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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
Related Questions: - The transverse nature of electromagnetic waves is proved by which of the following?
- Light of two different frequencies whose photons have energies 1eV
- Light of wavelength λ is incident on slit of width d. The resulting diffraction
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Topics: Ray Optics
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The transverse nature of electromagnetic waves is proved by which of the following?
- Light of two different frequencies whose photons have energies 1eV
- Light of wavelength λ is incident on slit of width d. The resulting diffraction
- A thermally insulated rigid container contains an ideal gas heated by a filament
- A copper disc of radius 0.1 m is rotated about its centre, with 10 revolutions
Topics: Ray Optics (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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