| ⇦ | 
| ⇨ |
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:
Options
(a) 10 cm, 10 cm
(b) 15 cm, 5 cm
(c) 18 cm, 2 cm
(d) 11 cm, 9 cm
Correct Answer:
18 cm, 2 cm
Explanation:
M.P. = 9 = f₀ / fₑ
f₀ = 9fₑ …(1) f₀ + fₑ = 20 …(2)
on solving
f₀ = 18 cm = focal length of the objective
fₑ = 2 cm = focal length of the eyepiece
Related Questions: - Two discs of moment of inertia l₁ and I₂ and angular speeds
- The displacement of particle is given by x=aₒ+(a₁t/2)-(a₂t²/3) what is acceleration?
- A body starts from rest with an acceleration of 2 m/s². After 5 second,
- A particle moves in a circle of radius 5 cm with constant speed and time period
- In Bohr model of hydrogen atom, the force on the electron depends on the principal
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two discs of moment of inertia l₁ and I₂ and angular speeds
- The displacement of particle is given by x=aₒ+(a₁t/2)-(a₂t²/3) what is acceleration?
- A body starts from rest with an acceleration of 2 m/s². After 5 second,
- A particle moves in a circle of radius 5 cm with constant speed and time period
- In Bohr model of hydrogen atom, the force on the electron depends on the principal
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply