The following three lenses L1, L2 and L3 are given here with the description given below-
L1 - Power 3 D, Aperture 8 cm
L2 - Power 6 D, Aperture 1 cm
L3 - Power 10 D, Aperture 1 cm
Which two of the above lenses will you select as objective and eyepiece for constructing best possible Telescope?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → L1 as objective, L3 as eye piece
A good astronomical telescope needs a large magnifying power and a large aperture for the objective, so that it gathers more light and resolves finer detail. The magnifying power (normal adjustment) is the ratio of the focal lengths, which is the ratio of the powers.
\( \displaystyle m = \frac{f_o}{f_e} = \frac{P_e}{P_o} \)
So the objective should be the lens of least power and largest aperture, L1 (3 D, 8 cm), and the eyepiece the lens of greatest power, L3 (10 D).
\( \displaystyle m = \frac{10\,\mathrm{D}}{3\,\mathrm{D}} \approx 3.3 \)
Why the other options are incorrect
Option (2) L2 as objective, L1 as eye piece — Incorrect. L2 has a 1 cm aperture, and with L1 as eyepiece \( m = 3/6 = 0.5 \), which is less than 1.
Option (3) L3 as objective, L1 as eye piece — Incorrect. With L3 as objective, \( m = 3/10 = 0.3 \), and its aperture is only 1 cm.
Option (4) L1 as objective, L2 as eye piece — Incorrect. L1 is the right objective, but L2 gives \( m = 6/3 = 2 \), smaller than the 3.3 obtained with L3.
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 \( \beta \) subtended at the eye by the final image to the angle \( \alpha \) which the object subtends at the lens or the eye. Hence
\( \displaystyle m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_e}\cdot\frac{f_o}{h} = \frac{f_o}{f_e} \qquad (9.46) \)”
