The dispersive powers of the materials of two | vedclass

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(A) For an achromatic combination of two thin lenses, the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$, which implies $\frac{f_1}{f_

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$-20 \ cm, 25 \ cm$

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(A) For an achromatic combination of two thin lenses, the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$, which implies $\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$.Given the ratio of dispersive powers $\frac{\omega_1}{\omega_2} = \frac{4}{3}$, we have $\frac{f_1}{f_2} = -\frac{4}{3}$, so $f_1 = -\frac{4}{3}f_2$.The effective focal length $F$ of the combination is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.Substituting $F = 60 \ cm$ and $f_1 = -\frac{4}{3}f_2$:$\frac{1}{60} = \frac{1}{-(4/3)f_2} + \frac{1}{f_2} = \frac{1}{f_2} (1 - \frac{3}{4}) = \frac{1}{f_2} (\frac{1}{4})$.Thus, $f_2 = 60 	imes \frac{1}{4} = 15 \ cm$.Then, $f_1 = -\frac{4}{3} 	imes 15 = -20 \ cm$.Therefore, the focal lengths are $-20 \ cm$ and $15 \ cm$.

The dispersive powers of the materials of two lenses forming an achromatic combination are in the ratio of $4:3$. If the effective focal length of the combination is $+60 \ cm$, then the focal lengths of the lenses should be:

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