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NEET PHYSICSMedium

An astronomical telescope has an objective and eyepiece of focal lengths 40 cm40\text{ cm} and 4 cm4\text{ cm} respectively. To view an object 200 cm200\text{ cm} away from the objective, the lenses must be separated by a distance of:

A

46.0 cm

B

50.0 cm

C

54.0 cm

D

37.3 cm

Step-by-Step Solution

  1. Image Formation by Objective: Using the lens formula for the objective lens: 1vo−1uo=1fo\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}. Given: fo=+40 cmf_o = +40\text{ cm}, uo=−200 cmu_o = -200\text{ cm}. Substitution: 1vo−1−200=140  ⟹  1vo=140−1200\frac{1}{v_o} - \frac{1}{-200} = \frac{1}{40} \implies \frac{1}{v_o} = \frac{1}{40} - \frac{1}{200}. Calculation: 1vo=5−1200=4200=150\frac{1}{v_o} = \frac{5 - 1}{200} = \frac{4}{200} = \frac{1}{50}.
  • Image distance vo=50 cmv_o = 50\text{ cm}.
  1. Separation of Lenses: For an astronomical telescope in normal adjustment (final image at infinity), the real image formed by the objective acts as a virtual object for the eyepiece and must lie at the focus of the eyepiece. Therefore, the distance between the intermediate image and the eyepiece is ue=fe=4 cmu_e = f_e = 4\text{ cm}. The total distance between the lenses (tube length) is the sum of the image distance from the objective and the focal length of the eyepiece: L=vo+feL = v_o + f_e. L=50 cm+4 cm=54 cmL = 50\text{ cm} + 4\text{ cm} = 54\text{ cm}.
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