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(B) Given: Focal length $f = -20\,cm$. The midpoint of the rod is at $40\,cm$ from the pole. Since the rod length is $10\,cm$,the ends $A$ and $B$ are

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$32$

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(B) Given: Focal length $f = -20\,cm$. The midpoint of the rod is at $40\,cm$ from the pole. Since the rod length is $10\,cm$,the ends $A$ and $B$ are at distances $u_A = -(40 + 5) = -45\,cm$ and $u_B = -(40 - 5) = -35\,cm$ from the pole.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $v = \frac{uf}{u-f}$.For end $A$: $v_A = \frac{(-45)(-20)}{-45 - (-20)} = \frac{900}{-25} = -36\,cm$.For end $B$: $v_B = \frac{(-35)(-20)}{-35 - (-20)} = \frac{700}{-15} = -\frac{140}{3}\,cm$.The length of the image is $|v_A - v_B| = |-36 - (-140/3)| = |-108/3 + 140/3| = |32/3|\,cm$.Comparing this with $\frac{x}{3}\,cm$,we get $x = 32$.

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