The intensity of sound of 50 dB (Take reference of intensity `10^-12` W/`m^2`)

Options

A. `10^-10W/m^2`
B. `210^-5 W/m^2`
C. `10^-12 W/m^2`
D. `10^-7 W/m^2`

Answer

To solve the problem of finding the sound intensity corresponding to a sound level of 50 dB, we can follow these steps:

### Step 1: Understand the formula for sound intensity level
The sound intensity level (in decibels) is given by the formula:
\[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where:
- \(\beta\) is the sound intensity level in decibels (dB),
- \(I\) is the sound intensity in watts per meter squared (W/m²),
- \(I_0\) is the reference intensity, which is \(10^{-12}\) W/m².

### Step 2: Substitute the known values into the formula
We know:
- \(\beta = 50\) dB,
- \(I_0 = 10^{-12}\) W/m².

Substituting these values into the formula gives:
\[ 50 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right) \]

### Step 3: Simplify the equation
Divide both sides by 10:
\[ 5 = \log_{10} \left( \frac{I}{10^{-12}} \right) \]

### Step 4: Eliminate the logarithm
To eliminate the logarithm, we can rewrite the equation in exponential form:
\[ 10^5 = \frac{I}{10^{-12}} \]

### Step 5: Solve for \(I\)
Now, multiply both sides by \(10^{-12}\):
\[ I = 10^5 \times 10^{-12} \]
Using the properties of exponents, we can simplify this:
\[ I = 10^{-7} \text{ W/m}^2 \]

### Final Answer
The sound intensity \(I\) corresponding to a sound level of 50 dB is:
\[ I = 10^{-7} \text{ W/m}^2 \]