$\therefore \mathrm{L}_{1}-\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right)-10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$

or, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}} \times \frac{\mathrm{I}_{0}}{\mathrm{I}_{2}}\right)$ or, $\Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$

or, $20=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ or, $2=\log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$

or, $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=10^{2}$ or, $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{100}$

$\Rightarrow$ Intensity decreases by a factor $100 .$