How much intense is 80 dB sound in comparision to 40 dB?

Options

A. `10^(2)`
B. `10^(4)`
C. 2
D. `1/2`

Answer

To determine how much more intense an 80 dB sound is compared to a 40 dB sound, we can use the formula for sound intensity level in decibels (dB):

1. **Understanding the Decibel Scale**:
The decibel level (B) is given by the formula:
\[ B = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \text{W/m}^2 \).

2. **Setting Up the Equations**:
For the two sound levels given:
- For \( B_1 = 80 \, \text{dB} \):
\[ 80 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]
- For \( B_2 = 40 \, \text{dB} \):
\[ 40 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \]

3. **Solving for Intensities**:
Rearranging both equations to solve for \( I_1 \) and \( I_2 \):
- From the first equation:
\[ 8 = \log_{10} \left( \frac{I_1}{I_0} \right) \implies \frac{I_1}{I_0} = 10^8 \implies I_1 = 10^8 I_0 \]
- From the second equation:
\[ 4 = \log_{10} \left( \frac{I_2}{I_0} \right) \implies \frac{I_2}{I_0} = 10^4 \implies I_2 = 10^4 I_0 \]

4. **Finding the Ratio of Intensities**:
Now we can find the ratio of the two intensities:
\[ \frac{I_1}{I_2} = \frac{10^8 I_0}{10^4 I_0} = \frac{10^8}{10^4} = 10^{8-4} = 10^4 \]

5. **Conclusion**:
Therefore, the intensity of the 80 dB sound is \( 10^4 \) times more intense than the intensity of the 40 dB sound.