Q.Draw a ray diagram of a refracting telescope when final image is formed at the least distance of the distinct vision. Write the formula of its magnifying power for this position. OR A compound microscope has an objective lens of focal length 2.0 cm and an eyepiece lens of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at the least distance of distinct vision (25 cm)? What is the magnifying power of the microscope?

(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).

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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 …