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(A) For an achromatic doublet,the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,which implies $\frac{P_1}{\omega_1} + \frac{P_2}{\ome

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$L_1$ is converging

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(A) For an achromatic doublet,the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,which implies $\frac{P_1}{\omega_1} + \frac{P_2}{\omega_2} = 0$,where $P_1$ and $P_2$ are the powers of the lenses.Since $\omega_2 For the combination to be converging,the net power $P = P_1 + P_2$ must be positive.Since $|P_1| > |P_2|$,the sign of the net power $P$ is determined by the lens with the larger power,which is $L_1$.Therefore,$L_1$ must be the converging lens (convex) and $L_2$ must be the diverging lens (concave).

It is desired to make an achromatic combination of two lenses ($L_1$ and $L_2$) made of materials having dispersive powers $\omega_1$ and $\omega_2$ $(< \omega_1)$. If the combination of lenses is converging,then:

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