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Question

# A concave mirror produces three times enlarged virtual image of an object placed at 10 cm in front of it. Calculate the radius of curvature of the mirror.

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Solution

## Given, Distance of the object 'u' = -10 cm Magnification 'm' = 3 We have to find the focal length 'f' and the radius of curvature 'R'. Using the magnification formula, we get $m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}3=\frac{-\mathrm{v}}{-10}\phantom{\rule{0ex}{0ex}}v=30\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{image}\text{'}\mathrm{v}\text{'}\mathrm{is}30\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}Now,u\mathrm{sin}gthe\mathrm{mirror}\mathrm{formula},weget\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{30}+\frac{1}{-10}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{f}=\frac{1}{30}-\frac{3}{30}=\frac{-2}{30}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{\mathrm{f}}=-\frac{1}{15}\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{f}=-15\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{mirror}\text{'}\mathrm{f}\text{'}\mathrm{is}-15\mathrm{cm}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}\text{'}\mathrm{R}\text{'}\mathrm{will}\mathrm{be}2f=2×-15\phantom{\rule{0ex}{0ex}}=-30\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}\mathrm{is}30\mathrm{cm}.$

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