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A concave mirror has a focal length 20 cm. The distance between the two positions of object for which the image size is double of object size.

$(a)\;15\;cm \\ (b)\;25\;cm \\ (c)\;35\;cm \\ (d)\;20\;cm $

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For real image
$u=u_1,v=-2u_1,f=20\;cm$
Substituting in $\large\frac{1}{v} +\frac{1}{u} =\frac{1}{f}$
Then, we get,
$\large\frac{1}{-2u}-\frac{1}{u_1}=\frac{-1}{20}$
or $u_1=30\;cm$
For virtual image $u=-u_2,v=2u_2$
$f= -20\;cm$
$u_2= 10\;cm$
$\therefore$ Distance between two position of object are $u_1-u_2$
or $30-10=20\;cm$
Hence d is the correct answer.
answered Feb 19, 2014 by meena.p
 

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