Question

Show that, two thin lenses kept in contact, form an achromatic doublet if they satisfy the condition:
$\frac{\omega }{f}+\frac{{\omega }^{\prime }}{f^{\prime }}=0$
where the terms have their usual meaning.

Answer 1

Let two thin lenses are kept in contact.
For first lens we have
$\frac{1}{f_v}=(n_v-1)(\frac{1}{R_1}-\frac{1}{R_2})$
and $\frac{1}{f_R}=(n_R-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f_V}-\frac{1}{f_R}=(n_V-n_R)(\frac{1}{R_1}-\frac{1}{R_2})$
$=\frac{(n_V-n_R)}{(n_Y-1)}(n_Y-1)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{\omega }{f_Y}$
or $\frac{1}{f_{V^{\prime }}}-\frac{1}{f_R}=\frac{\omega }{f}$ $(Let{f}_y=f)$ ..... (1)
Similarly for second lens
$\frac{1}{f_{V^{\prime }}}-\frac{1}{f_{R^{\prime }}}=\frac{{\omega }^{\prime }}{f^{\prime }}$ ......... (2)
Adding equations (1) and (2)
$(\frac{1}{f_V}+\frac{1}{f_{V^{\prime }}})-(\frac{1}{f_R}+\frac{1}{f_{R^{\prime }}})=\frac{\omega }{f}+\frac{{\omega }^{\prime }}{f^{\prime }}$ ....... (3)
If the focal length of this lens combination for violet and red rays be $F_V$ and $F_R$ respectively then
$\frac{1}{f_V}+\frac{1}{f_{V^{\prime }}}=\frac{1}{F_V}$
and $\frac{1}{f_R}+\frac{1}{f_{R^{\prime }}}=\frac{1}{F_R}$
So equation (3) becomes
$\frac{1}{F_V}-\frac{1}{F_R}=\frac{\omega }{f}+\frac{{\omega }^{\prime }}{f^{\prime }}$
But for achromatism of the lens combination, the focal length must be same for all colours of light so, $F_V=F_R$
So, $\frac{\omega }{f}+\frac{{\omega }^{\prime }}{f^{\prime }}=0$
i.e., $\frac{\omega }{f}+\frac{{\omega }^{\prime }}{f^{\prime }}=0$