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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
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Topics: Ray Optics
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Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Moseley’s law √ν=a(z-b), the values of the screening constant for K-series
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- At 0⁰C, 15 gm of ice melts to form water at 0⁰C. The change in entropy is
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- In the half wave rectifier circuit operating from 50 Hz mains frequency,
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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