(i) The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation `I = I_0 e^(-alphax)`, where `I_0` is the intensity at ` x = 0 ` and `alpha` is the attenuation constant.
Show that the intensity reduces by 75 percent after a distance of `(ln 4)/(alpha)`
(ii) Attenuation of a signal can be expressed in decibel (dB) according to the relation
dB `= 10log_10 (I//I_0).` What is the attenuation in `dB//km` for an optical fibre in which the intensity falls by 50 percent over a distance of 50 km?

Options

A. 0.7
B. 0.75
C. 0.8
D. 0.85

Answer

### Step-by-Step Solution

#### Part (i)

1. **Understanding the Given Relation**: We start with the relation for intensity:
\[ I = I_0 e^{-\alpha x} \]
where \( I_0 \) is the initial intensity, \( \alpha \) is the attenuation constant, and \( x \) is the distance.

2. **Expressing the Ratio of Intensities**: We can express the ratio of the intensity at distance \( x \) to the initial intensity as:
\[ \frac{I}{I_0} = e^{-\alpha x} \]

3. **Finding the Distance for 75% Reduction**: We want to find the distance \( x \) at which the intensity reduces by 75%. This means that \( I \) is 25% of \( I_0 \):
\[ I = 0.25 I_0 \]
Therefore, we can write:
\[ \frac{I}{I_0} = 0.25 \]

4. **Setting Up the Equation**: Substituting this into our ratio gives:
\[ 0.25 = e^{-\alpha x} \]

5. **Taking the Natural Logarithm**: Taking the natural logarithm of both sides:
\[ \ln(0.25) = -\alpha x \]

6. **Calculating \( \ln(0.25) \)**: We know that \( 0.25 = \frac{1}{4} \), so:
\[ \ln(0.25) = \ln\left(\frac{1}{4}\right) = \ln(1) - \ln(4) = -\ln(4) \]
Thus, we have:
\[ -\ln(4) = -\alpha x \]

7. **Solving for \( x \)**: Rearranging gives:
\[ x = \frac{\ln(4)}{\alpha} \]

8. **Conclusion**: This shows that the intensity reduces by 75% after a distance of \( \frac{\ln(4)}{\alpha} \).

#### Part (ii)

1. **Understanding the Decibel Relation**: The attenuation in decibels (dB) is given by:
\[ \text{dB} = 10 \log_{10}\left(\frac{I}{I_0}\right) \]

2. **Finding the Intensity Ratio for 50% Reduction**: If the intensity falls by 50%, then:
\[ I = 0.5 I_0 \]
Hence:
\[ \frac{I}{I_0} = 0.5 \]

3. **Substituting into the Decibel Formula**: We substitute this into the decibel formula:
\[ \text{dB} = 10 \log_{10}(0.5) \]

4. **Calculating \( \log_{10}(0.5) \)**: We know that \( 0.5 = \frac{1}{2} \), so:
\[ \log_{10}(0.5) = \log_{10}(1) - \log_{10}(2) = -\log_{10}(2) \]
Thus:
\[ \text{dB} = 10 \cdot (-\log_{10}(2)) = -10 \log_{10}(2) \]

5. **Finding the Value of \( \log_{10}(2) \)**: The approximate value of \( \log_{10}(2) \) is about \( 0.3010 \):
\[ \text{dB} = -10 \cdot 0.3010 = -3.010 \text{ dB} \]

6. **Calculating Attenuation per Kilometer**: Given that this attenuation occurs over a distance of 50 km, the attenuation per kilometer is:
\[ \text{Attenuation per km} = \frac{-3.010 \text{ dB}}{50 \text{ km}} = -0.0602 \text{ dB/km} \]

7. **Conclusion**: The attenuation in dB/km for the optical fiber is approximately \( -0.0602 \text{ dB/km} \).