What is the magnification produced by a convex mirror of focal length f when the object is placed at a distance 2f from its pole?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \frac{1}{3} \)
The mirror equation \( 1/v+1/u=1/f \) gives the image distance, and the magnification is \( m=-v/u \). With the Cartesian sign convention, a convex mirror has a positive focal length \( +f \), and the object in front of it is at \( u=-2f \).
The image distance is:
\( \begin{aligned} \frac{1}{v} &= \frac{1}{f}-\frac{1}{u} \\ &= \frac{1}{f}+\frac{1}{2f} \\ &= \frac{3}{2f} \end{aligned} \)
\( \displaystyle v=\frac{2f}{3} \)
The positive \( v \) places the image behind the mirror. The magnification is:
\( \begin{aligned} m &= -\frac{v}{u} \\ &= -\frac{2f/3}{-2f} \\ &= \frac{1}{3} \end{aligned} \)
Therefore, the image is virtual, erect and one-third the size of the object.