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NEET2026PhysicsChapterActual

An astronomical telescope is set in normal adjustment for viewing a distant object. The magnifying power of the telescope is 14 and the distance between the objective lens and the eyepiece is 105 cm . The focal lengths of the objective and the eyepiece are respectively:

Options

  1. A100 cm and 5 cm
  2. B7 cm and 98 cm
  3. C98 cm and 7 cm
  4. D91 cm and 14 cm

Correct answer

C. 98 cm and 7 cm

Step-by-step solution

For an astronomical telescope in normal adjustment, the magnifying power is given by m = f_o f_e and the length of the telescope tube is L = f_o + f_e . Given m = 14 and L = 105 cm . f_o f_e = 14 f_o = 14 f_e Substituting this into the length equation: 14 f_e + f_e = 105 15 f_e = 105 f_e = 7 cm f_o = 14 7 = 98 cm The focal lengths of the objective and the eyepiece are 98 cm and 7 cm respectively. Answer: 98 cm and 7 cm

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