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An achromatic convergent doublet of two lense in contact has a power of $+5 \mathrm{D}$. The power of converging lens is $+6 \mathrm{D}$. The ratio of the dispersive power of the convergent and divergent lenses is
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$1: 5$
$1: 5$
- 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, $\quad P_1+P_2=4 \mathrm{D}$
but, Power of converging lens,
$$
P_1=5 \mathrm{D}
$$
$\because$ Power of diverging lens
$$
\begin{aligned}
P_2 & =4 \mathrm{D}-P_1 \\
& =4 \mathrm{D}-5 \mathrm{D}=-\mathrm{D}
\end{aligned}
$$
[From ii]
$\therefore$ From Eq. (i), we have
$$
\frac{W_1}{W_2}=-\frac{P_2}{P_1}=\frac{-(-D)}{5 D}=\frac{1}{5} \Rightarrow \frac{W_1}{W_2}=\frac{1}{5}
$$
$$
\begin{array}{rlrl}
\Rightarrow & W_1 P_1 & =-W_2 P_2 \\
\Rightarrow & & \frac{W_1}{W_2} & =-\frac{P_2}{P_1}
\end{array}
$$
Now, $\quad P_1+P_2=4 \mathrm{D}$
but, Power of converging lens,
$$
P_1=5 \mathrm{D}
$$
$\because$ Power of diverging lens
$$
\begin{aligned}
P_2 & =4 \mathrm{D}-P_1 \\
& =4 \mathrm{D}-5 \mathrm{D}=-\mathrm{D}
\end{aligned}
$$
[From ii]
$\therefore$ From Eq. (i), we have
$$
\frac{W_1}{W_2}=-\frac{P_2}{P_1}=\frac{-(-D)}{5 D}=\frac{1}{5} \Rightarrow \frac{W_1}{W_2}=\frac{1}{5}
$$
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