A sound increases its decibel reading from `20 "to" 40 dB`. This means that the intensity of the sound

Options

A. is doubled
B. is `20 `times greater
C. is `100` times greater
D. is the old intensity `20`

Explanation

To solve the problem of how the intensity of sound changes when its decibel level increases from 20 dB to 40 dB, we can follow these steps:

### Step-by-Step Solution:

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

2. **Setting Up the Equations**:
For the initial sound level of 20 dB:
\[ 20 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]
For the final sound level of 40 dB:
\[ 40 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \]

3. **Dividing the Two Equations**:
Subtract the first equation from the second:
\[ 40 - 20 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) - 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]
This simplifies to:
\[ 20 = 10 \left( \log_{10} \left( \frac{I_2}{I_0} \right) - \log_{10} \left( \frac{I_1}{I_0} \right) \right) \]

4. **Using Logarithm Properties**:
Using the property of logarithms that states \( \log_{10} a - \log_{10} b = \log_{10} \left( \frac{a}{b} \right) \):
\[ 20 = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \]

5. **Isolating the Logarithm**:
Divide both sides by 10:
\[ 2 = \log_{10} \left( \frac{I_2}{I_1} \right) \]

6. **Exponentiating Both Sides**:
To eliminate the logarithm, exponentiate both sides:
\[ 10^2 = \frac{I_2}{I_1} \]
This gives:
\[ \frac{I_2}{I_1} = 100 \]

7. **Conclusion**:
Therefore, the intensity of the sound increases by a factor of 100 when the decibel level increases from 20 dB to 40 dB.

### Final Answer:
The intensity of the sound is 100 times greater.

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