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(B) For a concave mirror, the focal length $f = -10 	ext{ cm}$.The rod is placed from $u_1 = -20 	ext{ cm}$ to $u_2 = -30 	ext{ cm}$ (since the len

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$5$

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(B) For a concave mirror, the focal length $f = -10 	ext{ cm}$.The rod is placed from $u_1 = -20 	ext{ cm}$ to $u_2 = -30 	ext{ cm}$ (since the length is $10 	ext{ cm}$ and the near end is at $20 	ext{ cm}$ from the pole).Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:For the near end at $u_1 = -20 	ext{ cm}$:$\frac{1}{v_1} - \frac{1}{20} = -\frac{1}{10} \implies \frac{1}{v_1} = -\frac{1}{10} + \frac{1}{20} = -\frac{1}{20} \implies v_1 = -20 	ext{ cm}$.For the far end at $u_2 = -30 	ext{ cm}$:$\frac{1}{v_2} - \frac{1}{30} = -\frac{1}{10} \implies \frac{1}{v_2} = -\frac{1}{10} + \frac{1}{30} = -\frac{2}{30} = -\frac{1}{15} \implies v_2 = -15 	ext{ cm}$.The length of the image is the distance between the images of the two ends:$	ext{Length of image} = |v_1 - v_2| = |-20 - (-15)| = |-5| = 5 	ext{ cm}$.

$A$ rod of length $10 \text{ cm}$ lies along the principal axis of a concave mirror of focal length $10 \text{ cm}$ as shown in the figure. The length of the image is . . . . . . $\text{cm}$.

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