The magnifying power of an astronomical teles | vedclass

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(B) For an astronomical telescope in normal adjustment,the magnifying power is given by $m = \frac{f_o}{f_e} = 8$. This implies $f_o = 8f_e$. The leng

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

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

The magnifying power of an astronomical telescope is $8$. If the distance between the two lenses is $54 \, cm$,find the focal lengths of the eyepiece and the objective lens.

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