Question

It is desired to make an achromatic combination of two lenses $(L_{1}and{L}_2$) made of materials having dispersive power ${\omega }_{1}and{\omega }_2$, ${\omega }_2<{\omega }_1$. If the combination of lenses is converging, then -

• A. $L_1$ is converging
  
• B. $L_2$ is converging
  
• C. Power of $L_1$ is greater than the power of $L_2$
• D. None of these

Answer 1

The correct option is B $L_2$ is converging


As combination of lens is converging then
$\frac{1}{f_1}+\frac{1}{f_2}=+ve.....\left(1\right)$
Condition for achromatism
$\frac{{\omega }_1}{f_1}+\frac{{\omega }_2}{f_2}=0\Rightarrow \frac{{\omega }_1}{f_1}=\frac{-{\omega }_2}{f_2}$
${\omega }_2<{\omega }_1\Rightarrow \left|f_2\right|<\left|f_1\right|......\left(2\right)$
From equation (1) and (2), we can conclude that $f_2$ should be positive, that means $L_2$ will be converging lens.