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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSEasy

Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal lengths of lenses are:

A

10 cm, 10 cm

B

15 cm, 5 cm

C

18 cm, 2 cm

D

11 cm, 9 cm

Step-by-Step Solution

  1. Magnification Formula: For an astronomical telescope in 'normal adjustment' (where the final image is at infinity and rays are parallel), the magnifying power (MM) is defined as the ratio of the focal length of the objective (fof_o) to the focal length of the eyepiece (fef_e). M=fofe=9    fo=9feM = \frac{f_o}{f_e} = 9 \implies f_o = 9f_e.
  2. Tube Length Formula: In normal adjustment, the distance between the objective lens and the eyepiece lens is the sum of their focal lengths. L=fo+fe=20 cmL = f_o + f_e = 20\text{ cm}.
  3. Calculation: Substitute the value of fof_o from the magnification equation into the length equation: 9fe+fe=209f_e + f_e = 20 10fe=2010f_e = 20 fe=2 cmf_e = 2\text{ cm}. Now, calculate fof_o: fo=9×2=18 cmf_o = 9 \times 2 = 18\text{ cm}.
  4. Conclusion: The focal lengths of the objective and eyepiece are 18 cm18\text{ cm} and 2 cm2\text{ cm}, respectively.

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 INSTRUMENTSmagnifyingtelescopeadjustedparalleldistance

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