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Image formed by a concave mirror of is three times of the object Focal length of the mirror is 6 cm. Then the object distance is

$(a)\;-4\;cm \\ (b)\;8\;cm \\ (c)\;6\;cm \\ (d)\;12\;cm $

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Since the image is 3 times the object magnification can be $\pm 3$
and focal length $=-6 \;cm$
We know $m= \large\frac{f}{f-u}$
For real image $m=-3$
$-3= \large\frac{-6}{-6-u}$
$u= -8 \;cm$
For virtual image $m=+3$
$3= \large\frac{-6}{-6-u}$
$u=-4 \;cm$
Hence a is the correct answer.
answered Jan 16, 2014 by meena.p
 

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