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If the focal length of the eye piece of a telescope is doubled, its magnifying power \((\mathrm{m})\) will be
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Verified Answer
The correct answer is:
\(\frac{\mathrm{m}}{2}\)
Hints: \(\mathrm{m}=\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
\(\mathrm{m}^{\prime}=\frac{\mathrm{m}}{2}\)
\(\mathrm{m}^{\prime}=\frac{\mathrm{m}}{2}\)
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