A) 18 cm, 2 cm
B) 11 cm, 9 cm
C) 10 cm, 10 cm
D) 11 cm, 5 cm
Correct Answer: A
Solution :
Magnifying power of telescope \[M=\frac{{{f}_{o}}}{{{f}_{e}}}\] \[9=\frac{{{f}_{o}}}{{{f}_{e}}}\] \[{{f}_{o}}=9{{f}_{o}}\] Length of telescope, \[L={{f}_{o}}+{{f}_{e}}\] \[20=9{{f}_{e}}+{{f}_{e}}\] \[{{f}_{e}}=2\,cm\] \[{{f}_{e}}=2\times 9=18\,cm\]
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