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A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects - Physics

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प्रश्न

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

  1. the telescope is in normal adjustment (i.e., when the final image is at infinity)?
  2. the final image is formed at the least distance of distinct vision (25 cm)?
संख्यात्मक
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उत्तर

Focal length of the objective lens, fo = 140 cm

Focal length of the eyepiece, fe = 5 cm

Least distance of distinct vision, d = 25 cm

(a) When the telescope is in normal adjustment, its magnifying power is given as:

`"m" = "f"_"o"/"f"_"e"`

= `140/5`

= 28

(b) When the final image is formed at d, the magnifying power of the telescope is given as:

`"f"_"o"/"f"_"e"  [1 + "f"_"e"/"d"]`

= `140/5 [1 + 5/25]`

= 28[1 + 0.2]

= 28 × 1.2

= 33.6

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Optical Instruments - Telescope
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अध्याय 9: Ray Optics and Optical Instruments - EXERCISES [पृष्ठ २५२]

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एनसीईआरटी Physics [English] Class 12
अध्याय 9 Ray Optics and Optical Instruments
EXERCISES | Q 9.27 | पृष्ठ २५२

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