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(D) Let the focal length of the convex lens be $f_1$ and the concave lens be $f_2$. Since the lens is concave,$f_2$ is negative.Given the ratio of mag

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$-15, 10$

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(D) Let the focal length of the convex lens be $f_1$ and the concave lens be $f_2$. Since the lens is concave,$f_2$ is negative.Given the ratio of magnitudes: $|f_1| / |f_2| = 2/3$,so $f_1 / (-f_2) = 2/3$,which implies $f_2 = -1.5 f_1$.The formula for the equivalent focal length $F$ of two thin lenses in contact is $1/F = 1/f_1 + 1/f_2$.Given $F = 30 \ cm$,we have $1/30 = 1/f_1 - 1/|f_2|$.Substituting $|f_2| = 1.5 f_1$: $1/30 = 1/f_1 - 1/(1.5 f_1) = (1.5 - 1) / (1.5 f_1) = 0.5 / (1.5 f_1) = 1 / (3 f_1)$.Thus,$3 f_1 = 30$,so $f_1 = 10 \ cm$.Then,$|f_2| = 1.5 	imes 10 = 15 \ cm$. Since it is a concave lens,$f_2 = -15 \ cm$.Therefore,the focal lengths are $10 \ cm$ and $-15 \ cm$.

$A$ convex lens is in contact with a concave lens. The magnitude of the ratio of their focal lengths is $2/3$. Their equivalent focal length is $30 \ cm$. What are their individual focal lengths?

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