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Question

# (a) Draw a diagram to represent a convex mirror. On this diagram mark principal axis, principal focus F and the centre of curvature C if the focal length of convex mirror is 3 cm. (b) An object 1 cm tall is placed 30 cm in front of a convex mirror of focal length 20 cm. Find the size and position of the image formed by the convex mirror.

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Solution

## (a) Ray Diagram- (b) Given, The mirror is a diverging mirror, i.e. convex mirror. Height of the object 'ho' = 1 cm Distance of the object from the mirror 'u' = -30 cm Focal length of the convex mirror 'f' = 20 cm We have to find the position of the image 'v', height of the image 'hi' and its magnification 'm'. Using the mirror formula, we get $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{20}=\frac{1}{v}+\frac{1}{-30}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{30}+\frac{1}{20}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{60}+\frac{3}{60}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{60}=\frac{1}{v}\phantom{\rule{0ex}{0ex}}⇒v=\frac{60}{5}=12\mathrm{cm}$ The image will be at a distance of 12 cm behind the mirror . Now, using the magnification formula, we get $m=\frac{-v}{u}\phantom{\rule{0ex}{0ex}}m=\frac{-12}{-30}=0.4$ Thus, the image is virtual, erect and smaller in size. Also, $m=\frac{{h}_{i}}{{h}_{o}}\phantom{\rule{0ex}{0ex}}⇒0.4=\frac{{h}_{i}}{1}\phantom{\rule{0ex}{0ex}}⇒{h}_{i}=0.4\mathrm{cm}$ Therefore, the height of the image is 0.4 cm.

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