| ⇦ | 
| ⇨ |
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 velocity of a particle performing simple harmonic motion, when it passes
- Across a metallic conductor of non-uniform cross-section, a constant potential
- If X=3YZ², find dimensions of Y if X and Z are the dimensions of capacity
- A particle is executing SHM along a straight line. Its velocities at distances
- A particle is kept at rest at the top of a sphere of diameter 42m.When disturbed slightly
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The velocity of a particle performing simple harmonic motion, when it passes
- Across a metallic conductor of non-uniform cross-section, a constant potential
- If X=3YZ², find dimensions of Y if X and Z are the dimensions of capacity
- A particle is executing SHM along a straight line. Its velocities at distances
- A particle is kept at rest at the top of a sphere of diameter 42m.When disturbed slightly
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