Discussion
The distance of an object placed in front of a concave mirror of radius of curvature 24 cm that gives its magnification as 3 is
Answer: A
The keyed distance is 8 cm.
The focal length is \(f = \frac{R}{2} = 12\) cm. A magnification of \(+3\) means an erect, virtual, enlarged image, formed when the object is inside the focus. With \(m = -\frac{v}{u} = 3\), \(v = -3u\), and the mirror formula with the sign convention gives
\[ \begin{aligned} \frac{1}{v} + \frac{1}{u} &= \frac{1}{f} \\ -\frac{1}{3u} + \frac{1}{u} = \frac{2}{3u} &= -\frac{1}{12} \;\Rightarrow\; u = -8\ \text{cm} \end{aligned} \]
The object is 8 cm in front of the mirror. If the magnification were \(-3\) (a real, inverted image), the distance would be 16 cm instead; the key takes the stated magnification 3 as positive.
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