Two thin lenses have a combined power of +9 D | vedclass

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(D) Let the powers of the two thin lenses be $P_1$ and $P_2$.Given, the combined power of the lenses in contact is $P = P_1 + P_2 = 9 D$ ---$(1)$When

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$3 D, 6 D$

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(D) Let the powers of the two thin lenses be $P_1$ and $P_2$.Given, the combined power of the lenses in contact is $P = P_1 + P_2 = 9 D$ ---$(1)$When the lenses are separated by a distance $d = 20 \,cm = 0.2 \,m$, the equivalent power $P_{eq}$ is given by the formula:$P_{eq} = P_1 + P_2 - d P_1 P_2$Substituting the given values:$\frac{27}{5} = 9 - 0.2 P_1 P_2$$5.4 = 9 - 0.2 P_1 P_2$$0.2 P_1 P_2 = 9 - 5.4 = 3.6$$P_1 P_2 = \frac{3.6}{0.2} = 18$ ---$(2)$We have the sum $P_1 + P_2 = 9$ and the product $P_1 P_2 = 18$.These are the roots of the quadratic equation $x^2 - (P_1 + P_2)x + P_1 P_2 = 0$, which is $x^2 - 9x + 18 = 0$.Solving the quadratic equation:$(x - 3)(x - 6) = 0$So, $x = 3$ or $x = 6$.Thus, the individual powers are $3 D$ and $6 D$.The correct option is $(D)$.

Two thin lenses have a combined power of $+9 D$. When they are separated by a distance of $20 \,cm$, their equivalent power becomes $+\frac{27}{5} D$. Their individual powers (in diopters) are respectively:

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