Four lenses of focal length \( \pm 5 \) cm and \( \pm 200 \) cm are available for making a telescope. To produce the largest magnification for making a telescope with the help of these lenses, the focal length of eye piece should be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → +5 cm
A refracting (astronomical) telescope uses two converging lenses: the objective forms a real image of the distant object, and the eyepiece magnifies it like a simple magnifier. Its magnifying power is
\( \displaystyle m = \frac{f_{o}}{f_{e}} \)
For the largest \( m \), the objective must have the largest focal length and the eyepiece the smallest focal length, both positive (converging).
\( f_{o} = +200\,\mathrm{cm} \)
\( f_{e} = +5\,\mathrm{cm} \)
\( m = 200/5 = 40 \)
Why the other options are incorrect
Option (1) +200 cm — Incorrect. +200 cm is the right choice for the objective, not the eyepiece; as the eyepiece it would give \( m = 5/200 \).
Option (2) -5 cm — Incorrect. A lens of \( -5 \) cm is diverging, and the eyepiece of a refracting telescope must be converging to act as a magnifier.
Option (3) -200 cm — Incorrect. \( -200 \) cm is a diverging lens with a large focal length, which cannot serve as the eyepiece.
NCERT (2026-27):
Telescope – “The telescope is used to provide angular magnification of distant objects (Fig. 9.25). It also has an objective and an eyepiece. But here, the objective has a large focal length and a much larger aperture than the eyepiece. … The magnifying power m is the ratio of the angle β subtended at the eye by the final image to the angle α which the object subtends at the lens or the eye. Hence
\( \displaystyle m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_e}\cdot\frac{f_o}{h} = \frac{f_o}{f_e} \qquad (9.46) \)”