Calculate the image distance for an object of height 12mm at a distance of 0.20m from a concave lens of focal length 0.30 m, and state the nature and size of the image.
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$h _1=12 mm=0.012 m$ u = -0.20m f = -0.03m$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-0.02}=\frac{1}{-0.03}$
$\frac{1}{\text{v}}=\frac{-1}{0.03}-\frac{1}{0.20}$
$\text{v}=-0.12\text{m}$
Image is virtual and erect.$\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{\text{v}}{\text{u}}$
$\frac{\text{h}_2}{0.012}=\frac{-0.12}{0.20}$
$\text{h}_2=0.0072\text{m}=7.2\text{mm}$
Image is 7.2mm high.
$\frac{1}{\text{v}}-\frac{1}{-0.02}=\frac{1}{-0.03}$
$\frac{1}{\text{v}}=\frac{-1}{0.03}-\frac{1}{0.20}$
$\text{v}=-0.12\text{m}$
Image is virtual and erect.$\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{\text{v}}{\text{u}}$
$\frac{\text{h}_2}{0.012}=\frac{-0.12}{0.20}$
$\text{h}_2=0.0072\text{m}=7.2\text{mm}$
Image is 7.2mm high.
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