| ⇦ | 
| ⇨ |
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 particle of mass m moves with constant speed along a circular path of radius r under
- In the phenomenon of electric discharge through gases at low pressure,
- Assertion(A) :The change in kinetic energy of a particle is equal to the work done
- A beam of electron passes undeflected through mutually perpendicular
- Copper and carbon wires are connected in series and the combined resistor is kept
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle of mass m moves with constant speed along a circular path of radius r under
- In the phenomenon of electric discharge through gases at low pressure,
- Assertion(A) :The change in kinetic energy of a particle is equal to the work done
- A beam of electron passes undeflected through mutually perpendicular
- Copper and carbon wires are connected in series and the combined resistor is kept
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