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# A 4.5 cm needle is placed at 12 cm away from a convex mirror of a focal length of 15cm. a) Give the location of the image and the magnification. b) Explain what will happen if the needle is moved farther away from the mirror.

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Solution

## a)Step 1: Given data. Height of object (ho) = 4.5 cm. Object distance (u) = -12 cm ( negative sign because of Cartesian sign convention) Focal length of the mirror (f) = 15 cm.Step 2: Finding the image distance. 1. According to mirror formulae ,$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\phantom{\rule{0ex}{0ex}}v\to imagedis\mathrm{tan}ce$ . $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}=\frac{1}{15}-\frac{1}{-12}\phantom{\rule{0ex}{0ex}}=\frac{1}{15}+\frac{1}{12}\phantom{\rule{0ex}{0ex}}=\frac{27}{180}\phantom{\rule{0ex}{0ex}}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}v=\frac{20}{3}\phantom{\rule{0ex}{0ex}}=6.7cm$Hence the location of the image is 20 cm in the back of the mirror.Step 3: Calculating the magnification and size of image. 1. $Magnification,M=\frac{{h}_{i}}{{h}_{o}}=-\frac{V}{U}\phantom{\rule{0ex}{0ex}}M=-\frac{6.7}{-12}\phantom{\rule{0ex}{0ex}}=0.56$ 2. $M=\frac{{h}_{i}}{{h}_{o}}=0.56\phantom{\rule{0ex}{0ex}}{h}_{i}=0.56×{h}_{o}\phantom{\rule{0ex}{0ex}}=0.56×4.5\phantom{\rule{0ex}{0ex}}=2.52cm$ Size of image = 2.52 cmb) Explanation: The image distance also increases with the object distance, and size of the image will reduce with an increase in object distance.  Suggest Corrections  0      Similar questions
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