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(A) For an achromatic combination of two thin lenses in contact,the condition is given by $\frac{\omega_{1}}{f_{1}} + \frac{\omega_{2}}{f_{2}} = 0$.Gi

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$5\, cm, -10\, cm$

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(A) For an achromatic combination of two thin lenses in contact,the condition is given by $\frac{\omega_{1}}{f_{1}} + \frac{\omega_{2}}{f_{2}} = 0$.Given the ratio of dispersive powers $\omega_{1} : \omega_{2} = 2 : 1$,we have $\omega_{1} = 2\omega_{2}$.Substituting this into the condition: $\frac{2\omega_{2}}{f_{1}} + \frac{\omega_{2}}{f_{2}} = 0$,which simplifies to $\frac{2}{f_{1}} + \frac{1}{f_{2}} = 0$,or $f_{1} = -2f_{2} \quad \dots(1)$.The net focal length of the combination is given by $\frac{1}{f_{	ext{net}}} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$.Given $f_{	ext{net}} = 10\, cm$,we have $\frac{1}{10} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \quad \dots(2)$.Substituting $f_{1} = -2f_{2}$ from equation $(1)$ into equation $(2)$:$\frac{1}{10} = \frac{1}{-2f_{2}} + \frac{1}{f_{2}} = \frac{-1 + 2}{2f_{2}} = \frac{1}{2f_{2}}$.Thus,$2f_{2} = 10$,which gives $f_{2} = 5\, cm$.Substituting $f_{2} = 5\, cm$ back into equation $(1)$: $f_{1} = -2(5) = -10\, cm$.Therefore,the individual focal lengths are $-10\, cm$ and $5\, cm$.

Two lenses in contact made of materials with dispersive powers in the ratio $2 : 1$,behave as an achromatic convex lens of focal length $10\, cm$. The individual focal lengths of the lenses are

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