The magnifying power of an astronomical teles | vedclass

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(B) For an astronomical telescope in normal adjustment,the magnifying power $M$ is given by $M = \frac{f_o}{f_e} = 100$,where $f_o$ is the focal lengt

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$100$

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(B) For an astronomical telescope in normal adjustment,the magnifying power $M$ is given by $M = \frac{f_o}{f_e} = 100$,where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.From this,we get $f_o = 100 f_e$.The distance between the objective and the eyepiece in normal adjustment is $L = f_o + f_e = 101 \,cm$.Substituting $f_o = 100 f_e$ into the equation,we get $100 f_e + f_e = 101 \,cm$.$101 f_e = 101 \,cm$,which gives $f_e = 1 \,cm$.Now,calculating the focal length of the objective: $f_o = 100 	imes 1 \,cm = 100 \,cm$.

The magnifying power of an astronomical telescope in the normal adjustment position is $100$. The distance between the objective and the eye piece is $101 \,cm$. Find the focal length of the objective lens in $cm$.

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