A) 95
B) 105
C) 120
D) 125
Correct Answer: B
Solution :
: For a telescope for relaxed eye, Magnifying power\[M=\frac{{{f}_{o}}}{{{f}_{e}}}\] Tube length\[L={{f}_{o}}+{{f}_{e}}\] \[\therefore \]\[20=\frac{1.0}{{{f}_{e}}}\]or\[{{f}_{e}}=\frac{1}{20}m=\frac{100}{20}cm=5\,cm\] Tube length\[L={{f}_{o}}+{{f}_{e}}=(100+5)=105\text{ }cm\].You need to login to perform this action.
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