| ⇦ | 
| ⇨ |
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: - In a mass spectrometer used for measuring the masses of ions, the ions
- Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
- A body of mass 3 kg acted upon by a constant force is displaced by S meter
- A boat taken 2h to travel 8km and back in still water.If the velocity of water 4 km/h
- A bubble is rising from bottom of a lake. The volume of bubble becomes
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a mass spectrometer used for measuring the masses of ions, the ions
- Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
- A body of mass 3 kg acted upon by a constant force is displaced by S meter
- A boat taken 2h to travel 8km and back in still water.If the velocity of water 4 km/h
- A bubble is rising from bottom of a lake. The volume of bubble becomes
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