Question

The magnifying power of a telescope in normal adjustment is 24, when the length of the telescope tube 1 meter. The focal length of the eye lens is

विकल्प

• 100 cm

• 24 cm

• 76 cm

• 4 cm

Answer 1

4 cm

Explanation:

m = `(fo)/(fe)` = 24, F₀ = 24fe 

Length of telescope tube fo + fe = 1 m

2Kfe + fe = 100 cm

25fe = 100 cm

fe = 100 cm / 25 = 4 cm