= Solution
Use the transmitted-flux form of the <Beer-Lambert law>. The <optical depth> falls from $0.25$ to $0.225$, so
$$
\frac{F'}{F}=\frac{e^{-0.225}}{e^{-0.25}}=e^{0.025},
\qquad\boxed{100\left(\frac{F'}F-1\right)=2.532\%\ \mathrm{increase}.}
$$
A ten-percent loss of <ozone> therefore does not imply a ten-percent increase of transmitted UV. The response depends on the initial <optical depth>, <wavelength> and path geometry. This calculation uses the stated effective <optical depth> and holds the incident flux and other absorbers/scatterers fixed.
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