An increase in intensity level of 1 dB implies an increase in intensity of (given anti `log_(10)0.1=1.2589`)

Options

A. `1%`
B. `3.01%`
C. `26%`
D. `0.1%`

Explanation

To solve the problem of how an increase in intensity level of 1 dB implies an increase in intensity, we can follow these steps:

### Step 1: Understand the relationship between intensity level and intensity
The intensity level \( L \) in decibels (dB) is given by the formula:
\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \]
where \( I \) is the intensity and \( I_0 \) is a reference intensity.

### Step 2: Set up the equations for the initial and increased intensity levels
Let \( L_1 \) be the initial intensity level and \( L_2 \) be the intensity level after an increase of 1 dB. Therefore, we have:
\[ L_1 = 10 \log_{10} \left(\frac{I_1}{I_0}\right) \quad \text{(Equation 1)} \]
\[ L_2 = L_1 + 1 = 10 \log_{10} \left(\frac{I_2}{I_0}\right) \quad \text{(Equation 2)} \]

### Step 3: Subtract the two equations
Subtract Equation 1 from Equation 2:
\[ L_2 - L_1 = 1 = 10 \log_{10} \left(\frac{I_2}{I_0}\right) - 10 \log_{10} \left(\frac{I_1}{I_0}\right) \]
This simplifies to:
\[ 1 = 10 \left( \log_{10} \left(\frac{I_2}{I_0}\right) - \log_{10} \left(\frac{I_1}{I_0}\right) \right) \]
Using the property of logarithms, we can combine the logs:
\[ 1 = 10 \log_{10} \left(\frac{I_2}{I_1}\right) \]

### Step 4: Solve for the ratio of intensities
Dividing both sides by 10 gives:
\[ 0.1 = \log_{10} \left(\frac{I_2}{I_1}\right) \]
Now, exponentiating both sides to eliminate the logarithm:
\[ \frac{I_2}{I_1} = 10^{0.1} \]

### Step 5: Calculate the value of \( 10^{0.1} \)
Using the given information that \( \text{anti} \log_{10}(0.1) = 1.2589 \):
\[ \frac{I_2}{I_1} = 1.2589 \]

### Step 6: Find the increase in intensity
This means:
\[ I_2 = 1.2589 I_1 \]
To find the increase in intensity, we calculate:
\[ \Delta I = I_2 - I_1 = 1.2589 I_1 - I_1 = (1.2589 - 1) I_1 = 0.2589 I_1 \]

### Step 7: Calculate the percentage increase
The percentage increase in intensity is given by:
\[ \text{Percentage Increase} = \left(\frac{\Delta I}{I_1}\right) \times 100\% = \left(0.2589\right) \times 100\% \approx 25.89\% \]
Rounding this gives approximately 26%.

### Final Answer
Thus, an increase in intensity level of 1 dB implies an increase in intensity of approximately **26%**.

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