| ⇦ | 
| ⇨ |
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: - A boy standing at the top of a tower of 20m height drops a stone
- In Young’s double slit experiment, the slit seperation is 1 mm and the screen
- Two waves are represented by the equations y₁ = a sin (?t + kx + 0.57) m
- As a result of interference of two coherent sources of light, energy is
- An insulated container of gas has two champers seperated by an insulating partition.
Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A boy standing at the top of a tower of 20m height drops a stone
- In Young’s double slit experiment, the slit seperation is 1 mm and the screen
- Two waves are represented by the equations y₁ = a sin (?t + kx + 0.57) m
- As a result of interference of two coherent sources of light, energy is
- An insulated container of gas has two champers seperated by an insulating partition.
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