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(D) The intensity level $L$ in $bel$ is given by $L = \log_{10} \left( \frac{I}{I_0} \right)$.For two waves with intensity levels $L_1 = 1 \, bel$ and

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$1:10^2$

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(D) The intensity level $L$ in $bel$ is given by $L = \log_{10} \left( \frac{I}{I_0} \right)$.For two waves with intensity levels $L_1 = 1 \, bel$ and $L_2 = 5 \, bel$:$L_2 - L_1 = \log_{10} \left( \frac{I_2}{I_0} \right) - \log_{10} \left( \frac{I_1}{I_0} \right) = \log_{10} \left( \frac{I_2}{I_1} \right)$.Substituting the values:$5 - 1 = \log_{10} \left( \frac{I_2}{I_1} \right) \implies 4 = \log_{10} \left( \frac{I_2}{I_1} \right)$.Therefore,$\frac{I_2}{I_1} = 10^4$.Since intensity $I$ is proportional to the square of the amplitude $a$ $(I \propto a^2)$:$\frac{I_2}{I_1} = \left( \frac{a_2}{a_1} \right)^2 = 10^4$.Taking the square root on both sides:$\frac{a_2}{a_1} = \sqrt{10^4} = 10^2 = 100$.Thus,the ratio of amplitudes $a_1 : a_2 = 1 : 10^2$.

The intensity level due to two waves of the same frequency in a given medium are $1 \, bel$ and $5 \, bel$. Then the ratio of amplitudes is

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