Question:

For large magnifying power of a telescope

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For a telescope $M=\frac{f_o}{f_e}$.
Updated On: Oct 1, 2026
  • object must be large
  • focal length of objective is large
  • focal length of eyepiece is large
  • focal length of objective and eyepiece must be large
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The Correct Option is B

Solution and Explanation

Step 1: Formula
Magnifying power of an astronomical telescope in normal adjustment is \(M=\frac{f_o}{f_e}\).

Step 2: Meaning
To make \(M\) large, \(f_o\) must be large and \(f_e\) small. Among the options, only "focal length of objective is large" fits. Option (B).

Final Answer:
A large objective focal length gives a large magnification, option (B). \[ \boxed{\text{(B)}} \]
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