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(B) Given: Focal length $f = -10 \, cm$,length of the rod $l = 5 \, cm$.One end of the rod is at $2f = 20 \, cm$ from the mirror. Let this end be $A$.

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$2$

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(B) Given: Focal length $f = -10 \, cm$,length of the rod $l = 5 \, cm$.One end of the rod is at $2f = 20 \, cm$ from the mirror. Let this end be $A$. The other end is at $u = 20 - 5 = 15 \, cm$ from the mirror.Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:For end $A$ at $u_A = 20 \, cm$,the image $v_A = 20 \, cm$ (at the center of curvature).For the other end at $u_B = 15 \, cm$:$\frac{1}{-10} = \frac{1}{v_B} + \frac{1}{-15}$$\frac{1}{v_B} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30}$So,$v_B = -30 \, cm$.The length of the image is $|v_A - v_B| = |20 - 30| = 10 \, cm$.Magnification $m = \frac{	ext{Length of image}}{	ext{Length of object}} = \frac{10 \, cm}{5 \, cm} = 2$.

$A$ thin rod of $5 \, cm$ length is kept along the axis of a concave mirror of $10 \, cm$ focal length such that its image is real and magnified. If one end of the rod is at a distance of $2f$ from the mirror,what will be its magnification?

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