A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm as shown in figure. The length of the image is cm. [figure]
- 2.5
- 5
- 7.5
- 7
Answer
(2) 5
Solution
[figure]
f=-10 cm ⇒ R=2f=-20 cm: the near end of the rod is at the centre of curvature C.
Rod: from u=-20 cm to u=-30 cm.
Near end (u=-20 cm, at C): image at C itself ⇒ v1=-20 cm
Far end (u=-30 cm): 1v+ 1u= 1f
⇒ 1v2=- 110+ 130=- 115 ⇒ v2=-15 cm
∴ Length of image =|v1-v2|=20-15=5 cm (real, inverted, between F and C)
Answer: (2)