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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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

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प्रश्न

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?

संख्यात्मक
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उत्तर

Focal length of the objective lens, fo = 144 cm

Focal length of the eyepiece, fe = 6.0 cm

The magnifying power of the telescope is given as:

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

= `144/6`

= 24

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

fo + 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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पाठ 9: Ray Optics and Optical Instruments - EXERCISES [पृष्ठ २५०]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 9 Ray Optics and Optical Instruments
EXERCISES | Q 9.13 | पृष्ठ २५०
एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 9 Ray Optics and Optical Instruments
Exercise | Q 9.13 | पृष्ठ ३४५

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