Find the position and size of the virtual image formed when an object 2cm tall is placed 20cm from:
- A diverging lens of focal length 40cm.
- A converging lens of focal length 40cm.
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$h _1=2 cm$
u = -20cm
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{-40}$
$\frac{1}{\text{v}}=-\frac{1}{40}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{-3}{40}$
$\text{v}=-13.33\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-13.33}{-20}=\frac{\text{h}_2}{2}$
$\text{h}_2=1.33\text{cm}$
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{40}$
$\frac{1}{\text{v}}=\frac{1}{40}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{-1}{40}$
$\text{v}=-40\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-40}{-20}=\frac{\text{h}_2}{2}$
$\text{h}_2=4\text{cm}$
u = -20cm
- f = -40cm (Diverging lens)
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{-40}$
$\frac{1}{\text{v}}=-\frac{1}{40}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{-3}{40}$
$\text{v}=-13.33\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-13.33}{-20}=\frac{\text{h}_2}{2}$
$\text{h}_2=1.33\text{cm}$
- f = 40cm (Diverging lens)
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{40}$
$\frac{1}{\text{v}}=\frac{1}{40}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{-1}{40}$
$\text{v}=-40\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-40}{-20}=\frac{\text{h}_2}{2}$
$\text{h}_2=4\text{cm}$
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