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(D) Let the focal length of the convex lens be $f_1$ and the concave lens be $f_2$. Given that the ratio of their magnitudes is $|f_1/f_2| = 2/3$.Sinc

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$10, -15$

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(D) Let the focal length of the convex lens be $f_1$ and the concave lens be $f_2$. Given that the ratio of their magnitudes is $|f_1/f_2| = 2/3$.Since the convex lens has a positive focal length and the concave lens has a negative focal length,we have $f_1 = 2x$ and $f_2 = -3x$.The formula for the equivalent focal length $F$ of two thin lenses in contact is $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.Given $F = 30 \, cm$,we substitute the values: $\frac{1}{30} = \frac{1}{2x} - \frac{1}{3x}$.Solving for $x$: $\frac{1}{30} = \frac{3-2}{6x} \implies \frac{1}{30} = \frac{1}{6x}$.Thus,$6x = 30$,which gives $x = 5$.Therefore,$f_1 = 2(5) = 10 \, cm$ and $f_2 = -3(5) = -15 \, cm$.

$A$ 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 $30 \, cm$. Their individual focal lengths in $cm$ will be

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