$$ P_1 = 2.5\,D,\quad f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4\,m $$

After an increase in optical power by 0.1 D:

$$ P_2 = 2.5\,D + 0.1\,D = 2.6\,D,\quad f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846\,m $$

The change in focal length is:

$$ \Delta f = f_1 - f_2 = 0.4\,m - 0.3846\,m \approx 0.0154\,m $$

The relative decrease in focal length is then calculated as:

$$ \text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} \approx 0.0385 \approx 0.04 $$

Thus, the relative decrease in focal length is approximately 0.04.