An achromatic convergent doublet of two lense | vedclass

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(A) For an achromatic doublet, the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$, which can be written in terms of power as $\omega_1

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$3: 5$

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(A) For an achromatic doublet, the condition is $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$, which can be written in terms of power as $\omega_1 P_1 + \omega_2 P_2 = 0$.Given: Total power $P = P_1 + P_2 = +2 	ext{ D}$.Power of convex lens $P_1 = +5 	ext{ D}$.Therefore, $P_2 = P - P_1 = 2 - 5 = -3 	ext{ D}$.Substituting the values into the achromatic condition: $\omega_1(5) + \omega_2(-3) = 0$.This implies $5\omega_1 = 3\omega_2$.The ratio of the dispersive powers is $\frac{\omega_1}{\omega_2} = \frac{3}{5}$.

An achromatic convergent doublet of two lenses in contact has a power $+2 \text{ D}$. The convex lens has power $+5 \text{ D}$. The ratio of the dispersive powers of the convergent and divergent lenses is (in magnitude)

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