| ⇦ | 
| ⇨ |
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: - The liquid-crystal phase of a matter is called
- Steam at 100⁰C is passed into 20 g of water at 10°C. When water acquires a temperature
- A beam of light contains two wavelengths 6500Å and 5200Å.
- The molar specific heats of an ideal gas at constant pressure and volume are denoted
- For a given velocity, a projectile has the same range R for two angles of projection.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The liquid-crystal phase of a matter is called
- Steam at 100⁰C is passed into 20 g of water at 10°C. When water acquires a temperature
- A beam of light contains two wavelengths 6500Å and 5200Å.
- The molar specific heats of an ideal gas at constant pressure and volume are denoted
- For a given velocity, a projectile has the same range R for two angles of projection.
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