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Q.

A square ABCD of side 1mm is kept at distance 15 cm in front of the concave mirror as shown in the figure. The focal length of the mirror is 10 cm. The length of the perimeter of its image(in mm) will be:

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answer is 12.

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Detailed Solution

$v = - 30, m = - \frac{v}{u} = - 2$

$∴ A' B' = C' D' = 2 \times 1 = 2 m m$

Now $\frac{B' C'}{B C} = \frac{A' D'}{A D} = \frac{v^{2}}{u^{2}} = 4$
$\Rightarrow B' C' = A' D' = 4 m m$
Length = 2 + 2 + 4 + 4 = 12 mm

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