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Question

# If a concave mirror has a focal length of 10 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.

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Solution

## Given, Focal length (f) of the concave mirror = -10 cm Case-1 The image is real, and its magnification (m) is -2. Using the magnification formula, we get $m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}⇒-2=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}⇒v=2u$ Now, using the mirror formula, we get $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{-10}=\frac{1}{2u}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{-10}=\frac{1}{2u}+\frac{2}{2u}=\frac{3}{2u}\phantom{\rule{0ex}{0ex}}⇒u=\frac{3×\left(-10\right)}{2}=-15\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{object}\mathrm{should}\mathrm{be}\mathrm{placed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}15\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{concave}\mathrm{mirror}.$ Case-2 The image is virtual and has a magnification 'm' of 2. Using the magnification formula, we get $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$ $⇒2=\frac{-\mathrm{v}}{\mathrm{u}}$ v=-2u Now, using the mirror formula, we get $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ $⇒\frac{1}{-10}=\frac{1}{-2\mathrm{u}}+\frac{1}{\mathrm{u}}$ $⇒\frac{1}{-10}=\frac{-1}{2u}+\frac{2}{2u}=\frac{1}{2u}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒u=\frac{-10}{2}=-5\mathrm{cm}$ Thus, the object should be placed at a distance of 5 cm in front of the concave mirror.

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