Question 23 of 24intermediate🔧 ApplyNumerical3 marks

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25 cm)?

(a)
Answer

Magnifying power = 28

(b)
Answer

Magnifying power = 33.6

Explanation

For a telescope in normal adjustment, the magnifying power is simply the ratio of focal lengths (f_o/f_e). When the final image is formed at the least distance of distinct vision, the eyepiece acts as a simple microscope, and the magnification increases by a factor of (1 + f_e/D). This is derived from the principle that for a simple microscope, when the image is at infinity, m = D/f, and when the image is at distance D, m = 1 + D/f. The ratio gives the additional magnification factor.