For a telescope in normal adjustment,the leng | vedclass

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(A) In normal adjustment,the magnifying power $m$ is given by $m = \frac{f_o}{f_e}$.Given $m = 8$,so $f_o = 8f_e$.The length of the telescope $L$ in n

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$24 \, cm, 3 \, cm$

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(A) In normal adjustment,the magnifying power $m$ is given by $m = \frac{f_o}{f_e}$.Given $m = 8$,so $f_o = 8f_e$.The length of the telescope $L$ in normal adjustment is $L = f_o + f_e$.Given $L = 27 \, cm$,we have $f_o + f_e = 27$.Substituting $f_o = 8f_e$ into the equation: $8f_e + f_e = 27$.$9f_e = 27$,which gives $f_e = 3 \, cm$.Now,$f_o = 8 	imes 3 = 24 \, cm$.Thus,the focal lengths are $24 \, cm$ and $3 \, cm$.

For a telescope in normal adjustment,the length of the telescope is found to be $27 \, cm$. If the magnifying power of the telescope at normal adjustment is $8$,the focal lengths of the objective and the eyepiece are respectively:

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