An object is placed $10\,\text{cm}$ in front of a converging mirror of radius of curvature $15\,\text{cm}$. What is the magnification of the image?
A$+3$
B$-3$
C$+\dfrac{1}{3}$
D$-\dfrac{1}{3}$
Answer & Solution
Correct answer: B. $-3$
A converging mirror is a concave mirror, so $R=-15\,\text{cm}$ and $f=R/2=-7.5\,\text{cm}$. With $u=-10\,\text{cm}$, the mirror formula gives $v=-30\,\text{cm}$. Hence $m=-\frac{v}{u}=-\frac{-30}{-10}=-3$. The negative sign indicates a real, inverted image.
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