Q. The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are
A) 11 cm, 9 cm
B) 10 cm, 10 cm
C) 15 cm, 5 cm
D) 18 cm, 2 cm
Explanation
[Solution from source] For normal adjustment, magnifying power = f₀/fₑ = 9 and tube length = f₀ + fₑ = 20 cm. Solving, fₑ = 2 cm and f₀ = 18 cm. 15 cm and 5 cm add to 20 cm but give a magnification of only 3. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.