The dispersive powers of the materials of two | vedclass

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

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$-20 ~cm, 25 ~cm$

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(A) For an achromatic combination of two thin lenses in contact,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 $\frac{\omega_1}{\omega_2} = \frac{4}{3}$,we have $\frac{f_1}{f_2} = -\frac{4}{3}$,so $f_1 = -\frac{4}{3}f_2$.The effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.Given $F = +60 ~cm$,we have $\frac{1}{60} = \frac{1}{f_1} + \frac{1}{f_2}$.Substituting $f_1 = -\frac{4}{3}f_2$ into the equation:$\frac{1}{60} = \frac{1}{-(4/3)f_2} + \frac{1}{f_2} = -\frac{3}{4f_2} + \frac{1}{f_2} = \frac{-3+4}{4f_2} = \frac{1}{4f_2}$.Thus,$4f_2 = 60$,which gives $f_2 = 15 ~cm$.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 two lenses is $+60 ~cm$,then the focal lengths of the lenses should be:

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