A small telescope has an objective lens of focal length 100 cm and an eye-piece of focal length 5 cm. The magnifying power of the telescope when the final image is formed at the least distance of distinct vision will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 24 \)
For an astronomical telescope with the final image at the least distance of distinct vision \( D \), the magnifying power is, in magnitude, the formula below. The least distance of distinct vision is taken as the standard value \( D = 25\,\mathrm{cm} \).
\( \displaystyle m = \frac{f_0}{f_e}\left(1 + \frac{f_e}{D}\right) = \frac{100}{5}\left(1 + \frac{5}{25}\right) = 20 \times 1.2 = 24 \)
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 – …
