(b) $\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
Also, by Cauchy’s formula
$\mu = A + \frac{B}{\lambda^{2}} + \frac{C}{\lambda ⁴} + \ldots.$
$\lambda_{blue} < \ \lambda_{red}$
$\mu_{blue} > \ \mu_{red}$
Hence, $f_{red} > f_{blue}$