The magnifying power of an astronomical teles | vedclass

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(A) For an astronomical telescope in normal adjustment,the length of the tube is $L = f_o + f_e = 54 \, cm$.The magnifying power is given by $m = \fra

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$6 \, cm$ and $48 \, cm$

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(A) For an astronomical telescope in normal adjustment,the length of the tube is $L = f_o + f_e = 54 \, cm$.The magnifying power is given by $m = \frac{f_o}{f_e} = 8$,which implies $f_o = 8f_e$.Substituting $f_o = 8f_e$ into the length equation: $8f_e + f_e = 54 \, cm$.$9f_e = 54 \, cm \Rightarrow f_e = 6 \, cm$.Now,calculating the objective focal length: $f_o = 8 	imes 6 \, cm = 48 \, cm$.Thus,the focal length of the eye lens is $6 \, cm$ and the objective lens is $48 \, cm$.

The magnifying power of an astronomical telescope is $8$ and the distance between the two lenses is $54 \, cm$. The focal lengths of the eye lens and the objective lens will be respectively:

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