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Question

(A) For an achromatic doublet,the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,where $\omega_1, \omega_2$ are dispersive powers and

Vedclass · Accepted Answer

$5\,cm, -10\,cm$

Vedclass · Answer

(A) For an achromatic doublet,the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,where $\omega_1, \omega_2$ are dispersive powers and $f_1, f_2$ are focal lengths.Given $\frac{\omega_1}{\omega_2} = 2$,we have $\omega_1 = 2\omega_2$.Substituting this into the condition: $\frac{2\omega_2}{f_1} + \frac{\omega_2}{f_2} = 0 \implies \frac{2}{f_1} = -\frac{1}{f_2} \implies f_1 = -2f_2$.The equivalent focal length $F$ of two lenses in contact is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.Given $F = 10\,cm$,we have $\frac{1}{10} = \frac{1}{-2f_2} + \frac{1}{f_2} = \frac{-1 + 2}{2f_2} = \frac{1}{2f_2}$.Thus,$2f_2 = 10 \implies f_2 = 5\,cm$.Then $f_1 = -2(5) = -10\,cm$.However,checking the options,if we assume the ratio $\frac{\omega_1}{\omega_2} = 2$ and the combination is $10\,cm$,the pair $(5\,cm, -10\,cm)$ satisfies $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$ if $\frac{2}{5} + \frac{1}{-10} 
eq 0$. Re-evaluating: $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0 \implies \frac{f_1}{f_2} = -\frac{\omega_1}{\omega_2} = -2$.Using $\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{10}$,and $f_1 = -2f_2$,we get $\frac{1}{-2f_2} + \frac{1}{f_2} = \frac{1}{10} \implies \frac{1}{2f_2} = \frac{1}{10} \implies f_2 = 5\,cm, f_1 = -10\,cm$. The correct option is $A$ $(5\,cm, -10\,cm)$.

Two lenses in contact made of materials with dispersive powers in the ratio $2 : 1$ behave as an achromatic lens of focal length $10\,cm$. The individual focal lengths of the lenses are:

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