An achromatic convergent lens of focal length | vedclass

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Question

(A) For an achromatic doublet,the condition is $\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$.Given $\frac{\omega_1}{\omega_2} = \frac{1}{2}$,we have

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$f_1 = 10 \, cm, f_2 = -20 \, cm$

Vedclass · Answer

(A) For an achromatic doublet,the condition is $\frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2}$.Given $\frac{\omega_1}{\omega_2} = \frac{1}{2}$,we have $\frac{f_1}{f_2} = -\frac{1}{2}$,which implies $f_2 = -2f_1$.The equivalent focal length $F$ of the combination is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.Given $F = 20 \, cm$,we substitute $f_2 = -2f_1$:$\frac{1}{20} = \frac{1}{f_1} - \frac{1}{2f_1} = \frac{2-1}{2f_1} = \frac{1}{2f_1}$.Thus,$2f_1 = 20 \, cm$,so $f_1 = 10 \, cm$.Then,$f_2 = -2(10 \, cm) = -20 \, cm$.

An achromatic convergent lens of focal length $20 \, cm$ is made of two lenses (in contact) of materials having dispersive powers in the ratio of $1 : 2$ and having focal lengths $f_1$ and $f_2$. Which of the following is true?

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