Question

The sound intensity increases from \(I_1=10^{-6}\,\mathrm{W/m^2}\) to \(I_2=10^{-4}\,\mathrm{W/m^2}\). The increase in sound level is

• A. \(10\,\mathrm{dB}\)
• B. \(20\,\mathrm{dB}\)
• C. \(30\,\mathrm{dB}\)
• D. \(40\,\mathrm{dB}\)

Hint:

Sound level difference: \[ \boxed{ \Delta L = 10\log_{10}\left(\frac{I_2}{I_1}\right) } \] A 100-fold increase in intensity corresponds to \[ \boxed{20\,\mathrm{dB}.} \]

Answer 1

The Correct Option is B

Step 1: Recall why sound level is measured in decibels.
Sound level is a logarithmic ratio scale, so a change from $I_1$ to $I_2$ does not scale directly with the intensity ratio, it scales with its logarithm.
Step 2: Work out the intensity ratio first. \[ \frac{I_2}{I_1} = \frac{10^{-4}}{10^{-6}} = 10^2 = 100 \]
Step 3: Convert this ratio to decibels. \[ \Delta L = 10\log_{10}(100) = 10 \times 2 = 20\,\mathrm{dB} \]
since $\log_{10}(100)=2$ directly, without needing to expand the exponents separately.
\[ \boxed{20\,\mathrm{dB}} \]