Physics - Geometrical Optics Question with Solution | TestHub

- The condition of achromatism is $W_1{P}_1+W_2{P}_2=0$ $$\begin{array}{rlrl}\Rightarrow & W_1{P}_1& =-W_2{P}_2& \\ \Rightarrow & & \frac{W_1}{W_2}& =-\frac{P_2}{P_1}\end{array}$$ Now, $P_1+P_2=4\mathrm{D}$ but, Power of converging lens, $$P_1=5\mathrm{D}$$ $\because$ Power of diverging lens $$\begin{array}{rl}{P}_2& =4\mathrm{D}-P_1\\ & =4\mathrm{D}-5\mathrm{D}=-\mathrm{D}\end{array}$$ [From ii] $\therefore$ From Eq. (i), we have $$\frac{W_1}{W_2}=-\frac{P_2}{P_1}=\frac{-(-D)}{5D}=\frac{1}{5}\Rightarrow \frac{W_1}{W_2}=\frac{1}{5}$$