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An object is placed in front of a concave mirror of focal length f. A virtual image of magnification 2 is formed.How for must the object be moved from original position so that a real image of same size is formed.

$(a)\;f \\ (b)\;\frac{f}{2} \\ (c)\;\frac{3f}{2} \\ (d)\;\frac{2f}{3} $

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Let the object be kept at x distance from concave mirror .
Since we have magnification of 2.
We have, $u=-x$
$v=+ 2x$
focal length $=-f$
$\large\frac{-1}{x} +\frac{1}{2x}= \frac{-1}{f}$
$\large\frac{1}{-2x} =\frac{-1}{f}$
$x=\large\frac{f}{2}$
Now Let the new position of image be y .
Since we want real image of magnification 2 $u=-y\; ;v=-2y$
focal length $=f$
$\large\frac{-1}{y} +\frac{1}{-2y} =\frac{-1}{f}$
$y= \large\frac {3}{2} $$f$
The object has to be moved by a distance
$y-x =-\large\frac{3}{2} $$f-\large\frac{f}{2}$
$\qquad= f$
Hence a is the correct answer.
answered Jan 16, 2014 by meena.p
 

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