c
We have, \(\mathrm{L}_{1}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right) ; \mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)\)

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

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 \(100 .\)