In the absence of the teacher, the noise level in a classroom...

Let the sound intensity produced by one student be $I_{s}$ and the reference intensity be $I_{0}$.

The total intensity produced by $50$ students is $I_{1} = 50I_{s}$.

The noise level is given by $L_{1} = 10 \log_{10} \left( \dfrac{I_{1}}{I_{0}} \right) = 50 \text{ dB}$.

When the number of students increases to $100$, the new intensity is $I_{2} = 100I_{s}$.

The new noise level is $L_{2} = 10 \log_{10} \left( \dfrac{I_{2}}{I_{0}} \right).<br><br>The difference in noise levels is$L_{2} - L_{1} = 10 \log_{10} \left( \dfrac{I_{2}}{I_{1}} \right)$.<br><br>Substituting the values,$L_{2} - 50 = 10 \log_{10} \left( \dfrac{100I_{s}}{50I_{s}} \right) = 10 \log_{10}(2)$.<br><br>Since$\log_{10}(2) \approx 0.301$, we get$L_{2} - 50 = 10 \times 0.301 \approx 3$.<br><br>$L_{2} = 53 \text{ dB}$.$