An astronomical telescope has objectives and eye-piece lenses of focal length 3.5m and 5cm respectively…

Analyze the question We need the magnifying power of an astronomical telescope in normal adjustment (the final image is at infinity). The focal lengths are given for the objective and the eye-piece.

Concept In normal adjustment, the objective and eyepiece are separated by a distance equal to the sum of their focal lengths. The magnifying power is the ratio of the focal lengths:

M = f₀ / fₑ

where f₀ is the focal length of the objective and fₑ is the focal length of the eye-piece.

Solution

• f₀ = 3.5 m = 350 cm (convert to the same units as the eyepiece)
• fₑ = 5 cm

M = 350 / 5 = 70

Without converting the units, 3.5 / 5 = 0.7 (option D). That is a common trap, because the two focal lengths must be in the same unit before dividing.

Answer: M = 70.0, which is option A.

Tip: Before dividing focal lengths, always convert them to the same unit. Also, a telescope's magnifying power should be greater than 1, so 0.7 can't be right. The objective always has the longer focal length, and the eyepiece the shorter one.