In a good FM radio receiver, the radio signal detected may be as much as 65 dB greater than the noise signal. What is the ratio of signal intensity to noise intensity ?

Explanation

To find the ratio of signal intensity to noise intensity given that the radio signal is 65 dB greater than the noise signal, we can follow these steps:

### Step-by-Step Solution:

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

2. **Set Up the Loudness Equations**:
Let \(I_R\) be the intensity of the radio signal and \(I_N\) be the intensity of the noise signal. The loudness of the radio signal (\(L_R\)) and the noise signal (\(L_N\)) can be expressed as:
\[ L_R = 10 \log_{10}\left(\frac{I_R}{I_0}\right) \]
\[ L_N = 10 \log_{10}\left(\frac{I_N}{I_0}\right) \]

3. **Use the Given Information**:
According to the problem, the difference in loudness is given as:
\[ L_R - L_N = 65 \, \text{dB} \]

4. **Substitute the Loudness Equations**:
Substitute the expressions for \(L_R\) and \(L_N\) into the equation:
\[ 10 \log_{10}\left(\frac{I_R}{I_0}\right) - 10 \log_{10}\left(\frac{I_N}{I_0}\right) = 65 \]

5. **Factor Out the 10**:
Dividing the entire equation by 10 gives:
\[ \log_{10}\left(\frac{I_R}{I_0}\right) - \log_{10}\left(\frac{I_N}{I_0}\right) = 6.5 \]

6. **Use Logarithmic Properties**:
Using the property of logarithms that \(\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)\):
\[ \log_{10}\left(\frac{I_R}{I_N}\right) = 6.5 \]

7. **Convert from Logarithmic to Exponential Form**:
Converting from logarithmic form to exponential form gives:
\[ \frac{I_R}{I_N} = 10^{6.5} \]

8. **Calculate the Ratio**:
Evaluating \(10^{6.5}\):
\[ 10^{6.5} = 10^{6} \times 10^{0.5} = 1,000,000 \times 3.162 = 3,162,000 \]
Thus, the ratio of signal intensity to noise intensity is:
\[ \frac{I_R}{I_N} \approx 3.16 \times 10^{6} \]

### Final Answer:
The ratio of signal intensity to noise intensity is approximately \(3.16 \times 10^{6}\).

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