A small telescope has an objective lens of focal length 120 cm and an eyepiece of focal length of 5.0 cm. The magnifying power of telescope for viewing distant objects in normal adjustment is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 24
In normal adjustment the final image of a telescope is at infinity, and the magnifying power is the ratio of the focal lengths of the objective and the eyepiece.
\( \displaystyle m = \frac{f_o}{f_e} = \frac{120\,\mathrm{cm}}{5.0\,\mathrm{cm}} = 24 \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
