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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSMedium

Question

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}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSastronomicaltelescopeobjectiveeyepiecelengths

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