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Question

# An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens.

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Solution

## Height of object (h) = 50 cm Height of image (h') = -20 cm (real and inverted) Distance of image from the lens (v) = 10 cm Distance of object from the lens (u) = ? Focal length of the lens (f)= ? We know, magnification (m) of the lens is given by: $\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}}$ Thus, substituting the values of v, h and h', we get: $\frac{10}{\mathrm{u}}=\frac{-20}{50}\phantom{\rule{0ex}{0ex}}\frac{10}{\mathrm{u}}=\frac{-20}{50}\phantom{\rule{0ex}{0ex}}\mathrm{u}=\frac{-5}{2}\mathrm{x}10\phantom{\rule{0ex}{0ex}}\therefore \mathrm{u}=-25\mathrm{cm}.$ Using the lens formula: $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ $\frac{1}{10}-\frac{1}{-25}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{10}+\frac{1}{25}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}⇒\frac{5+2}{50}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{50}=\frac{1}{\mathrm{f}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{f}=\frac{50}{7}\phantom{\rule{0ex}{0ex}}⇒\mathrm{f}=7.14\mathrm{cm}.$

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