| ⇦ |
| ⇨ |
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: - Two rigid bodies A and B rotate with rotational kinetic energies Eᴀ and Eʙ
- A satellite A of mass m is at a distance r from the centre of earth. Another satellite
- The minimum wavelength of X-rays emitted by X-ray tube is 0.4125Å
- A concave lens of focal length f forms an image which is 1/3 times the size
- Three unequal resistors in parallel are equivalent to a resistance 1 Ω.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two rigid bodies A and B rotate with rotational kinetic energies Eᴀ and Eʙ
- A satellite A of mass m is at a distance r from the centre of earth. Another satellite
- The minimum wavelength of X-rays emitted by X-ray tube is 0.4125Å
- A concave lens of focal length f forms an image which is 1/3 times the size
- Three unequal resistors in parallel are equivalent to a resistance 1 Ω.
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