A 2.0cm tall object is placed 40 cm from a diverging lens of focal length 15cm. Find the position and size of the image.
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$h_1=2 cm u =-40 cm f =-15 cm \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{\text{v}}=-\frac{1}{15}-\frac{1}{40}$
$\frac{1}{\text{v}}=\frac{-11}{120}$
$\text{v}=-10.90\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-10.09}{-40}=\frac{\text{h}_2}{2}$
$\text{h}_2=0.54\text{cm}$
$\frac{1}{\text{v}}=-\frac{1}{15}-\frac{1}{40}$
$\frac{1}{\text{v}}=\frac{-11}{120}$
$\text{v}=-10.90\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-10.09}{-40}=\frac{\text{h}_2}{2}$
$\text{h}_2=0.54\text{cm}$
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