Ray Optics

JEE Main level, single correct, medium

19 of 295

Question 19 of 295

A rod of length 1010 cm lies along the principle axis of a concave mirror of focal length 1010 cm as shown in figure. The length of the image is _____ cm.

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Answer: (B) 55

For the given concave mirror, the focal length is f=−10f = -10 cm.

The rod lies along the principal axis. From the figure, the near end of the rod is at a distance of 2020 cm from the pole of the mirror. Since the length of the rod is 1010 cm, the far end is at a distance of 20+10=3020 + 10 = 30 cm from the pole.

Let the near end be AA and the far end be BB.
Object distance for end AA, uA=−20u_A = -20 cm.
Object distance for end BB, uB=−30u_B = -30 cm.

Using the mirror formula, 1v+1u=1f\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}

For end AA:
1vA+1−20=1−10\dfrac{1}{v_A} + \dfrac{1}{-20} = \dfrac{1}{-10}
1vA=120−110=−120\dfrac{1}{v_A} = \dfrac{1}{20} - \dfrac{1}{10} = -\dfrac{1}{20}
vA=−20v_A = -20 cm

For end BB:
1vB+1−30=1−10\dfrac{1}{v_B} + \dfrac{1}{-30} = \dfrac{1}{-10}
1vB=130−110=1−330=−230=−115\dfrac{1}{v_B} = \dfrac{1}{30} - \dfrac{1}{10} = \dfrac{1 - 3}{30} = -\dfrac{2}{30} = -\dfrac{1}{15}
vB=−15v_B = -15 cm

The length of the image is the distance between the images of the two ends:
Length of image=∣vA−vB∣=∣−20−(−15)∣=∣−5∣=5\text{Length of image} = |v_A - v_B| = |-20 - (-15)| = |-5| = 5 cm.

Answer: 55

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