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If the tube length of astronomical telescope is 105 cm and magnifying power for normal setting is 20, the focal length of objective is 

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\begin{array}{l} \text { Tube length, } L=f_{o}+f_{e}=105 \mathrm{cm} \\ \text { Magnification, } m=\frac{f_{o}}{f_{e}}=20 \\ \Rightarrow f_{o}=20 f_{e} \\ 20 f_{e}+f_{e}=105 \\ \Rightarrow 21 f_{e}=105 \\ \Rightarrow f_{e}=\frac{105}{21} \\ \Rightarrow f_{e}=5 \mathrm{cm} \\ \therefore f_{o}=20 \times 5=100 \mathrm{cm} \end{array}

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