Question

# An object is placed 15 cm from (a) a converging mirror, and (b) a diverging mirror, of radius of curvature 20 cm. Calculate the image position and magnification in each case.

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Solution

## Case -1 It is a converging mirror, i.e. concave mirror. Distance of the object 'u' = -15 cm Radius of curvature of the mirror 'R' = -20 cm Focal length of the mirror 'f' = $\frac{\mathrm{R}}{2}$ = -10 cm We have to find the position of the image 'v' and its magnification 'M'. Using the mirror formula, we get $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{-10}=\frac{1}{-15}+\frac{1}{\mathrm{v}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{v}}=\frac{1}{-10}+\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{v}}=-\frac{1}{10}+\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{v}}=\frac{-15}{150}+\frac{10}{150}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{v}}=\frac{-5}{150}\phantom{\rule{0ex}{0ex}}⇒\mathrm{v}=-30\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{form}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}30\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{converging}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}\mathrm{magnification}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-\left(-30\right)}{-15}\phantom{\rule{0ex}{0ex}}m=-2\phantom{\rule{0ex}{0ex}}\mathrm{magnification}=-2$ Thus the image is real, inverted and large in size. Case -2 Mirror is converging mirror i.e. convex mirror. Distance of the object u = -15 cm Radius of curvature of the mirror R = 20 cm Focal length of the mirror f = $\frac{\mathrm{R}}{2}$ = 10 cm We have to find the position of the image v = ? Magnification = ? $U\mathrm{sing}the\mathrm{mirror}\mathrm{formula},weget\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{f}=\frac{1}{-15}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{10}=\frac{1}{-15}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{10}+\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{15}{150}+\frac{10}{150}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{25}{150}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒v=6\mathrm{cm}\phantom{\rule{0ex}{0ex}}There\mathrm{fore},\mathrm{the}\mathrm{image}\mathrm{willl}\mathrm{form}6\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}.\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\mathrm{m}=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-6}{-15}\phantom{\rule{0ex}{0ex}}m=0.4$ Thus, the image is virtual, erect and small in size.

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