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(A) When lenses are in contact,the equivalent power is $P = P_1 + P_2 = 10 \, D$. Since $P = 1/f$,we have $1/f_1 + 1/f_2 = 10$ ... $(i)$.When they are

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$0.125$ and $0.5$

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(A) When lenses are in contact,the equivalent power is $P = P_1 + P_2 = 10 \, D$. Since $P = 1/f$,we have $1/f_1 + 1/f_2 = 10$ ... $(i)$.When they are at a distance $d = 0.25 \, m$ apart,the equivalent power is $P' = P_1 + P_2 - d P_1 P_2 = 6 \, D$.Substituting $P_1 + P_2 = 10$ and $d = 0.25$,we get $10 - 0.25(1/f_1)(1/f_2) = 6$.$0.25/(f_1 f_2) = 4 \Rightarrow f_1 f_2 = 0.25/4 = 1/16 = 0.0625$ ... $(ii)$.From $(i)$,$(f_1 + f_2)/(f_1 f_2) = 10 \Rightarrow f_1 + f_2 = 10 	imes 0.0625 = 0.625$ ... $(iii)$.We have the sum $S = 0.625$ and product $P = 0.0625$. The quadratic equation $x^2 - Sx + P = 0$ gives the focal lengths.$x^2 - 0.625x + 0.0625 = 0$.Using the quadratic formula,$x = [0.625 \pm \sqrt{0.625^2 - 4(0.0625)}] / 2 = [0.625 \pm \sqrt{0.390625 - 0.25}] / 2 = [0.625 \pm \sqrt{0.140625}] / 2 = [0.625 \pm 0.375] / 2$.$x_1 = 1/2 = 0.5 \, m$ and $x_2 = 0.25/2 = 0.125 \, m$.

Two thin lenses,when in contact,produce a combination of power $+10 \, D$. When they are $0.25 \, m$ apart,the power reduces to $+6 \, D$. The focal lengths of the lenses (in $m$) are

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