Magnifying power of an astronomical telescope | vedclass

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(A) For an astronomical telescope in normal adjustment,the magnifying power $m$ is given by $m = \frac{f_o}{f_e} = 10$,where $f_o$ and $f_e$ are the f

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) For an astronomical telescope in normal adjustment,the magnifying power $m$ is given by $m = \frac{f_o}{f_e} = 10$,where $f_o$ and $f_e$ are the focal lengths of the objective and eyepiece lenses respectively.Thus,$f_o = 10 f_e$.The length of the telescope tube in normal adjustment is $L = f_o + f_e = 110 \ cm$.Substituting $f_o = 10 f_e$ into the length equation: $10 f_e + f_e = 110 \implies 11 f_e = 110 \implies f_e = 10 \ cm$.Then,$f_o = 100 \ cm$.When the image is formed at the near point $(D = 25 \ cm)$,the magnifying power is given by $m = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}ight)$.Substituting the values: $m = \frac{100}{10} \left(1 + \frac{10}{25}ight) = 10 	imes \left(1 + 0.4ight) = 10 	imes 1.4 = 14$.

Magnifying power of an astronomical telescope for normal adjustment is $10$ and the length of the telescope is $110 \ cm$. The magnifying power of the same telescope,when the image is formed at the near point,is:

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