A diverging mirror of radius of curvature 40cm forms an image which is half the height of the object. Find the object and image positions.
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R = 40cm$\text{f}=\frac{\text{R}}{2}=\frac{40}{2}=20\text{cm}$
Image is half the height of the object.$\text{i.e.}\ \text{m}=-\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\Rightarrow \text{u}=-2\text{v}$
know,$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{(-2\text{v})}=\frac{1}{20}$
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{2\text{v}}=\frac{1}{20}$
$\Rightarrow\frac{1}{2\text{v}}=\frac{1}{20}$
$\therefore \text{v}=10\text{cm}$
$\text{u}=-2\text{v}=-2\times10=-20\text{cm}$
So, the object is palced 20cm in front of the mirror and the image is formed 10cm the mirror.
Image is half the height of the object.$\text{i.e.}\ \text{m}=-\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\Rightarrow \text{u}=-2\text{v}$
know,$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{(-2\text{v})}=\frac{1}{20}$
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{2\text{v}}=\frac{1}{20}$
$\Rightarrow\frac{1}{2\text{v}}=\frac{1}{20}$
$\therefore \text{v}=10\text{cm}$
$\text{u}=-2\text{v}=-2\times10=-20\text{cm}$
So, the object is palced 20cm in front of the mirror and the image is formed 10cm the mirror.
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