When an object is moved closer to a concave mirror from $15\,\text{cm}$ to $9\,\text{cm}$, the image becomes 3 times the magnification of the first case. What is the focal length of the mirror?
A$-6\,\text{cm}$
B$+6\,\text{cm}$
C$-12\,\text{cm}$
D$+12\,\text{cm}$
Answer & Solution
Correct answer: A. $-6\,\text{cm}$
Using the relation shown for this situation, in the first case $m=\frac{f}{f+15}$ and in the second case $m'=\frac{f}{f+9}$. Given $m'=3m$, we get $\frac{f}{f+9}=3\cdot\frac{f}{f+15}$. Cancelling $f$ and solving gives $f+15=3f+27$, so $f=-6\,\text{cm}$. The negative sign is consistent with a concave mirror.
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