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# An object is placed at a distance of 10 cm from a convex mirror of focal length 8 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 $f = 8 cm$.

**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 $f = 8 cm$

$\frac{1}{\mathrm{v}}-\frac{1}{10}=\frac{1}{8}$

$\frac{1}{\mathrm{v}}=\frac{1}{8}+\frac{1}{10}=\frac{18}{80}$

$\mathrm{v}= \frac{40}{9} \mathrm{cm}$

Hence image will be formed $\frac{40}{9} \mathrm{cm}$ beyond the mirror. **So it is a virtual image.**

magnification $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{- \frac{40}{9}}{-10}= \frac{4}{9}$

Hence the image will be erect and diminished.

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