The intensity level of two sounds are 100 dB and 50 dB. What is the ratio of their intensities?

Options

A. `10^1`
B. `10^3`
C. `10^5`
D. `10^10`

Explanation

To find the ratio of the intensities of two sounds with intensity levels of 100 dB and 50 dB, we can follow these steps:

### Step 1: Understand the formula for intensity level
The intensity level \( B \) in decibels (dB) 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.

### Step 2: Set up the equations for both sounds
Let \( B_1 \) be the intensity level of the first sound (100 dB) and \( B_2 \) be the intensity level of the second sound (50 dB). We can write:
\[ B_1 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \quad \text{(for 100 dB)} \]
\[ B_2 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \quad \text{(for 50 dB)} \]

### Step 3: Substitute the values of \( B_1 \) and \( B_2 \)
Substituting the values:
\[ 100 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]
\[ 50 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \]

### Step 4: Simplify the equations
Dividing both equations by 10 gives:
\[ 10 = \log_{10} \left( \frac{I_1}{I_0} \right) \]
\[ 5 = \log_{10} \left( \frac{I_2}{I_0} \right) \]

### Step 5: Convert logarithmic form to exponential form
Converting from logarithmic to exponential form:
\[ \frac{I_1}{I_0} = 10^{10} \]
\[ \frac{I_2}{I_0} = 10^{5} \]

### Step 6: Find the ratio of the intensities
Now, we can find the ratio of the intensities \( \frac{I_1}{I_2} \):
\[ \frac{I_1}{I_2} = \frac{\frac{I_1}{I_0}}{\frac{I_2}{I_0}} = \frac{10^{10}}{10^{5}} = 10^{10 - 5} = 10^{5} \]

### Conclusion
Thus, the ratio of the intensities \( I_1 \) to \( I_2 \) is:
\[ \frac{I_1}{I_2} = 10^{5} \]