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(D) Given,focal length of the concave mirror $f = -10\, cm$.For end $A$ of the rod,the object distance is $u_A = -20\, cm$.Using the mirror formula $\

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$5$

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(D) Given,focal length of the concave mirror $f = -10\, cm$.For end $A$ of the rod,the object distance is $u_A = -20\, cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v_A} + \frac{1}{-20} = \frac{1}{-10}$$\frac{1}{v_A} = -\frac{1}{10} + \frac{1}{20} = -\frac{1}{20}$$v_A = -20\, cm$.For end $B$ of the rod,the object distance is $u_B = -(20 + 10) = -30\, cm$.Using the mirror formula:$\frac{1}{v_B} + \frac{1}{-30} = \frac{1}{-10}$$\frac{1}{v_B} = -\frac{1}{10} + \frac{1}{30} = -\frac{2}{30} = -\frac{1}{15}$$v_B = -15\, cm$.The length of the image is the distance between the image positions of the two ends:$L = |v_A - v_B| = |-20 - (-15)| = |-5| = 5\, cm$.

$A$ rod of length $10\, cm$ lies along the principal axis of a concave mirror of focal length $10\, cm$ in such a way that its end closer to the pole is $20\, cm$ away from the mirror. The length of the image is ......$cm$.

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