Question

An intensity level change of \(+1.00 \mathrm{dB}\) corresponds to what percentage change in intensity?

Short answer

Answer: The percentage change in intensity for a change of 1 dB in intensity level is approximately 25.89%.

Step-by-step solution

Recall the decibel formula

The decibel formula helps us relate the intensity level (in dB) to the intensity (in W/m²). The formula is: $$ L = 10 \log_{10} \frac{I}{I_0} $$ where \(L\) is the intensity level in dB, \(I\) is the intensity in W/m², and \(I_0\) is the reference intensity, typically taken as \(10^{-12}\) W/m².

Find the initial and new intensity levels

We know that the intensity level change is \(+1.00\) dB. Let's assume the initial intensity level is \(L_1\) and the new intensity level is \(L_2\). Since the change in intensity level is \(+1.00\) dB, we can write: $$ L_2 = L_1 + 1 $$ Now, let's write the decibel formula for both the initial and new intensity levels: $$ L_1 = 10 \log_{10} \frac{I_1}{I_0} $$ $$ L_2 = 10 \log_{10} \frac{I_2}{I_0} $$

Find the ratio between the initial and new intensities

As we can see from the formulas above, the reference intensity \(I_0\) will cancel out when we find the ratio of the new intensity to the initial intensity. Subtract the equation for \(L_1\) from the equation for \(L_2\): $$ L_2 - L_1 = 10 \log_{10} \frac{I_2}{I_0} - 10 \log_{10} \frac{I_1}{I_0} $$ Substitute \(L_2 = L_1 + 1\) into the equation: $$ 1 = 10 \log_{10} \frac{I_2}{I_1} $$

Solve for the ratio between the initial and new intensities

To find the ratio between the initial and new intensities, \( \frac{I_2}{I_1}\), we need to isolate it from the equation: $$ \frac{1}{10} = \log_{10} \frac{I_2}{I_1} $$ Using the properties of logarithms, we can solve for the ratio: $$ \frac{I_2}{I_1} = 10^{\frac{1}{10}} $$ Evaluate this expression to find the ratio: $$ \frac{I_2}{I_1} \approx 1.2589 $$

Calculate the percentage change in intensity

Now that we have the ratio between the initial and new intensities, we can calculate the percentage change in intensity using the following formula: Percentage Change in Intensity \(= \frac{I_2 - I_1}{I_1} \times 100\) Substitute the ratio into the formula: Percentage Change in Intensity \(= \frac{1.2589I_1 - I_1}{I_1} \times 100\) Factor out \(I_1\): Percentage Change in Intensity \(= \frac{(1.2589-1)I_1}{I_1} \times 100\) Cancel out \(I_1\): Percentage Change in Intensity \(= (1.2589-1) \times 100 \approx 25.89 \%\) Thus, an intensity level change of \(+1.00\) dB corresponds to a percentage change in intensity of approximately \(25.89 \%\).