A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5.0 cm. The magnifying power of the telescope is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30
When the magnifying power of a telescope is asked without any further detail, it is taken for normal adjustment, in which the final image is formed at infinity. Then the magnifying power is the ratio of the focal lengths of the objective and the eyepiece.
\( \displaystyle m = \frac{f_o}{f_e} \)
\( f_o = 150\,\mathrm{cm} \)
\( f_e = 5.0\,\mathrm{cm} \)
\( \displaystyle m = \frac{150}{5.0} = 30 \)
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 – …
