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An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Given: An object is placed at a distance of 10 cm from a convex mirror of focal length 8 cm. So $u = -10\ cm$ and $f =15\ cm$.
To find: The position and nature of the image
Solution:
We know, mirror formula is:
$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
We have $u = -10\ cm$ and $f =15\ cm$
$\frac{1}{\mathrm{v}}-\frac{1}{10}=\frac{1}{15}$
$\frac{1}{\mathrm{v}}=\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$
$\mathrm{v}=6\ cm$
Hence image will be formed $6\ \mathrm{cm}$ beyond the mirror. So it is a virtual image.
magnification $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{- 6}{-10}=0.6$
Hence the image will be erect and diminished.
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