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Question

# A bright object 50 mm high stands on the axis of a concave mirror of focal length 100 mm and at a distance of 300 mm from the concave mirror. How big will the image be?

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Solution

## Distance of the object from the mirror 'u' = -300 mm Height of the object 'ho' = 50 mm Focal length of the mirror 'f' = -100 mm We have to find the distance of the image 'v' and the height of the image 'hi'. $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{-100}=\frac{1}{v}+\frac{1}{-300}\phantom{\rule{0ex}{0ex}}-\frac{1}{100}=\frac{1}{v}-\frac{1}{300}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{300}-\frac{1}{100}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{300}-\frac{1}{100}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{-2}{300}\phantom{\rule{0ex}{0ex}}v=-150\mathrm{mm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}150\mathrm{mm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}m=\frac{{h}_{i}}{{h}_{o}}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}\frac{{h}_{i}}{50}=\frac{-\left(-150\right)}{-300}=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}{h}_{i}=-\frac{50}{2}=-25\mathrm{mm}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{size}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}25\mathrm{mm}.$

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