| ⇦ | 
| ⇨ |
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: - When radiation is incident on a photoelectron emitter,
- Equal charge q each are placed at the vertices AQ and B of an equilateral triangle
- An electron moving in a uniform magnetic field of induction of intensity B,
- An isolated sphere has a capacitance of 50pF. What would be the radius of the sphere?
- In an astronomical telescope in normal adjustment a straight black line of length
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When radiation is incident on a photoelectron emitter,
- Equal charge q each are placed at the vertices AQ and B of an equilateral triangle
- An electron moving in a uniform magnetic field of induction of intensity B,
- An isolated sphere has a capacitance of 50pF. What would be the radius of the sphere?
- In an astronomical telescope in normal adjustment a straight black line of length
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