| ⇦ | 
| ⇨ |
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 round disc of moment of inertia i₂ about its axis perpendicular to its plane
- The frequencies of X-rays,γ-rays and ultraviolet rays are respectively p,q and r
- The angle between the two vectors A = 3i+4j+5k and B = 3i+4j-5k will be
- The surface tension of a liquid is 5 newton per metre. If a film
- A particle under the action of a constant force moves from rest upto 20 seconds
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A round disc of moment of inertia i₂ about its axis perpendicular to its plane
- The frequencies of X-rays,γ-rays and ultraviolet rays are respectively p,q and r
- The angle between the two vectors A = 3i+4j+5k and B = 3i+4j-5k will be
- The surface tension of a liquid is 5 newton per metre. If a film
- A particle under the action of a constant force moves from rest upto 20 seconds
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