Q.The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the eye lens and the objective is 20 cm. Determine the focal lengths of both the lenses.

(c) Both I and II are true …

The concave mirror is the objective; it forms the first, real image of a distant object.

In a reflecting type astronomical telescope (e.g. the Cassegrain design), a large concave (parabolic) mirror is used in place of a large objective lens. This concave mirror gathers light from a distant object and brings it to focus, forming a real image, which is then viewed/magnified by an eyepiece (often after being redirected by a small secondary mirror) …

For an astronomical telescope in normal adjustment, tube length $= f_o + f_e$, magnification $= f_o/f_e$, and the final image is always inverted.

Given objective focal length $f_o = 20\ \text{m} = 2000\ \text{cm}$ and eyepiece focal length $f_e = 2\ \text{cm}$.

Length of tube (normal adjustment, final image at infinity): $L = f_o + f_e = 2000\ \text{cm} + 2\ \text{cm} = 2002\ \text{cm} = 20.02\ \text{m}$ ✓ matches (a).

Magnifying power: $m = \dfrac{f_o}{f_e} = \dfrac{2000}{2} = 1000$ ✓ matches (b).

…

A telescope's objective must have a much larger focal length than its eyepiece to give large magnifying power and gather more light.

For an astronomical telescope in normal adjustment, the magnifying power is MP = fo/fe. For MP to be appreciably greater than 1 (the whole purpose of a telescope), the objective's focal length fo must be much GREATER than the eyepiece's focal le …