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(C) For a virtual image, magnification $m = +2$. Using the magnification formula for a spherical mirror, $m = -v/u$. Since the image is virtual, it is

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$-10 \,cm$

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(C) For a virtual image, magnification $m = +2$. Using the magnification formula for a spherical mirror, $m = -v/u$. Since the image is virtual, it is formed behind the mirror, so $v$ is positive. Let the object distance be $u$ (where $u$ is negative by sign convention). The distance between the object and the image is $|v - u| = 15 \,cm$. Since the object is in front of the mirror $(u  0)$, the distance is $v - u = 15$. Substituting $v = -mu = -2u$: $-2u - u = 15 -3u = 15 
u = -5 \,cm$. Then, $v = -2(-5) = 10 \,cm$. Using the mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$: $\frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = \frac{1 - 2}{10} = -\frac{1}{10}$. Therefore, $f = -10 \,cm$.

If the distance between an object and its two times magnified virtual image produced by a curved mirror is $15 \,cm$, the focal length of the mirror must be:

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