A concave lens has a focal length of 20cm. At what distance from the lens a 5cm tall object be placed so that it forms an image at 15cm from the lens? Also calculate the size of the image formed.
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f = -20cm $h _1=5 cm$ v = -15cm (Concave lens forms virtual image)$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{-15}-\frac{1}{\text{u}}=\frac{1}{-20}$
$\frac{1}{\text{u}}=-\frac{1}{15}+\frac{1}{20}$
$\frac{1}{\text{u}}=\frac{-1}{60}$
$\text{u}=-60\text{cm}$
Object should be placed 60cm to the left of the lens.$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-15}{-60}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-15}{-60}=\frac{\text{h}_2}{5}$
$\text{h}_2=1.25\text{cm}$
Image formed is 1.25cm high.
$\frac{1}{-15}-\frac{1}{\text{u}}=\frac{1}{-20}$
$\frac{1}{\text{u}}=-\frac{1}{15}+\frac{1}{20}$
$\frac{1}{\text{u}}=\frac{-1}{60}$
$\text{u}=-60\text{cm}$
Object should be placed 60cm to the left of the lens.$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-15}{-60}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-15}{-60}=\frac{\text{h}_2}{5}$
$\text{h}_2=1.25\text{cm}$
Image formed is 1.25cm high.
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