Show that the axial chromatic aberration $(f_{r} - f_{v})$ for a convex lens is equal to the product of its mean focal length $(f)$ and dispersive power $(ω)$ of its material i.e. Prove:
$f_{r} - f_{v} = ω f$

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Solution

We know $\frac{1}{f_{r}} = (n_{r} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\frac{1}{f_{v}} = (n_{v} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\frac{1}{f_{v}} - \frac{1}{f_{r}} = (n_{v} - n_{r}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$= \frac{n_{v} - n_{r}}{(n_{y} - 1)} (n_{y} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
but $(n_{y} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{1}{f}$ , where $f$ is the mean focal length.
So, $\frac{1}{f_{v}} - \frac{1}{f_{r}} = \frac{n_{v} - n_{r}}{n_{y} - 1} \times \frac{1}{f}$
$\frac{f_{r} - f_{v}}{f_{v} f_{r}} = \frac{n_{v} - n_{r}}{n_{y} - 1} \times \frac{1}{f}$
$\frac{f_{r} - f_{v}}{f^{2}} = \frac{n_{v} - n_{r}}{n_{y} - 1} \times \frac{1}{f}$
$f_{r} - f_{v} = ω f$

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