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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSEasy

Question

A small telescope has an objective of focal length 140 cm140\text{ cm} and an eyepiece of focal length 5.0 cm5.0\text{ cm}. The magnifying power of the telescope for viewing a distant object is:

A

28

B

17

C

32

D

34

Step-by-Step Solution

  1. Magnifying Power Formula: For an astronomical telescope in normal adjustment (viewing a distant object with the final image formed at infinity), the magnifying power (mm) is given by the ratio of the focal length of the objective lens (fof_o) to the focal length of the eyepiece (fef_e). m=fofem = \frac{f_o}{f_e}
  2. Given Data: Focal length of objective, fo=140 cmf_o = 140\text{ cm} Focal length of eyepiece, fe=5.0 cmf_e = 5.0\text{ cm}
  3. Calculation: m=1405.0=28m = \frac{140}{5.0} = 28
  4. Conclusion: The magnifying power of the telescope is 28.

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 INSTRUMENTStelescopeobjectivelengtheyepiecelength

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