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A rod of length 10 cm10\text{ cm} lies along the principal axis of a concave mirror of focal length 10 cm10\text{ cm} in such a way that its end closer to the pole is 20 cm20\text{ cm} away from the mirror. The length of the image is:

A

10 cm

B

15 cm

C

2.5 cm

D

5 cm

Step-by-Step Solution

  1. Given Data: Focal length of the concave mirror, f=−10 cmf = -10\text{ cm} (using sign convention). Length of the rod =10 cm= 10\text{ cm}. Distance of the nearer end from the pole, u1=−20 cmu_1 = -20\text{ cm}. Distance of the farther end from the pole, u2=−(20+10) cm=−30 cmu_2 = -(20 + 10)\text{ cm} = -30\text{ cm}.

  2. Image of the Nearer End (v1v_1): Using the mirror formula: 1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f} 1v1+1−20=1−10\frac{1}{v_1} + \frac{1}{-20} = \frac{1}{-10} 1v1=120−110=1−220=−120\frac{1}{v_1} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = -\frac{1}{20} v1=−20 cmv_1 = -20\text{ cm}.

  • (Note: The object is at the center of curvature C=2f=20 cmC = 2f = 20\text{ cm}, so the image is also formed at CC.)

  1. Image of the Farther End (v2v_2): Using the mirror formula for u2=−30 cmu_2 = -30\text{ cm}: 1v2+1−30=1−10\frac{1}{v_2} + \frac{1}{-30} = \frac{1}{-10} 1v2=130−110=1−330=−230=−115\frac{1}{v_2} = \frac{1}{30} - \frac{1}{10} = \frac{1 - 3}{30} = -\frac{2}{30} = -\frac{1}{15} v2=−15 cmv_2 = -15\text{ cm}.

  2. Length of the Image: The length of the image is the distance between the images of the two ends. Limage=∣v1−v2∣=∣−20−(−15)∣=∣−5∣=5 cmL_{image} = |v_1 - v_2| = |-20 - (-15)| = |-5| = 5\text{ cm}.

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