Given, fe=2 cm, and fo=20 m=2000 cm For normal adjustment, Magnification power Length of telescope =fe−fo=2−2000=−1000=fo+fe=2000+2=2002 cm=20.02 m The image formed is inverted and magnified, and the aperture of objective is larger than eye piece of the telescope. (fo & fe are focal lengths of objective & eye piece respectively.)
An astronomical refracting telescope is being used by an observer to observe planets in normal adjustment. The focal lengths of the objective and eye…
NEET UG 2022 — Physics Optics
2022mcqhard
An astronomical refracting telescope is being used by an observer to observe planets in normal adjustment. The focal lengths of the objective and eye piece used in the construction of the telescope are 20 m and 2 cm respectively. Consider the following statements about the telescope: (a) The distance between the objective and eye piece is 20.02 m (b) The magnification of the telescope is (−)1000 (c) The image of the planet is erect and diminished (d) The aperture of eye piece is smaller than that of objectie The correct statements are:
Official previous-year question
Held on 30 Apr 2022 · Verified 9 Jul 2026.
Options
- A
(a), (b) and (c)
- B
(b), (c) and (d)
- C
(c), (d) and (a)
- D
(a), (b) and (d)
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