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An object of length 3 cm is placed on the principle axis of a concave mirror of focal length f at a distance of 4f. The length of the image will be :

$(a)\;-1 \;cm \\ (b)\;2\;cm \\ (c)\;8\;cm \\ (d)\;12\;cm $

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magnification $=\large\frac{I}{0}=\frac{f}{f-u}$
=> $\large\frac{I}{3} =\frac{-f}{-f-(-4f)}$
=> $ \large\frac{I}{3} =\large\frac{-f}{+3 f}$
$I= -1 \;cm$
Hence a is the correct answer.

 

answered Jan 16, 2014 by meena.p
 

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