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Ashok0210
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Asked: 2021-03-21T03:50:06+00:00 In:

A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

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Class 12 Physics
CBSE and UP Board
Ray Optics and Optical Instruments
Chapter-9 Exercise 9.13
NCERT Solutions for Class 12 Physics Chapter 9 Question-13

2020-2021cbse and up boardclass 12 chapter-9 physicsclass 12 physicsncert class 12 ray optics and optical instruments
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      Focal length of the objective lens, f0 = 144 cm

      Focal length of the eyepiece, fe = 6.0 cm

      The magnifying power of the telescope is given as:

      m = f0 /fe =  144/6 = 24

      The separation between the objective lens and the eyepiece is calculated as:

      f0 + fe     = 144 + 6 = 150 cm

      Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

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