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(D) Let the rod be $AB$ with length $L = f/3$. The end $B$ is at distance $u_B = 5f/3$ from the pole,and end $A$ is at $u_A = 2f$. Using the mirror fo

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$ - 3/2 $

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(D) Let the rod be $AB$ with length $L = f/3$. The end $B$ is at distance $u_B = 5f/3$ from the pole,and end $A$ is at $u_A = 2f$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with sign convention (all distances negative): For end $B$ $(u_B = -5f/3)$: $\frac{1}{v_B} + \frac{1}{-5f/3} = \frac{1}{-f} \Rightarrow \frac{1}{v_B} = -\frac{1}{f} + \frac{3}{5f} = -\frac{2}{5f} \Rightarrow v_B = -2.5f$. For end $A$ $(u_A = -2f)$: $\frac{1}{v_A} + \frac{1}{-2f} = \frac{1}{-f} \Rightarrow \frac{1}{v_A} = -\frac{1}{f} + \frac{1}{2f} = -\frac{1}{2f} \Rightarrow v_A = -2f$. The length of the image is $|v_B - v_A| = |-2.5f - (-2f)| = 0.5f = f/2$. The longitudinal magnification $m = \frac{	ext{length of image}}{	ext{length of object}} = \frac{f/2}{f/3} = 1.5$. Since the image is inverted,the magnification is $m = -3/2$.

$A$ thin rod of length $f/3$ lies along the axis of a concave mirror of focal length $f$. One end of its magnified image touches an end of the rod. The longitudinal magnification of the image is

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