The magnifying power of a telescope is 9. When it is adjusted for parallel rays

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

admin:

Related Questions

  1. A small object of uniform density rolls up a curved surface with an initial velocity
  2. What is the energy of photon whose wavelength is 6840 Å?
  3. In adiabatic expansion, product of PV
  4. Work of 3.0×10⁻⁴ joule is required to be done in increasing the size of a soap film
  5. Curie is the unit of