rightarrow -\frac{11.1}{(-25)}=\frac{\text{h}_2}{2.5}$
${\text{h}_2}=\frac{11.1\times2.5}{25}=1.11\text{cm}$
The image is virtual, erect and 1.11cm tall.
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