A telescope has an eye piece of focal length 5.0 cm. The objective lens has focal length 12 times that of the eye-piece. In normal adjustment, what is the separation between the two lenses?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 65 cm
In normal adjustment an astronomical telescope forms the final image at infinity, so the image formed by the objective lies at the focus of the eyepiece. The separation of the lenses is the sum of the two focal lengths.
\( f_e = 5.0\,\mathrm{cm},\quad f_o = 12\times5.0\,\mathrm{cm} = 60\,\mathrm{cm} \)
\( L = f_o + f_e = 60\,\mathrm{cm}+5\,\mathrm{cm} = 65\,\mathrm{cm} \)
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 – …
