Intensity level at a point is 100 dB. How much is the actual intensity of sound falling at that point ? Given threshold intensity of sound is `10^(-12) "Wm"^(-2)`.

Answer

To find the actual intensity of sound at a point where the intensity level is 100 dB, we can use the formula that relates intensity level (in decibels) to actual intensity. The formula is given by:

\[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]

where:
- \(\beta\) is the intensity level in decibels (dB),
- \(I\) is the actual intensity of the sound (in watts per meter squared, W/m²),
- \(I_0\) is the reference intensity, which is the threshold intensity of sound, given as \(10^{-12} \, \text{W/m}^2\).

### Step 1: Substitute the known values into the formula
Given that \(\beta = 100 \, \text{dB}\) and \(I_0 = 10^{-12} \, \text{W/m}^2\), we can substitute these values into the formula:

\[ 100 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right) \]

### Step 2: Simplify the equation
Divide both sides of the equation by 10:

\[ 10 = \log_{10} \left( \frac{I}{10^{-12}} \right) \]

### Step 3: Remove the logarithm by exponentiating
To eliminate the logarithm, we can rewrite the equation in exponential form:

\[ 10^{10} = \frac{I}{10^{-12}} \]

### Step 4: Solve for \(I\)
Now, multiply both sides by \(10^{-12}\) to isolate \(I\):

\[ I = 10^{10} \times 10^{-12} \]

### Step 5: Simplify the expression
Using the properties of exponents, we can combine the powers:

\[ I = 10^{10 - 12} = 10^{-2} \, \text{W/m}^2 \]

### Step 6: Final result
Thus, the actual intensity of sound falling at that point is:

\[ I = 0.01 \, \text{W/m}^2 \]

### Conclusion
The actual intensity of sound at the point where the intensity level is 100 dB is \(0.01 \, \text{W/m}^2\).

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