Concave and convex lenses are placed touching | vedclass

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Question

(B) Power $(P) = \frac{1}{f} \dots (i)$Given,the ratio of magnitudes of power is $\frac{|P_{	ext{concave}}|}{|P_{	ext{convex}}|} = \frac{2}{3} \dots

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$10 \ cm, -15 \ cm$

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(B) Power $(P) = \frac{1}{f} \dots (i)$Given,the ratio of magnitudes of power is $\frac{|P_{	ext{concave}}|}{|P_{	ext{convex}}|} = \frac{2}{3} \dots (ii)$Let the focal length of the convex lens be $f_{	ext{convex}} = f$ (positive) and the focal length of the concave lens be $f_{	ext{concave}} = -f'$ (negative).Since $P = \frac{1}{f}$,the ratio of powers is $\frac{1/f'}{1/f} = \frac{f}{f'} = \frac{2}{3}$,which implies $f' = \frac{3}{2}f$.Thus,$f_{	ext{convex}} = f$ and $f_{	ext{concave}} = -\frac{3}{2}f$.Using the formula for equivalent focal length of two lenses in contact: $\frac{1}{f_{eq}} = \frac{1}{f_{	ext{convex}}} + \frac{1}{f_{	ext{concave}}}$$\frac{1}{30} = \frac{1}{f} - \frac{1}{\frac{3}{2}f} = \frac{1}{f} - \frac{2}{3f} = \frac{3-2}{3f} = \frac{1}{3f}$$\frac{1}{30} = \frac{1}{3f} \Rightarrow 3f = 30 \Rightarrow f = 10 \ cm$.Therefore,$f_{	ext{convex}} = 10 \ cm$ and $f_{	ext{concave}} = -\frac{3}{2}(10) = -15 \ cm$.

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

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