| ⇦ | 
| ⇨ |
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: - A bus is moving with a speed of 10 ms⁻¹ on a straight road
- An electron of mass Mₑ, initially at rest, moves through a certain distance in a uniform
- A certain number of spherical dropes of a liquid of radius r coalesce to form a single
- Copper of fixed volume V is drawn into wire of length l. When this wire is subjected
- The following four wires are made of the same material. Which of these will have
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A bus is moving with a speed of 10 ms⁻¹ on a straight road
- An electron of mass Mₑ, initially at rest, moves through a certain distance in a uniform
- A certain number of spherical dropes of a liquid of radius r coalesce to form a single
- Copper of fixed volume V is drawn into wire of length l. When this wire is subjected
- The following four wires are made of the same material. Which of these will have
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