The objective glass of a telescope is an achr | vedclass

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Question

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

Vedclass · Accepted Answer

$30\,cm, -45\,cm$

Vedclass · Answer

(A) For an achromatic doublet,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 $\omega_1 = 0.024$ and $\omega_2 = 0.036$,we have $\frac{f_1}{f_2} = -\frac{0.024}{0.036} = -\frac{2}{3}$,so $f_1 = -\frac{2}{3}f_2$.The combined focal length is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{90}$.Substituting $f_1 = -\frac{2}{3}f_2$ into the equation: $\frac{1}{f_2} - \frac{3}{2f_2} = \frac{1}{90}$.$-\frac{1}{2f_2} = \frac{1}{90}$,which gives $f_2 = -45\,cm$.Then $f_1 = -\frac{2}{3}(-45) = 30\,cm$.Thus,the focal lengths are $30\,cm$ and $-45\,cm$.

The objective glass of a telescope is an achromat of focal length $90\,cm$. The magnitudes of dispersive powers of the two lenses are $0.024$ and $0.036$. The focal lengths of the two lenses are:

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