हिंदी

The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.

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प्रश्न

The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.

संख्यात्मक
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उत्तर

Given: D = 150 cm, m = 2.9

fo = ?

Since, `m = f_o/f_e`

D = `f_o + f_e`

`2.9 = f_o/f_e`

`f_o = f_e xx 2.9` .....(i)

`f_o + f_e = 150` cm

From equation (i),

`2.9 f_e + f_e = 150`

`3.9f_e = 150`

fe = 38.46 cm

fo = 2.9 × 38.46

fo = 111.54 cm

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2022-2023 (March) Delhi Set 1

संबंधित 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)?

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L2 6 1
L3 10 1

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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. Find the magnifying power of the telescope for viewing distant objects when

  1. the telescope is in normal adjustment,
  2. the final image is formed at the least distance of distinct vision.

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