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The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of telescope is 110 cm. The magnifying power of the telescope when image is formed at the least distance of distinct vision is

Option: 1

14


Option: 2

48
 


Option: 3

28

 


Option: 4

52


Answers (1)

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Answer (1)

M_{\infty }=\frac{-f_{0}}{f_{e}}

10=\frac{-f_{0}}{f_{e}}

L_{\infty }=f_{0}+f_{e}

110=10f_{e}+f_{e}

f_{e}=10\; cm

M_{D}=\frac{-f_{0}}{f_{e}}\left ( 1+\frac{f_{e}}{D} \right )

        =10\left ( 1+\frac{10}{25} \right )

        =\frac{10 \times 35}{25}

       =14\; cm

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chirag

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