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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
A
5 cm
B
10 cm
C
15 cm
D
2.5 cm
Detailed Solution
Near end: $u_1 = -20$ cm, which is the centre of curvature (2f), so its image forms at the same place: $v_1 = -20$ cm
Far end: $u_2 = -30$ cm. Mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$: $\frac{1}{-10} = \frac{1}{v_2} + \frac{1}{-30}$
$\frac{1}{v_2} = -\frac{1}{10} + \frac{1}{30} = -\frac{1}{15} \Rightarrow v_2 = -15$ cm
Length of image $= |v_1| - |v_2| = 20 - 15 = 5$ cm