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
- A100 cm and 5 cm
- B7 cm and 98 cm
- C98 cm and 7 cm
- 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