$(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.

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