The dispersive powers of the glasses of lense | vedclass

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

Vedclass · Accepted Answer

Convex,$9 \ cm$

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(A) For an achromatic doublet,the condition is given by $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$,which implies $\frac{\omega_1}{\omega_2} = -\frac{f_1}{f_2}$.Given the ratio of dispersive powers $\frac{\omega_1}{\omega_2} = \frac{5}{3}$.Let the concave lens be the first lens,so $f_1 = -15 \ cm$.Substituting the values: $\frac{5}{3} = -\frac{-15}{f_2}$.$\frac{5}{3} = \frac{15}{f_2} \Rightarrow 5f_2 = 45 \Rightarrow f_2 = 9 \ cm$.Since $f_2$ is positive,the second lens must be a convex lens.Therefore,the other lens is convex with a focal length of $9 \ cm$.

The dispersive powers of the glasses of lenses used in an achromatic pair are in the ratio $5 : 3$. If the focal length of the concave lens is $15 \ cm$,then the nature and focal length of the other lens would be:

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