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(B) The object is placed such that its farthest end is at the center of curvature $C$ (distance $2f$ from the mirror). Since the image of the farthest

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$f/2$

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(B) The object is placed such that its farthest end is at the center of curvature $C$ (distance $2f$ from the mirror). Since the image of the farthest end is formed at the same position,the farthest end is at $u = -2f$,and its image is at $v = -2f$.The nearest end of the object is at a distance $u' = -(2f - f/3) = -5f/3$ from the mirror.Using the mirror formula $\frac{1}{f} = \frac{1}{v'} + \frac{1}{u'}$:$\frac{1}{-f} = \frac{1}{v'} + \frac{1}{-5f/3}$$\frac{1}{v'} = \frac{3}{5f} - \frac{1}{f} = \frac{3-5}{5f} = \frac{-2}{5f}$$v' = -2.5f = -5f/2$.The length of the image is the distance between the image of the farthest end (at $2f$) and the image of the nearest end (at $2.5f$):Length $= |v'| - |v| = 2.5f - 2f = 0.5f = f/2$.

An object of length $f/3$ is placed on the principal axis of a concave mirror of focal length $f$. The image of the end of the object farthest from the mirror is formed at the same position. Find the length of the image.

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