Q.a) State Huygens' principle for propagation of light wave. b) Show with the help of a ray diagram how the image is formed in an astronomical telescope. c) The magnifying power of a telescope in normal adjustment is 20, and the focal length of the eye-piece is 5×10⁻² m. What is the magnifying power obtained when the system is adjusted so that the final image of a distant object is formed 25×10⁻² m away from the eye-piece? OR

(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 …