If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$
For $n = 1.5,\; \Delta n = 10^{-3}$ and $f = 20\,\mathrm{cm}$ ,the value of $|\Delta f|$ will be 0.02 cm (rounded off to 2nd decimal place).
The relation between $\;\; \frac{\Delta f}{f}$ and $\;\; \frac{\Delta n}{n}$ remains unchanged if both the convex surfaces arereplaced by concave surface of the same radius of curvature.
$\left|\;\; \frac{\Delta f}{f}\right| < \left|\;\; \frac{\Delta n}{n}\right|$

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