| ⇦ | 
| ⇨ |
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: - In a diffraction pattern due to a singleslit of width a, the first minimum is observed
- A small part of the rim of a fly wheel breaks off while it is rotating at a constant
- Graph of force per unit length between two long parallel current carrying
- In the middle of the depletion layer of a reverse-biased P-N junction, the
- The nuclear fusion reaction between deutrium and tritium takes place
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a diffraction pattern due to a singleslit of width a, the first minimum is observed
- A small part of the rim of a fly wheel breaks off while it is rotating at a constant
- Graph of force per unit length between two long parallel current carrying
- In the middle of the depletion layer of a reverse-biased P-N junction, the
- The nuclear fusion reaction between deutrium and tritium takes place
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