The dispersive powers of the materials of two lenses forming an achromatic combination are in the ratio of…

The dispersive powers of the materials of two lenses forming an achromatic combination are in the ratio of $4: 3$. Effective focal length of the two lenses is $+60 \mathrm{~cm}$ then the focal lengths of the lenses should be
  1. $-20 \mathrm{~cm}, 25 \mathrm{~cm}$
  2. $20 \mathrm{~cm},-25 \mathrm{~cm}$
  3. $-15 \mathrm{~cm}, 20 \mathrm{~cm}$
  4. $15 \mathrm{~cm},-20 \mathrm{~cm}$

Solution

$\begin{aligned} P=P_1+P_2 & =\frac{q}{f_1}+\frac{q}{f_2} \\ \frac{1}{60} & =\frac{q}{f_1}+\frac{q}{f_2} \\ \frac{1}{60} & =\frac{f_1+f_2}{f_1 f_2}\end{aligned}$ According to question, $\frac{\omega_1}{\omega_2}=\frac{f_1}{f_2} \Rightarrow \frac{f_1}{f_2}=\frac{-4}{3}$ or $f_1=-\frac{4}{3} f_2$ From Eq. (i) $\frac{1}{60}=\frac{\frac{4}{3} f_2+f_2}{\left(f_2\right)^2 \frac{4}{3}} \Rightarrow \frac{-\frac{1}{3}}{-\frac{4}{3} f_2}$ $\begin{aligned} f_2 & =15 \mathrm{~cm} \\ \therefore \quad f_1 & =-\frac{4}{3} \times 15 \\ & =-20 \mathrm{~cm}\end{aligned}$

Asked in: AP EAMCET 2012

Practice more Ray Optics questions on Aicharya