Question:
The focal length of objective and eye piece of a telescope are $ {{f}_{0}} $ and $ {{f}_{e}} $ respectively. Then its magnification will be:
The focal length of objective and eye piece of a telescope are $ {{f}_{0}} $ and $ {{f}_{e}} $ respectively. Then its magnification will be:
Updated On: Jul 28, 2022
- $ {{f}_{0}}+{{f}_{e}} $
- $ {{f}_{0}}\times {{f}_{e}} $
- $ \frac{{{f}_{0}}}{{{f}_{e}}} $
- $ \frac{{{f}_{e}}}{{{f}_{0}}} $
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The Correct Option is C
Solution and Explanation
Magnifying power of telescope for relaxed eye
$M=-\frac{f_{o}}{f_{e}}=\frac{f_{o}}{f_{e}} $ (numerically)
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Concepts Used:
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In the single-slit diffraction experiment, we can examine the bending phenomenon of light or diffraction that causes light from a coherent source to hinder itself and produce an extraordinary pattern on the screen called the diffraction pattern.
Read More: Difference Between Diffraction and Interference
Central Maximum
