A) 100cm
B) 5cm
C) 80cm
D) 95cm
Correct Answer: A
Solution :
For normal adjustment, magnification is given by \[m=\frac{{{f}_{o}}}{{{f}_{e}}}\]and the length of the telescope tube is\[{{f}_{0}}+{{f}_{e}}\]. Let the focal length of objective and eye lenses are\[{{f}_{o}}\]and\[{{f}_{e}}\]respectively. Therefore, magnification\[m=\frac{{{f}_{o}}}{{{f}_{e}}},\] \[20=\frac{{{f}_{o}}}{{{f}_{e}}}\Rightarrow {{f}_{o}}=20{{f}_{e}}\] And \[{{f}_{o}}+{{f}_{e}}=105,\] On solving,\[{{f}_{o}}=100\]cm and\[{{f}_{e}}~=5cm\].
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