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(D) Let the focal lengths of the two lenses be $f_1$ and $f_2$. Given that the lenses are in contact,the equivalent focal length $F$ is given by $\fra

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$12$

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(D) Let the focal lengths of the two lenses be $f_1$ and $f_2$. Given that the lenses are in contact,the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Given $F = 4 \ cm$,so $\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{4}$. This simplifies to $\frac{f_1 + f_2}{f_1 f_2} = \frac{1}{4}$. We are given $f_1 + f_2 = 18 \ cm$. Substituting this into the equation: $\frac{18}{f_1 f_2} = \frac{1}{4}$,which gives $f_1 f_2 = 72 \ cm^2$. We have the sum $f_1 + f_2 = 18$ and the product $f_1 f_2 = 72$. These are the roots of the quadratic equation $x^2 - 18x + 72 = 0$. Factoring the equation: $(x - 6)(x - 12) = 0$. Thus,the focal lengths are $6 \ cm$ and $12 \ cm$. Power $P$ is inversely proportional to focal length $(P = \frac{1}{f})$. The lens with the lower power will have the larger focal length. Therefore,the focal length of the lens with lower power is $12 \ cm$.

Two thin convex lenses are kept in contact coaxially. If the focal length of the combination of the lenses is $4 \ cm$ and the sum of the focal lengths of the two lenses is $18 \ cm$,then the focal length of the lens of lower power is (in $cm$)

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