Question:

A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is :

Updated On: Dec 4, 2024
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  • 28
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The Correct Option is B

Solution and Explanation

The magnifying power $M$ of a telescope for a distant object is given by the formula:
$M = \frac{f_o}{f_e}$,
where:
- $f_o = 140 \, \text{cm}$ is the focal length of the objective lens.
- $f_e = 5.0 \, \text{cm}$ is the focal length of the eyepiece.
Substituting the values:
$M = \frac{140}{5.0} = 28$.
Thus, the magnifying power of the telescope is 28.

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