A) \[2\,m\]
B) \[3\,m\]
C) \[\frac{m}{2}\]
D) \[4m\]
Correct Answer: C
Solution :
We know the magnifying power (m) is given by \[m=-\frac{{{f}_{o}}}{{{f}_{e}}}.\]Now if the focal length of the Ie eye-piece \[({{f}_{e}})\] is doubled, then the new magnification would be equal to \[m=-\frac{{{f}_{o}}}{2{{f}_{e}}}=\frac{m}{2}\]You need to login to perform this action.
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