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(A) In the first case,the image is virtual and magnified,so magnification $m = +2$. Let the object distance be $u_1 = -x$.Using the magnification form

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$f$

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(A) In the first case,the image is virtual and magnified,so magnification $m = +2$. Let the object distance be $u_1 = -x$.Using the magnification formula $m = -v/u$,we get $2 = -v_1 / (-x)$,which implies $v_1 = 2x$.Using the mirror formula $1/v + 1/u = 1/f$:$1/(2x) - 1/x = -1/f$$-1/(2x) = -1/f$,so $x = f/2$.In the second case,the image is real and magnified,so magnification $m = -2$. Let the object distance be $u_2 = -y$.Using $m = -v/u$,we get $-2 = -v_2 / (-y)$,which implies $v_2 = -2y$.Using the mirror formula $1/v + 1/u = 1/f$:$1/(-2y) - 1/y = -1/f$$-3/(2y) = -1/f$,so $y = 3f/2$.The distance the object must be moved is $\Delta d = y - x = 3f/2 - f/2 = f$.

An object is placed in front of a concave mirror of focal length $f$. $A$ virtual image is formed with a magnification of $2$. To obtain a real image of the same magnification,the object has to be moved by a distance of:

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