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(B) For an achromatic combination,the condition is $\frac{f_1}{\omega_1} + \frac{f_2}{\omega_2} = 0$,which implies $\frac{f_1}{f_2} = -\frac{\omega_1}

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$30 \ cm$ and $-45 \ cm$

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(B) For an achromatic combination,the condition is $\frac{f_1}{\omega_1} + \frac{f_2}{\omega_2} = 0$,which implies $\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$.Given $\omega_1 = 0.024$ and $\omega_2 = 0.036$,we have $\frac{f_1}{f_2} = -\frac{0.024}{0.036} = -\frac{2}{3}$,so $f_2 = -1.5 f_1$.The equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{90}$.Substituting $f_2 = -1.5 f_1$ into the equation: $\frac{1}{f_1} - \frac{1}{1.5 f_1} = \frac{1}{90}$.$\frac{1.5 - 1}{1.5 f_1} = \frac{1}{90} \implies \frac{0.5}{1.5 f_1} = \frac{1}{90} \implies \frac{1}{3 f_1} = \frac{1}{90}$.Thus,$f_1 = 30 \ cm$ and $f_2 = -1.5(30) = -45 \ cm$.

The focal length of the field lens (which is an achromatic combination of two lenses) of a telescope is $90 \ cm$. The dispersive powers of the two lenses in the combination are $0.024$ and $0.036$. The focal lengths of the two lenses are:

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