A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length 10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the wire is

4 c m^{2}

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6 c m^{2}

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16 c m^{2}

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36 c m^{2}

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Solution

The correct option is A $4 c m^{2}$
By using $m^{2} = \frac{A_{i}}{A_{0}}$ ; where $m = \frac{f}{f - u}$
Hence from given values $m = \frac{- 10}{- 10 - (- 25)} = \frac{- 2}{3}$ and $A_{0} = 9 c m^{2} ∴ A_{i} = {(\frac{- 2}{3})}^{2} \times 9 = 4 c m^{2}$

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