Two sounds differ in sound level by$\mathbf{\text{1.00dB}}$. What is the ratio of the greater intensity to the smaller intensity?

Change in sound level,$\text{Δ}\beta =1.0\text{dB}$

The sound level$\mathbf{\beta }\mathbf{}$in decibels (dB) can be defined as,

$\mathbf{\beta }\mathbf{=}\text{10}\left(\text{dB}\right)\text{log}(\mathbf{I}\mathbf{/}{\mathbf{I}}_0)$

where,${\mathbf{I}}_0\mathbf{}$is the reference intensity level to which all intensities are compared. For every multiple of 10 of intensity, 10 dB is added to the sound level.

To find the ratio of the greater intensity to the lower intensity, consider the original intensity$I_1$and final intensity$I_2$. Also, the original and final sound level can be given as,

$$\begin{gathered} {\beta }_1=10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right) \\ {\beta }_2=10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right) \\ \begin{array}{c}\text{Δ}\beta ={\beta }_2-{\beta }_1\\ =1.\text{0dB}\end{array} \end{gathered}$$

Thus, the above equation becomes,

$\begin{array}{l}10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right)=10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right)+1.0\text{dB}\\ 10\left(\text{dB}\right)\log \left(\frac{I_2}{I_0}\right)-10\left(\text{dB}\right)\log \left(\frac{I_1}{I_0}\right)=1.0\text{dB}\end{array}$

Dividing the above equation by 10 dB and using identity,

$\log \left(\frac{I_2}{I_0}\right)-\log \left(\frac{I_1}{I_0}\right)=\log \left(\frac{I_2}{I_1}\right)$

Therefore,

$\log \left(\frac{I_2}{I_1}\right)=0.1\text{dB}$

Now, using each side as exponent of 10.Therefore,

$$\begin{gathered} {10}^{\log \left(\frac{I_2}{I_1}\right)}=\left(\frac{I_2}{I_1}\right) \\ \left(\frac{I_2}{I_1}\right)={10}^{0.1} \\ \left(\frac{I_2}{I_1}\right)=1.26 \end{gathered}$$

Therefore, the ratio of greater intensity and smaller intensity is$1.26$