A convex lens of focal length 0.10m is used to form a magnified image of an object of height 5mm placed at a distance of 0.08m from the lens. Calculate the position, nature and size of the image.
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$h _1=5 mm=0.005 m$ u = -0.08m Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{\text{v}}-\frac{1}{-0.08}=\frac{1}{0.10}$
$\frac{1}{\text{v}}=\frac{1}{0.10}-\frac{1}{0.08}$
$\text{v}=-0.4\text{m}$
Image is formed 0.40m in front of the convex lens.$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-0.4}{-0.08}=\frac{\text{h}_2}{0.005}$
$h_2$ = 0.025m = 25mm Size of image is 25mm Image is virtual and erect.
$\frac{1}{\text{v}}=\frac{1}{0.10}-\frac{1}{0.08}$
$\text{v}=-0.4\text{m}$
Image is formed 0.40m in front of the convex lens.$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-0.4}{-0.08}=\frac{\text{h}_2}{0.005}$
$h_2$ = 0.025m = 25mm Size of image is 25mm Image is virtual and erect.
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