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Geometrical Optics question

2024 · Q188
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Geometrical Optics question

2024 · Q188

NEETPhysicsGeometrical OpticsMCQ+4 / −1

A small telescope has an objective of focal length 140 cm140 \mathrm{~cm}140 cm and an eye piece of focal length 5.0 cm5.0 \mathrm{~cm}5.0 cm. The magnifying power of telescope for viewing a distant object is:

  1. A
    34
  2. B
    28
  3. C
    17
  4. D
    32
View written solutionFree

Correct answer: B

The magnifying power of a telescope when viewing distant objects can be calculated using the formula:

$$ \text{Magnifying Power} (M) = \frac{\text{Focal Length of Objective Lens} (f_o)}{\text{Focal Length of Eyepiece Lens} (f_e)} $$

Given in the problem:

  • Focal Length of Objective Lens, $f_o = 140 \mathrm{~cm}$
  • Focal Length of Eyepiece Lens, $f_e = 5.0 \mathrm{~cm}$

Substituting these values into the formula gives:

$$ M = \frac{140 \mathrm{~cm}}{5.0 \mathrm{~cm}} = 28 $$

Thus, the magnifying power of the telescope is 28. The correct answer is Option B.

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