A lens maker has to design an achromatic comb | vedclass

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(B) For an achromatic combination,the condition is $\frac{\omega_A}{f_A} + \frac{\omega_B}{f_B} = 0$. Given $\frac{\omega_A}{\omega_B} = \frac{3}{5}$,

Vedclass · Accepted Answer

$-80$

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(B) For an achromatic combination,the condition is $\frac{\omega_A}{f_A} + \frac{\omega_B}{f_B} = 0$. Given $\frac{\omega_A}{\omega_B} = \frac{3}{5}$,we have $\frac{3}{f_A} + \frac{5}{f_B} = 0$,which implies $\frac{f_A}{f_B} = -\frac{3}{5}$,or $f_B = -\frac{5}{3} f_A$. The effective power $P = P_A + P_B = -0.5 \, D$. Since $P = \frac{100}{f(cm)}$,we have $\frac{100}{f_A} + \frac{100}{f_B} = -0.5$. Substituting $f_B = -\frac{5}{3} f_A$: $\frac{100}{f_A} - \frac{100 	imes 3}{5 f_A} = -0.5$ $\frac{100}{f_A} - \frac{60}{f_A} = -0.5$ $\frac{40}{f_A} = -0.5$ $f_A = \frac{40}{-0.5} = -80 \, cm$.

$A$ lens maker has to design an achromatic combination of effective power $-0.5 \, D$. The two types of glasses to be used (say $A$ and $B$) have dispersive powers in the ratio $\omega_A : \omega_B = 3 : 5$. Then the lens made from glass $A$ will have a focal length of ....... $cm$.

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