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A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from pole of mirror , The size of image is approximate equal to :

$(a)\;\frac{bf}{u-f} \\ (b)\;\frac{b^2 f}{u-f} \\ (c)\;\frac{bf}{(u-f)^2} \\ (d)\;b\bigg ( \frac{f}{u-f}\bigg)^2 $

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From mirror formula,
$\large\frac{1}{v}+\frac{1}{u} =\frac{1}{f}$ (f=constant) -----(1)
$-v^{-2}dv -u^{-2}du=0$
or $|dv|=\bigg| \large\frac{v^2}{u^2} \bigg|$$ |du|$ -----(2)
Here , $|dv|$ = size of image
$|du| $= size of object (short lying along axis=b)
Further, from equation (1) we can find,
$\large\frac{v^2}{w^2} =\bigg( \large\frac{f}{u-f}\bigg)^2$
Substituting in equation (2) we get,
Size of image $=b \bigg( \large\frac{f}{u-f} \bigg)^2$
Hence d is the correct answer.
answered Feb 19, 2014 by meena.p
 

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