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(D) For an astronomical telescope in normal adjustment,the length of the telescope is $L = f_o + f_e = 10 \ cm$. The magnifying power is given by $M =

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${L_4}, {L_1}$

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(D) For an astronomical telescope in normal adjustment,the length of the telescope is $L = f_o + f_e = 10 \ cm$. The magnifying power is given by $M = \frac{f_o}{f_e} = 4$,which implies $f_o = 4f_e$. Substituting $f_o = 4f_e$ into the length equation: $4f_e + f_e = 10 \ cm$,so $5f_e = 10 \ cm$,which gives $f_e = 2 \ cm$. Then,$f_o = 4 	imes 2 \ cm = 8 \ cm$. Comparing these with the given focal lengths: $f_o = 8 \ cm$ corresponds to lens ${L_4}$ and $f_e = 2 \ cm$ corresponds to lens ${L_1}$. Thus,the objective lens is ${L_4}$ and the eye lens is ${L_1}$.

In a laboratory,four convex lenses ${L_1}, {L_2}, {L_3},$ and ${L_4}$ of focal lengths $2 \ cm, 4 \ cm, 6 \ cm,$ and $8 \ cm$ respectively are available. Two of these lenses form a telescope of length $10 \ cm$ and magnifying power $4.$ The objective and eye lenses are:

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