| ⇦ |
| ⇨ |
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: - If two slits in Young’s experiment are 0.4 mm apart and fringe width on a screen
- If a charge on the body is 1nC, then how many electrons are present on the body?
- The drift velocity of free electrons in a conductor is v, when a current
- A polished metal plate has a rough and black-spot. It is heated to 1400 K
- In common base mode of a transistor, the collector current is 5.488 mA
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If two slits in Young’s experiment are 0.4 mm apart and fringe width on a screen
- If a charge on the body is 1nC, then how many electrons are present on the body?
- The drift velocity of free electrons in a conductor is v, when a current
- A polished metal plate has a rough and black-spot. It is heated to 1400 K
- In common base mode of a transistor, the collector current is 5.488 mA
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