Concave and convex lenses are placed touching | vedclass

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(B) Let the power of the concave lens be $P_1$ and the convex lens be $P_2$. Given the ratio of magnitudes of power is $|P_1| : |P_2| = 2 : 3$. Since

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$-15 \ cm, 10 \ cm$

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(B) Let the power of the concave lens be $P_1$ and the convex lens be $P_2$. Given the ratio of magnitudes of power is $|P_1| : |P_2| = 2 : 3$. Since the concave lens has negative power and the convex lens has positive power,let $P_1 = -2k$ and $P_2 = 3k$. The equivalent power of the system is $P_{eq} = P_1 + P_2 = -2k + 3k = k$. The equivalent focal length is $F_{eq} = 30 \ cm = 0.3 \ m$. Thus,$P_{eq} = \frac{1}{F_{eq}} = \frac{1}{0.3} = \frac{10}{3} \ D$. So,$k = \frac{10}{3}$. The power of the concave lens is $P_1 = -2 	imes \frac{10}{3} = -\frac{20}{3} \ D$. The focal length of the concave lens is $f_1 = \frac{1}{P_1} = -\frac{3}{20} \ m = -15 \ cm$. The power of the convex lens is $P_2 = 3 	imes \frac{10}{3} = 10 \ D$. The focal length of the convex lens is $f_2 = \frac{1}{P_2} = \frac{1}{10} \ m = 10 \ cm$. Therefore,the individual focal lengths are $-15 \ cm$ and $10 \ cm$.

Concave and convex lenses are placed touching each other. The ratio of the magnitudes of their power is $2:3$. The focal length of the system is $30 \ cm$. Then the focal lengths of the individual lenses are:

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